## Big Ideas Math Geometry Answers Chapter 10 Circles

Big Ideas Math Book Geometry Chapter 10 Circles Answers are provided here. Students who have been looking for the BIM Geometry Chapter 10 Circles Answers can read the following sections. The high school students can find a direct link to download Big Ideas Math Geometry Answers Chapter 10 Circles pdf for free of cost. With the help of this answer key, you can prepare well for the exam.

## Big Ideas Math Book Geometry Answer Key Chapter 10 Circles

The different chapters included in Big Ideas Math Geometry Solutions are Lines and Segments That Intersect Circles, Finding Arc Measures, Inscribed Angles and Polygons, Angle Relationships in Circles, Segment Relationships in Circles, Circles in the Coordinate Plane, and Using Chords. Students have to practise all the questions from Big Ideas Math Textbook Geometry Chapter 10 Circles.

This Big Ideas Math Book Geometry Answer Key Chapter 10 Circles helps the students while doing the assignments. Get the solutions for all the questions through the quick links provided in the following sections. Test your skills through performance task, chapter review, and maintaining mathematical proficiency.

### Circles Maintaining Mathematical Proficiency

Find the Product.

Question 1.
(x + 7) (x + 4)

(x + 7) (x + 4) = x² + 14x + 28

Explanation:
(x + 7) (x + 4) = x(x + 7) + 7(x + 4)
= x² + 7x + 7x + 28
= x² + 14x + 28

Question 2.
(a + 1) (a – 5)

(a + 1) (a – 5) = a² – 4a – 5

Explanation:
(a + 1) (a – 5) = a(a – 5) + 1(a – 5)
= a² – 5a + a – 5
= a² – 4a – 5

Question 3.
(q – 9) (3q – 4)

(q – 9) (3q – 4) = 3q² – 31q + 32

Explanation:
(q – 9) (3q – 4) = q(3q – 4) – 9(3q – 4)Exercise 10.3 Using Chords

= 3q² – 4q – 27q + 32
= 3q² – 31q + 32

Question 4.
(2v – 7) (5v + 1)

(2v – 7) (5v + 1) = 10v² – 33v – 7

Explanation:
(2v – 7) (5v + 1) = 2v(5v + 1)- 7(5v + 1)
= 10v² + 2v – 35v – 7
= 10v² – 33v – 7

Question 5.
(4h + 3) (2 + h)

(4h + 3) (2 + h) = 4h² + 11h + 6

Explanation:
(4h + 3) (2 + h) = 4h(2 + h) + 3(2 + h)
= 8h + 4h² + 6 + 3h
= 4h² + 11h + 6

Question 6.
(8 – 6b) (5 – 3b)

(8 – 6b) (5 – 3b) = 18b² – 54b + 40

Explanation:
(8 – 6b) (5 – 3b) = 8(5 – 3b) – 6b(5 – 3b)
= 40 – 24b – 30b + 18b²
= 18b² – 54b + 40

Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary.

Question 7.
x2 – 2x = 5

The solutions are x = √6 + 1, x = 1 – √6

Explanation:
x² – 2x = 5
x² – 2x + 1² = 5 + 1²
(x – 1)² = 6
x – 1 = ±√6
x = ±√6 + 1
The solutions are x = √6 + 1, x = -√6 + 1

Question 8.
r2 + 10r = -7

The solutions are r = √18 – 5, r = 5 – √18

Explanation:
r2 + 10r = -7
r² + 10r + 5² = -7 + 5²
(r + 5)² = -7 + 25 = 18
r + 5 = ±√18
r = ±√18 – 5
The solutions are r = √18 – 5, r = 5 – √18

Question 9.
w2 – 8w = 9

The solutions are w = 9, w = -1

Explanation:
w2 – 8w = 9
w2 – 8w + 4² = 9 + 4²
(w – 4)² = 9 + 16 = 25
w – 4 = ±5
w = 5 + 4, w = -5 + 4
w = 9, w = -1
The solutions are w = 9, w = -1

Question 10.
p2 + 10p – 4 = 0

The solutions are p = √29 – 5, p = 5 – √29

Explanation:
p2 + 10p = 4
p² + 10p + 5² = 4 + 5²
(p + 5)² = 4 + 25
(p + 5)² = 29
p + 5 = ±√29
p = ±√29 – 5
The solutions are p = √29 – 5, p = 5 – √29

Question 11.
k2 – 4k – 7 = 0

The solutions are k = √11 + 2, k = 2 – √11

Explanation:
k² – 4k= 7
k² – 4k + 2² = 7 + 4
(k – 2)² = 11
k – 2 = ±√11
k = √11 + 2, k = 2 – √11
The solutions are k = √11 + 2, k = 2 – √11

Question 12.
– z2 + 2z = 1

The solutions are z = 1

Explanation:
-z² + 2z = 1
z² – 2z = -1
z² – 2z + 1 = -1 + 1
(z – 1)² = 0
z = 1
The solutions are z = 1

Question 13.
ABSTRACT REASONING
write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning.

Let us take two consecutive odd integers are x and (x + 2)
The product of two consecutive odd integers is x • (x + 2)
x(x + 2) = x² + 2x

Circles Mathematical Practices

Monitoring progress

Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers.

Question 1.
Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning.

Question 2.
Draw ⊙A, ⊙B, and OC so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D less than, greater than, or equal to 6? Explain.

### 10.1 Lines and Segments that Intersect Circles

Exploration 1

Lines and Line Segments That Intersect Circles

Work with a partner: The drawing at the right shows five lines or segments that intersect a circle. Use the relationships shown to write a definition for each type of line or segment. Then use the Internet or some other resource to verify your definitions.
Chord: _________________
Secant: _________________
Tangent: _________________
Diameter: _________________

Chord: A chord of a circle is a straight line segment whose endpoints both lie on a circular arc.
Secant: A straight line that intersects a circle in two points is called a secant line.
Tangent: Tangent line is a line that intersects a curved line at exactly one point.
Radius: It is the distance from the centre of the circle to any point on the circle.
Diameter: It the straight that joins two points on the circle and passes through the centre of the circle.

Exploration 2

Using String to Draw a Circle

Work with a partner: Use two pencils, a piece of string, and a piece of paper.

a. Tie the two ends of the piece of string loosely around the two pencils.

b. Anchor one pencil of the paper at the center of the circle. Use the other pencil to draw a circle around the anchor point while using slight pressure to keep the string taut. Do not let the string wind around either pencil.

c. Explain how the distance between the two pencil points as you draw the circle is related to two of the lines or line segments you defined in Exploration 1.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.

Question 3.
What are the definitions of the lines and segments that intersect a circle?

Question 4.
Of the five types of lines and segments in Exploration 1, which one is a subset of another? Explain.

Question 5.
Explain how to draw a circle with a diameter of 8 inches.

### Lesson 10.1 Lines and Segments that Intersect Circles

Monitoring progress

Question 1.
In Example 1, What word best describes $$\overline{A G}$$? $$\overline{C B}$$?

$$\overline{A G}$$ is secant because it is a line that intersects the circle at two points.
$$\overline{C B}$$ is the radius as it is the distance from the centre to the point of a circle.

Question 2.
In Example 1, name a tangent and a tangent segment.

$$\overline{D E}$$ is the tangent of the circle
$$\overline{D E}$$ is the tangent segment of the circle.

Tell how many common tangents the circles have and draw them. State whether the tangents are external tangents or internal tangents.

Question 3.

4 tangents.

A tangent is a line segment that intersects the circle at exactly one point. Internal tangents are the lines that intersect the segments joining the centres of two circles. External tangents are the lines that do not cross the segment joining the centres of the circles.
Blue lines represent the external tangents and red lines represent the internal tangents.

Question 4.

One tangent.

One external tangent.

Question 5.

No tangent.
It is not possible to draw a common tangent for this type of circle.

Question 6.
Is $$\overline{D E}$$ tangent to ⊙C?

Use the converse of Pythagorean theorem i.e 2² = 3² + 4²
4 = 9 + 16
By the tangent line to the circle theorem, $$\overline{D E}$$ is not a tangent to ⊙C

Question 7.
$$\overline{S T}$$ is tangent to ⊙Q.

The radius of ⊙Q is 7 units.

Explanation:
By using the Pythagorean theorem
(18 + r)² = r² + 24²
324 + 36r + r² = r² + 576
36r = 576 – 324
36r = 252
r = 7

Question 8.
Points M and N are points of tangency. Find the value(s) of x.

The values of x are 3 or -3.

Explanation:
x² = 9
x = ±3

### Exercise 10.1 Lines and Segments that Intersect Circles

Vocabulary and Core Concept Check

Question 1.
WRITING
How are chords and secants alike? How are they different?

Question 2.
WRITING
Explain how you can determine from the context whether the words radius and diameter are referring to segments or lengths.

Radius and diameter are the lengths of the line segments that pass through the centre of a circle. Radius is half of the diameter.

Question 3.
COMPLETE THE SENTENCE
Coplanar circles that have a common center are called ____________ .

Question 4.
WHICH ONE DOESNT BELONG?
Which segment does not belong with the other three? Explain your reasoning.

A chord, a radius and a diamter are segments and they intersect a circle in two points. A tangent is a line that intersects a circle at one point.

Monitoring Progress and Modeling with Mathematics

In Exercises 5 – 10, use the diagram.

Question 5.
Name the circle.

Question 6.

The name of the two radii is CD and AC.

Question 7.
Name two chords.

Question 8.
Name a diameter.

The name of diameter is AD

Question 9.
Name a secant.

Question 10.
Name a tangent and a point of tangency

GE is the tangent and F is the point of tangency.

In Exercises 11 – 14, copy the diagram. Tell how many common tangents the circles have and draw them.

Question 11.

Question 12.

No common tangent because two circles do not intersect at one point.

Question 13.

Question 14.

One common tangent.

In Exercises 15 – 18, tell whether the common tangent is internal or external.

Question 15.

Question 16.

The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles.

Question 17.

Question 18.

The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles.

In Exercises 19 – 22, tell whether $$\overline{A B}$$ is tangent to ⊙C. Explain your reasoning.

Question 19.

Question 20.

Use the converse of the Pythagorean theorem
18² _____________ 15² + 9²
324 _____________ 225 + 81
324 ≠ 304
△ ACB is not a right angled triangle.
So, $$\overline{A B}$$ is not tangent to ⊙C at B.

Question 21.

Question 22.

Use the converse of the Pythagorean theorem
8² _____________ 12² + 16²
64 _____________ 144 + 256
64 ≠ 400
△ ACB is not a right angled triangle.
So, $$\overline{A B}$$ is not tangent to ⊙C at B.

In Exercises 23 – 26, point B is a point of tangency. Find the radius r of ⊙C.

Question 23.

Question 24.

(r + 6)² = r² + 9²
r² + 12r + 36 = r² + 81
12r = 81 – 36
12r = 45
r = $$\frac { 15 }{ 4 }$$
Therefore, the radius of ⊙C is $$\frac { 15 }{ 4 }$$

Question 25.

Question 26.

(r + 18)² = r² + 30²
r² + 36r + 324 = r² + 900
36r = 900 – 324
36r = 576
r = 16
Therefore, the radius of ⊙C is 16

CONSTRUCTION
In Exercises 27 and 28. construct ⊙C with the given radius and point A outside of ⊙C. Then construct a line tangent to ⊙C that passes through A.

Question 27.
r = 2 in.

Question 28.
r = 4.5 cm

In Exercises 29 – 32, points B and D are points of tangency. Find the value(s) of x.

Question 29.

Question 30.

3x + 10 = 7x – 6
7x – 3x = 10 + 6
4x = 16
x = 4

Question 31.

Question 32.

2x + 5 = 3x² + 2x – 7
3x² = 5 + 7
3x² = 12
x² = 4
x = ±2

Question 33.
ERROR ANALYSIS
Describe and correct the error in determining whether $$\overline{X Y}$$ is tangent to ⊙Z.

Question 34.
ERROR ANALYSIS
Describe and correct the error in finding the radius of ⊙T.

39² = 36² + 15²
So, 15 is the diameter.
The radius is $$\frac { 15 }{ 2 }$$.

Question 35.
ABSTRACT REASONING
For a point outside of a circle, how many lines exist tangent to the circle that pass through the point? How many such lines exist for a point on the circle? inside the circle? Explain your reasoning.

Question 36.
CRITICAL THINKING
When will two lines tangent to the same circle not intersect? Justify your answer.

Using tangent line to circle theorem, it follow that the angle between tangent and radius is a right angle. Let’s draw these tangents at the two ends of the same diameter. We can observe a diameter AD like a transverzal of these tangents.

Question 37.
USING STRUCTURE
Each side of quadrilateral TVWX is tangent to ⊙Y. Find the perimeter of the quadrilateral.

Question 38.
LOGIC
In ⊙C, radii $$\overline{C A}$$ and $$\overline{C B}$$ are perpendicular. are tangent to ⊙C.

a. Sketch ⊙C, $$\overline{C A}$$, $$\overline{C B}$$, .

Question 39.
MAKING AN ARGUMENT
Two hike paths are tangent to an approximately circular pond. Your class is building a nature trail that begins at the intersection B of the bike paths and runs between the bike paths and over a bridge through the center P of the pond. Your classmate uses the Converse of the Angle Bisector Theorem (Theorem 6.4) to conclude that the trail must bisect the angle formed by the bike paths. Is your classmate correct? Explain your reasoning.

Question 40.
MODELING WITH MATHEMATICS
A bicycle chain is pulled tightly so that $$\overline{M N}$$ is a common tangent of the gears. Find the distance between the centers of the gears.

height h = 4.3 – 1.8
h = 2.5
x² = MN² + h²
x² = 17.6² + 2.5²
x² = 316.01
x = 17.8
Therefore, the distance between the centre of the gear is 17.8 in.

Question 41.
WRITING
Explain why the diameter of a circle is the longest chord of the circle.

Question 42.
HOW DO YOU SEE IT?
In the figure, $$\vec{P}$$A is tangent to the dime. $$\vec{P}$$C is tangent to the quarter, and $$\vec{P}$$B is a common internal tangent. How do you know that $$\overline{P A} \cong \overline{P B} \cong \overline{P C}$$

Question 43.
PROOF
In the diagram, $$\overline{R S}$$ is a common internal tangent to ⊙A and ⊙B. Prove that $$\frac{\Lambda C}{B C}=\frac{R C}{S C}$$

Question 44.
THOUGHT PROVOKING
A polygon is circumscribed about a circle when every side of the polygon is tangent to the circle. In the diagram. quadrilateral ABCD is circumscribed about ⊙Q. Is it always true that AB + CD = AD + BC? Justify your answer.

Question 45.
MATHEMATICAL CONNECTIONS

Question 46.
PROVING A THEOREM
Prove the External Tangent Congruence Theorem (Theorem 10.2).

Given $$\overline{S R}$$ and $$\overline{S T}$$ are tangent to ⊙P.
Prove $$\overline{S R} \cong \overline{S T}$$

∠PRS and ∠PTS are the right angles. So the legs of circles are congruent.
Therefore, $$\overline{S R} \cong \overline{S T}$$

Question 47.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Tangent Line to Circle Theorem (Theorem 10.1 ).

a. Prove indirectly that if a line is tangent to a circle, then it is perpendicular to a radius. (Hint: If you assume line m is not perpendicular to $$\overline{Q P}$$, then the perpendicular segment from point Q to line m must intersect line m at some other point R.)
Ghen Line m is tangent to ⊙Q at point P.
Prove m ⊥ $$\overline{Q P}$$
b. Prove indirectly that if a line is perpendicular to a radius at its endpoint, then the line is tangent to the circle.
Gien m ⊥ $$\overline{Q P}$$
Prove Line m is tangent to ⊙Q.

Question 48.
REASONING
In the diagram, AB = AC = 12, BC = 8, and all three segments are Langent to ⊙P. What is the radius of ⊙P? Justify your answer.

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 49.
m∠JKM

Question 50.
AB

AC = AB + BC
10 = AB + 7
AB = 10 – 7
AB = 3

### 10.2 Finding Arc Measures

Exploration 1

Measuring Circular Arcs

Work with a partner: Use dynamic geometry software to find the measure of $$\widehat{B C}$$. Verify your answers using trigonometry.

a.

Points
A(0, 0)
B(5, 0)
C(4, 3)

30 degrees

b.

Points
A(0, 0)
B(5, 0)
C(3, 4)

60 degrees

c.

Points
A(0, 0)
B(4, 3)
C(3, 4)

15 degrees

d.

Points
A(0, 0)
B(4, 3)
C(- 4, 3)

90 degrees

Question 2.
How are circular arcs measured?

Question 3.
Use dynamic geometry software to draw a circular arc with the given measure.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technological tools to explore and deepen your understanding of concepts.
a. 30°

b. 45°

c. 60°

d. 90°

### Lesson 10.2 Finding Arc Measures

Monitoring Progress

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 1.
$$\widehat{T Q}$$

$$\widehat{T Q}$$ is a minor arc.
$$\widehat{T Q}$$ = 120°

Question 2.
$$\widehat{Q R T}$$

$$\widehat{Q R T}$$

Question 3.
$$\widehat{T Q R}$$ is a major arc.
$$\widehat{Q R T}$$ = QR + RS + ST
RS = 360° – (60 + 120 + 80)
= 360 – 260 = 100°
So, $$\widehat{Q R T}$$ = 60° + 100° + 80°
$$\widehat{Q R T}$$ = 240°

Question 4.
$$\widehat{Q S}$$

$$\widehat{Q S}$$ = QR + RS
= 60 + 100 = 160°
Therefore, $$\widehat{Q S}$$ = 160° and it is a minor arc.

Question 5.
$$\widehat{T S}$$

$$\widehat{T S}$$ = 80° and it is a minor arc.

Question 6.
$$\widehat{R S T}$$

$$\widehat{R S T}$$ = 100 + 80 = 180
Therefore, $$\widehat{R S T}$$ = 180° and it is a minor arc.

Tell whether the red arcs are congruent. Explain why or why not.

Question 7.

$$\widehat{A B}$$, $$\widehat{C D}$$ are congruent as they measure same radius and same arc length.

Question 8.

$$\widehat{M N}$$, $$\widehat{P Q}$$ are not congruent as they measure different radius.

### Exercise 10.2 Finding Arc Measures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Copy and complele:
If ∠ACB and ∠DCE are congruent central angles of ⊙C, then $$\widehat{A B}$$ and $$\widehat{D E}$$ arc.

Question 2.
WHICH ONE DOESNT BELONG?
Which circle does not belong with the other three? Explain your reasoning.

We know that 1 ft = 12 in
So, the fourth circle does not belong to the other three as its diameter is different.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, name the red minor arc and find its measure. Then name the blue major arc and find its measure.

Question 3.

Question 4.

The minor arc $$\widehat{E F}$$ = 68° and major arc $$\widehat{F G E}$$ = 360° – 68° = 292°.

Question 5.

Question 6.

The minor arc is $$\widehat{M N}$$ = 170°, major arc $$\widehat{N P M}$$ = 360° – 170° = 190°.

In Exercises 7 – 14. identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 7.
$$\widehat{B C}$$

Question 8.
$$\widehat{D C}$$

$$\widehat{D C}$$ is a minor arc and it has a measure of 65°.

Question 9.
$$\widehat{E D}$$

Question 10.
$$\widehat{A E}$$

$$\widehat{A E}$$ is a minor arc and it has a measure of 70°.

Question 11.
$$\widehat{E A B}$$

Question 12.
$$\widehat{A B C}$$

$$\widehat{A B C}$$ is a semicircle and it has a measure of 180°.

Question 13.
$$\widehat{B A C}$$

Question 14.
$$\widehat{E B D}$$

$$\widehat{E B D}$$ is a major arc and it has a measure of 315°.

In Exercises 15 and 16, find the measure of each arc.

Question 15.

a. $$\widehat{J L}$$

b. $$\widehat{K M}$$

c. $$\widehat{J L M}$$

d. $$\widehat{J M}$$

Question 16.

a. $$\widehat{R S}$$

$$\widehat{R S}$$ = $$\widehat{Q R S}$$ – $$\widehat{Q R}$$
= 180 – 42
= 138°
So, $$\widehat{R S}$$ = 138°

b. $$\widehat{Q R S}$$

$$\widehat{Q R S}$$ = 180°

c. $$\widehat{Q S T}$$

$$\widehat{Q S T}$$ = $$\widehat{Q R S}$$ + $$\widehat{S T}$$
= 180 + 42 = 222
So, $$\widehat{Q S T}$$ = 222°

d. $$\widehat{Q T}$$

$$\widehat{Q T}$$ = 360 – (42 + 138 + 42)
= 360 – (222)
= 138°
$$\widehat{Q T}$$ = 138°

Question 17.
MODELING WITH MATHEMATICS
A recent survey asked high school students their favorite type of music. The results are shown in the circle graph. Find each indicated arc measure.

a. m$$\widehat{A E}$$
b. m$$\widehat{A C E}$$
c. m$$\widehat{G D C}$$
d. m$$\widehat{B H C}$$
e. m$$\widehat{F D}$$
f. m$$\widehat{F B D}$$

Question 18.
ABSTRACT REASONING
The circle graph shows the percentages of students enrolled in fall Sports at a high school. Is it possible to find the measure of each minor arc? If so, find the measure 0f the arc for each category shown. If not, explain why it is not possible.

Soccer angle = 30% of 360 = 108°
Volleyball angle = 15% of 360 = 54°
Cross-country angle = 20% of 360 = 72°
None angle = 15% of 360 = 54°
Football angle = 20% of 360 = 72°

In Exercises 19 – 22, tell whether the red arcs are congruent. Explain why or why not.

Question 19.

Question 20.

$$\widehat{L P}$$ and $$\widehat{M N}$$ are not congruet because they are not in the same circle.

Question 21.

Question 22.

$$\widehat{R S Q}$$, $$\widehat{F G H}$$ are not congruent because those two circles have different radii.

MATHEMATICAL CONNECTIONS
In Exercises 23 and 24. find the value of x. Then find the measure of the red arc.

Question 23.

Question 24.

4x + 6x + 7x + 7x = 360
24x = 360
x = 15
m$$\widehat{R S T}$$ = 6(15) + 7(15)
= 90 + 105 = 195°
So, m$$\widehat{R S T}$$ = 195°

Question 25.
MAKING AN ARGUMENT
Your friend claims that any two arcs with the same measure are similar. Your cousin claims that an two arcs with the same measure are congruent. Who is correct? Explain.

Question 26.
MAKING AN ARGUMENT
Your friend claims that there is not enough information given to find the value of x. Is your friend correct? Explain your reasoning.

My friend is wrong.
4x + x + x + 4x = 360
10x = 360
x = 36°

Question 27.
ERROR ANALYSIS
Describe and correct the error in naming the red arc.

Question 28.
ERROR ANALYSIS
Describe and correct the error in naming congruent arc.

$$\widehat{J K}$$, $$\widehat{N P}$$ are not congruent because those two arcs are from different circles.

Question 29.
ATTENDING TO PRECISION
Two diameters of ⊙P are $$\widehat{A B}$$ and $$\widehat{C D}$$. Find m$$\widehat{A C D}$$ and m$$\widehat{A C}$$ when m$$\widehat{A D}$$ = 20°.

Question 30.
REASONING
In ⊙R, m$$\widehat{A B}$$ = 60°, m$$\widehat{B C}$$ = 25°. m$$\widehat{C D}$$ = 70°, and m$$\widehat{D E}$$ = 20°. Find two possible measures of $$\widehat{A E}$$.

$$\widehat{A E}$$ = 360 – ($$\widehat{A B}$$ + $$\widehat{B C}$$ + $$\widehat{C D}$$ + $$\widehat{D E}$$)
= 360 – (60 + 25 + 70 + 20)
= 360 – (175)
= 185
$$\widehat{A E}$$ = $$\widehat{A B}$$ + $$\widehat{B C}$$ + $$\widehat{C D}$$ + $$\widehat{D E}$$
= 60 + 25 + 70 + 20 = 175
So, the two possibilities of $$\widehat{A E}$$ are 185°, 175°

Question 31.
MODELING WITH MATHEMATICS
On a regulation dartboard, the outermost circle is divided into twenty congruent sections. What is the measure of each arc in this circle?

Question 32.
MODELING WITH MATHEMATICS
You can use the time zone wheel to find the time in different locations across the world. For example, to find the time in Tokyo when it is 4 P.M. in San Francisco, rotate the small wheel until 4 P.M. and San Francisco line up, as shown. Then look at Tokyo to see that it is 9 A.M. there.

a. What is the arc measure between each time zone 0n the wheel?

As the circle is divided into 24 sectors, each time zone angle = $$\frac { 360 }{ 24 }$$ = 15°

b. What is the measure of the minor arc from the Tokyo zone to the Anchorage zone?

The measure of the minor arc from the Tokyo zone to the Anchorage zone = 15 + 15 + 15 + 15 + 15 + 15
= 90°

c. If two locations differ by 180° on the wheel, then it is 3 P.M. at one location when it is _________ at the other location.

Kuwaiti city.

Question 33.
PROVING A THEOREM
Write a coordinate proof of the Similar Circles Theorem (Theorem 10.5).
Given ⊙O with center O(0, 0) and radius r.
⊙A with center A(a, 0) and radius s
Prove ⊙O ~ ⊙A

Question 34.
ABSTRACT REASONING
Is there enough information to tell whether ⊙C ≅ ⊙D? Explain your reasoning.

Both circles ⊙C and ⊙D have the same radius so those circles are congruent.

Question 35.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Circles Theorem (Theorem 10.3).

a. Given $$\overline{A C} \cong \overline{B D}$$
Prove ⊙A ≅ ⊙B
b. Given ⊙A ≅ ⊙B
prove $$\overline{A C} \cong \overline{B D}$$

Question 36.
HOW DO YOU SEE IT?
Are the circles on the target similar or congruent? Explain your reasoning.

Question 37.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Central Angles Theorem (Theorem 10.4).

a. Given ∠ABC ≅ ∠DAE
Prove $$\widehat{B C}$$ ≅ $$\widehat{D E}$$
b. Given $$\widehat{B C}$$ ≅ $$\widehat{D E}$$
Prove ∠ABC ≅ ∠DAE

Question 38.
THOUGHT PROVOKING
Write a formula for the length of a circular arc. Justify your answer.

The formula to find the length of a circular arc is radius x angle.

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 39.

Question 40.

x² = 13² + 13²
= 169 + 169
= 338
x = 13√2

Question 41

Question 42.

14² = x² + 10²
196 = x² + 100
x² = 196 – 100
x² = 96
x = 4√6

### 10.3 Using Chords

Exploration 1

Drawing Diameters

Work with a partner: Use dynamic geometry software to construct a circle of radius 5 with center at the origin. Draw a diameter that has the given point as an endpoint. Explain how you know that the chord you drew is a diameter.
a. (4, 3)
b. (0, 5)
c. (-3, 4)
d. (-5, 0)

Exploration 2

Work with a partner. Use dynamic geometry software to construct a chord $$\overline{B C}$$ of a circle A. Construct a chord on the perpendicular bisector of $$\overline{B C}$$. What do you notice? Change the original chord and the circle several times. Are your results always the same? Use your results to write a conjecture.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.

Exploration 3

A Chord Perpendicular to a Diameter

Work with a partner. Use dynamic geometry software to construct a diameter $$\overline{B C}$$ of a circle A. Then construct a chord $$\overline{D E}$$ perpendicular to $$\overline{B C}$$ at point F. Find the lengths DF and EF. What do you notice? Change the chord perpendicular to $$\overline{B C}$$ and the circle several times. Do you always get the same results? Write a conjecture about a chord that is perpendicular to a diameter of a circle.

Question 4.
What are two ways to determine when a chord is a diameter of a circle?

If a chord passes through the centre of the circle, then it is the diameter of a circle.
The longest chord of the circle is the diameter of a circle.

### Lesson 10.3 Using Chords

Monitoring Progress

In Exercises 1 and 2, use the diagram of ⊙D.

Question 1.
If m$$\widehat{A B}$$ = 110°. find m$$\widehat{B C}$$.

Because AB and BC are congruent chords in congruent circles, the corresponding minor arcs $$\widehat{A B}$$, $$\widehat{B C}$$ are congruent by the congruent corresponding chords theorem.
So, $$\widehat{A B}$$ = $$\widehat{B C}$$
$$\widehat{B C}$$ = 110°

Question 2.
If m$$\widehat{A C}$$ = 150° find m$$\widehat{A B}$$.

$$\widehat{A C}$$ = 360 – ($$\widehat{A B}$$ + $$\widehat{B C}$$)
150 = 360 – 2($$\widehat{A B}$$)
2($$\widehat{A B}$$) = 360 – 150 = 210
$$\widehat{A B}$$ = 105°

In Exercises 3 and 4. find the indicated length or arc measure.

Question 3.
CE

CE = 5 + 5
= 10 units

Question 4.
m$$\widehat{C E}$$

m$$\widehat{C E}$$ = 9x + 180 – x = 180 – 8x
m$$\widehat{C E}$$ = 180 – 8x

Question 5.
In the diagram, JK = LM = 24, NP = 3x, and NQ = 7x – 12. Find the radius of ⊙N

### Exercise 10.3 Using Chords

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe what it means to bisect a chord.

Question 2.
WRITING
Two chords of a circle are perpendicular and congruent. Does one of them have to be a diameter? Explain your reasoning.

Imagine a line segment of length 3 units, AB.

A second congruent segment of length 3 that is perpendicular to AB called CD.

Circumscribe both these line segments and note that AB and CD are now chords.
While both chords are perpendicular and congruent, neither chord is a diameter. Thus, it is possible to have two chords of this type with neither one diameter of the circle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the measure of the red arc or chord in ⊙C.

Question 3.

Question 4.

Arc length = radius x angle
= 5 x 34 = 170

Question 5.

Question 6.

As the two circles radius is the same and the angle is the same so the chord length is 11 units.

In Exercise 7-10, find the value of x.

Question 7.

Question 8.

By the perpendicular bisector theorem RS = ST
x = 40°

Question 9.

Question 10.

5x + 2 = 7x – 12
7x – 5x = 2 + 12
2x = 14
x = 7

Question 11.
ERROR ANALYSIS
Describe and correct the error in reasoning.

Question 12.
PROBLEM SOLVING
In the cross section of the submarine shown, the control panels are parallel and the same length. Describe a method you can use to find the center of the cross section. Justify your method.

In Exercises 13 and 14, determine whether $$\overline{A B}$$ is a diameter of the circle. Explain your reasoning.

Question 13.

Question 14.

5² = 3² + x²
25 = 9 + x²
x² = 25 – 9
x = 4
So, AB is not diameter of the circle.

In Exercises 15 and 16, find the radius of ⊙Q.

Question 15.

Question 16.

4x + 4 = 6x – 6
6x – 4x = 4 + 6
2x = 10
x = 5
BC = 6(5) – 6 = 30 – 6 = 24
QC² = 5² + 12²
= 25 + 144 = 169
QC = 13

Question 17.
PROBLEM SOLVING
An archaeologist finds part of a circular plate. What was the diameter of the plate to the nearest tenth of an inch? Justify your answer.

Question 18.
HOW DO YOU SEE IT?
What can you conclude from each diagram? Name a theorem that justifies your answer.
a.

Perpendicular chord bisector converse theorem.

b.

Congruent Corresponding Chords theorem

c.

Perpendicular chord bisector theorem

d.

Equidistant chords theorem

Question 19.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Corresponding Chords Theorem (Theorem 10.6).

a. Given $$\overline{A B}$$ and $$\overline{C D}$$ are congruent chords.
Prove $$\widehat{A B} \cong \widehat{C D}$$
b. Given $$\widehat{A B} \cong \widehat{C D}$$
Prove $$\overline{A B}$$ ≅ $$\overline{C D}$$

Question 20.
MATHEMATICAL CONNECTIONS
In ⊙P, all the arcs shown have integer measures. Show that x must be even.

Question 21.
REASONING
In ⊙P. the lengths of the parallel chords are 20, 16, and 12. Find m$$\widehat{A B}$$. Explain your reasoning.

Question 22.
PROVING A THEOREM
Use congruent triangles to prove the Perpendicular Chord Bisector Theorem (Theorem 10.7).

Given $$\overline{E G}$$ is a diameter of ⊙L.
$$\overline{E G}$$ ⊥ $$\overline{D F}$$
Prove $$\overline{D C}$$ ≅ $$\overline{F C}$$, $$\widehat{D G} \cong \widehat{F G}$$

Let L be the centre of the circle
draw any chord DF on the circle
As DF passes through LG.
The length of DC is the same as FC.

Question 23.
PROVING A THEOREM
Write a proof of the Perpendicular Chord Bisector Converse (Theorem 10.8).

Given $$\overline{Q S}$$ is a perpendicular bisector of $$\overline{R T}$$.
Prove $$\overline{Q S}$$ is a diameter of the circle L.
(Hint: Plot the center L and draw △LPT and △LPR.)

Question 24.
THOUGHT PROVOKING
Consider two chords that intersect at point P. Do you think that $$\frac{A P}{B P}=\frac{C P}{D P}$$? Justify your answer.

Question 25.
PROVING A THEOREM
Use the diagram with the Equidistant Chords Theorem (Theorem 10.9) to prove both parts of the biconditional of this theorem.

$$\overline{A B}$$ ≅ $$\overline{C D}$$ if and only if EF = EG

Question 26.
MAKING AN ARGUMENT
A car is designed so that the rear wheel is only partially visible below the body of the car. The bottom edge of the panel is parallel [o the ground. Your friend claims that the point where the tire touches the ground bisects $$\widehat{A B}$$. Is your friend correct? Explain your reasoning.

Maintaining Mathematical Proficiency

Find the missing interior angle measure.

Question 27.
Quadrilateral JKLW has angle measures m∠J = 32°, m∠K = 25°, and m∠L = 44°. Find m∠M.

Question 28.
Pentagon PQRST has angle measures m∠P = 85°, m∠Q = 134°, m∠R = 97°, and m∠S =102°.
Find m∠T.

The sum of interior angles of a pentagon = 540°
m∠T = 540 – (85 + 134 + 97 + 102)
= 540 – 418 = 122
m∠T = 122°.

### 10.1 – 10.3 Quiz

In Exercises 1 – 6, use the diagram. (Section 10.1)

Question 1.
Name the circle.

The circle has a chord, diameter and tangent.

Question 2.

NP is the radius of the circle.

Question 3.
Name a diameter.

KN is the diameter of the circle.

Question 4.
Name a chord.

JL is the chord

Question 5.
Name a secant.

SN is the secant

Question 6.
Name a tangent.

QR is the tangent.

Find the value of x.

Question 7.

(9 + x)² = x² + 15²
81 + 18x + x² = x² + 225
18x = 225 – 81
18x = 144
x = 8

Question 8.

6x – 3 = 3x + 18
6x – 3x = 18 + 3
3x = 21
x = 7

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 9.
$$\widehat{A E}$$

$$\widehat{A E}$$ = 180 – 36
= 144
So, $$\widehat{A E}$$ = 144°

Question 10.
$$\widehat{B C}$$

$$\widehat{B C}$$ = 180 – (67 + 70)
= 180 – 137 = 43
So, $$\widehat{B C}$$ = 43°

Question 11.
$$\widehat{A C}$$

$$\widehat{A C}$$ = 43 + 67 = 110°

Question 12.
$$\widehat{A C D}$$

$$\widehat{A C D}$$ = 180°

Question 13.
$$\widehat{A C E}$$

$$\widehat{A C E}$$ = 180 + 36 = 216°

Question 14.
$$\widehat{B E C}$$

$$\widehat{B E C}$$ = 70 + 36 + 43 = 149°

Tell whether the red arcs are congruent. Explain why or why not.

Question 15.

As two chords pass through the centre of the circle. Those two red arcs are congruent.

Question 16.

Red arcs are not congruent because the radius of the two circles is different.

Question 17.
Find the measure of the red arc in ⊙Q.

Question 18.
In the diagram. AC = FD = 30, PG = x + 5, and PJ = 3x – 1. Find the radius of ⊙P.

Question 19.
A circular clock can be divided into 12 congruent sections.

a. Find the measure of each arc in this circle.
The measure of each arc = $$\frac { 360 }{ 12 }$$ = 30°

b. Find the measure of the minor arc formed by the hour and minute hands when the times is 7:00.
When the time is 7:00 the minute hand is at 12 and hour hand is at 7 and so the minor arc is subtended by 12 – 7 = 5 of these sections and so the angle subtended is 30 x 5 = 150°

c. Find a time at which the hour and minute hands form an arc that is congruent to the arc in part (b).
A minor arc is equal to 150° can be formed by multiple placements of the hour and the minute hand. One of them can be the time 5:00 when the minute hand is at 12 and the hour hand is at 5.

### 10.4 Inscribed Angles and Polygons

Exploration 1

Inscribed Angles and Central Angles

work with a partner: Use dynamic geometry software.

Sample

a. Construct an inscribed angle in a circle. Then construct the corresponding central angle.

b. Measure both angles. How is the inscribed angle related to its intercepted arc?

c. Repeat parts (a) and (b) several times. Record your results in a table. Write a conjecture about how an inscribed angle is related to its intercepted arc.
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely with others.

Exploration 2

work with a partner: Use dynamic geometry software.

Sample

a. Construct a quadrilateral with each vertex on a circle.

b. Measure all four angles. What relationships do you notice?

c. Repeat parts (a) and (b) several times. Record your results in a table. Then write a conjecture that summarizes the data.

Question 3.
How are inscribed angles related to their intercepted arcs? How are the angles of an inscribed quadrilateral related to each other?

Question 4.
Quadrilateral EFGH is inscribed in ⊙C. and m ∠ E = 80°. What is m ∠ G? Explain.

m ∠ E + m ∠ H = 80 + 80 = 160°
m ∠ E + m ∠ H + m ∠ G + m ∠ F = 360
160° + m ∠ G + m ∠ F = 360
m ∠ G + m ∠ F = 360 – 160 = 200
m ∠ G = 100°

### Lesson 10.4 Inscribed Angles and Polygons

Monitoring Progress

Find the measure of the red arc or angle.

Question 1.

m∠G = $$\frac { 90 }{ 2 }$$ = 45°

Question 2.

$$\widehat{T V}$$ = 2 • 38 = 76°

Question 3.

m∠W = 72°

Find the value of each variable.

Question 4.

x° = 90°
y° = 180 – (40 + 90) = 180 – 130
y° = 50°

Question 5.

∠B + ∠D = 180
∠B + 82 = 180
x° = 98°
∠C + ∠A = 180
68 + y° = 180
y° = 112°

Question 6.

∠S + ∠U = 180°
c + 2c – 6 = 180
3c = 186
c = 62°
∠T + ∠V = 180°
10x + 8x = 180
18x = 180
x = 10°

Question 7.
In Example 5, explain how to find locations where the left side of the statue is all that appears in your camera’s field of vision.

### Exercise 10.4 Inscribed Angles and Polygons

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
If a circle is circumscribed about a polygon, then the polygon is an ___________ .

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different?

Find m∠ABC.
m∠ABC = 60°

Find m∠AGC.
m∠AGC = 180 – (25 + 25)
= 180 – 50 = 130°

Find m∠AEC.
m∠AEC = 180 – (50 + 50)
= 180 – 100 = 80°

m∠ADC = 180 – (25 + 50)
= 180 – 75 = 105°

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find the indicated measure.

Question 3.
m∠A

Question 4.
m∠G

m∠G = 360 – (70 + 120)
= 360 – 190 = 170°

Question 5.
m ∠ N

Question 6.
m$$\widehat{R S}$$

m$$\widehat{R S}$$ = 2 • 67 = 134°

Question 7.
m$$\widehat{V U}$$

Question 8.
m$$\widehat{W X}$$

m$$\widehat{W X}$$ = $$\frac { 75 }{ 2 }$$ = 37.5

In Exercises 9 and 10, name two pairs of congruent angles.

Question 9.

Question 10.

m∠W = m∠Z, m∠X = m∠Y

In Exercises 11 and 12, find the measure of the red arc or angle.

Question 11.

Question 12.

$$\widehat{P S}$$ = 2 • 40 = 80

In Exercises 13 – 16, find the value of each variable.

Question 13.

Question 14.

m∠E + m∠G = 180
m + 60 = 180
m = 120°
m∠D + m∠F = 180
60 + 2k = 180
k = 60°

Question 15.

Question 16.

3x° = 90°
x° = 30°
2y° + 90° + 34° = 180°
2y° + 124° = 180°
2y° = 56°
y° = 28°

Question 17.
ERROR ANALYSIS
Describe and correct the error in finding m$$\widehat{B C}$$.

Question 18.
MODELING WITH MATHEMATICS
A carpenter’s square is an L-shaped tool used to draw right angles. You need to cut a circular piece of wood into two semicircles. How can you use the carpenter’s square to draw a diameter on the circular piece of wood?

Recall that when a right triangle is inscribed in a circle, then the hypotenuse is the diameter of the circle. Simply use the carpenter’s square to inscribe it into the circle. The hypotenuse formed by both legs of the square should provide a diameter for the circle.

MATHEMATICAL CONNECTIONS
In Exercises 19 – 21, find the values of x and y. Then find the measures of the interior angles of the polygon.

Question 19.

Question 20.

∠B + ∠C = 180
14x + 4x = 180
18x = 180°
x = 10°
∠A + ∠D = 180
9y + 24y = 180
33y = 180°
y = 5.45°
∠A = 130.9°, ∠B = 40°, ∠C = 140°, ∠D = 49°

Question 21.

Question 22.
MAKING AN ARGUMENT
Your friend claims that ∠PTQ ≅ ∠PSQ ≅ ∠PRQ. Is our friend correct? Explain your reasoning.

Yes, my friend is correct.
∠PTQ ≅ ∠PSQ ≅ ∠PRQ is correct according to the inscribed angles of a circle theorem.

Question 23.
CONSTRUCTION
Construct an equilateral triangle inscribed in a circle.

Question 24.
CONSTRUTION
The side length of an inscribed regular hexagon is equal to the radius of the circumscribed circle. Use this fact to construct a regular hexagon inscribed in a circle.

As the side length is equal to the radius. Draw a line representing the radius and draw a chord different chords in the form of hexagons of the radius of the circle.

REASONING
In Exercises 25 – 30, determine whether a quadrilateral of the given type can always be inscribed inside a circle. Explain your reasoning.

Question 25.
Square

Question 26.
rectangle
yes, angles are right angles.

Question 27.
parallelogram

Question 28.
kite
No.

Question 29.
rhombus

Question 30.
isosceles trapezoid
Yes, the opposite angles are always supplementary.

Question 31.
MODELING WITH MATHEMATICS
Three moons, A, B, and C, are in the same circular orbit 1,00,000 kilometers above the surface of a planet. The planet is 20,000 kilometers in diameter and m∠ABC = 90°. Draw a diagram of the situation. How far is moon A from moon C?

Question 32.
MODELING WITH MATHEMATICS
At the movie theater. you want to choose a seat that has the best viewing angle, so that you can be close to the screen and still see the whole screen without moving your eyes. You previously decided that seat F7 has the best viewing angle, but this time someone else is already sitting there. Where else can you sit so that your seat has the same viewing angle as seat F7? Explain.

Question 33.
WRITING
A right triangle is inscribed in a circle, and the radius of the circle is given. Explain how to find the length of the hypotenuse.

Question 34.
HOW DO YOU SEE IT?
Let point Y represent your location on the soccer field below. What type of angle is ∠AYB if you stand anywhere on the circle except at point A or point B?

Question 35.
WRITING
Explain why the diagonals of a rectangle inscribed in a circle are diameters of the circle.

Question 36.
THOUGHT PROVOKING
The figure shows a circle that is circumscribed about ∆ABC. Is it possible to circumscribe a circle about any triangle? Justify your answer.

Yes.

Question 37.
PROVING A THEOREM
If an angle is inscribed in ⊙Q. the center Q can be on a side of the inscribed angle, inside the inscribed angle, or outside the inscribed angle. Prove each case of the Measure of an Inscribed Angle Theorem (Theorem 10. 10).

a. Case 1
Given ∠ABC is inscribed in ⊙Q
Let m∠B = x°
Center Q lies on $$\overline{B C}$$.
Prove m∠ABC = $$\frac{1}{2}$$m$$\widehat{A C}$$
(Hint: Show that ∆AQB is isosceles. Then write m$$\widehat{A C}$$ in terms of x.)

b. Case 2
Use the diagram and auxiliary line to write Given and Prove statements for Case 2. Then write a proof

c. Case 3
Use the diagram and auxiliary line to write Given and Prove statements for Case 3. Then write a proof.

Question 38.
PROVING A THEOREM
Write a paragraph proof of the Inscribed Angles of a Circle Theorem (Theorem 10.11). First, draw a diagram and write Given and Prove statements.
If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

Question 39.
PROVING A THEOREM
The Inscribed Right Triangle Theorem (Theorem 10.12) is written as a conditional statement and its converse. Write a plan for proof for each statement.

Question 40.
PROVING A THEOREM
Copy and complete the paragraph proof for one part of the Inscribed Quadrilateral Theorem (Theorem 10. 13).
Given ⊙C with inscribed quadrilateral DEFG
Prove m ∠ D + m ∠ F = 180°,
m ∠ E + m ∠ G = 180°

By the Arc Addition Postulate (Postulate 10. 1),
m$$\widehat{E F G}$$ + ________ = 360° and m$$\widehat{F G D}$$ + m$$\widehat{D E F}$$ = 360°.
Using the ___________ Theorem. m$$\widehat{E D G}$$ = 2m ∠ F, m$$\widehat{E F G}$$ = 2m∠D, m$$\widehat{D E F}$$ = 2m∠G, and m$$\widehat{F G D}$$ = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + ________ = 360°, So _________ . Similarly, __________ .
m$$\widehat{E F G}$$ + m$$\widehat{E D F}$$ = 360° and m$$\widehat{F G D}$$ + m$$\widehat{D E F}$$ = 360°.
Using the the measure of an inscribed angle Theorem. m$$\widehat{E D G}$$ = 2m ∠ F, m$$\widehat{E F G}$$ = 2m∠D, m$$\widehat{D E F}$$ = 2m∠G, and m$$\widehat{F G D}$$ = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + 2m∠G = 360°.

Question 41.
CRITICAL THINKING
In the diagram, ∠C is a right angle. If you draw the smallest possible circle through C tangent to $$\overline{A B}$$, the circle will intersect $$\overline{A C}$$ at J and $$\overline{B C}$$ at K. Find the exact length of $$\overline{J K}$$.

Question 42.
CRITICAL THINKING
You are making a circular cutting board. To begin, you glue eight 1-inch boards together, as shown. Then you draw and cut a circle with an 8-inch diameter from the boards.

a. $$\overline{F H}$$ is a diameter of the circular cutting board. Write a proportion relating GJ and JH. State a theorem in to justify your answer.
Each board is 1 inch and FJ spans 6 boards.
$$\overline{F H}$$ = 6 inches

b. Find FJ, JH, and GJ. What is the length of the cutting board seam labeled $$\overline{G K}$$?
Each board is 1 inch and JH spans 2 boards.
JH = 2 inches
Equation is $$\frac { JH }{ GJ }$$ = $$\frac { GJ }{ FJ }$$
$$\frac { 2 }{ GJ }$$ = $$\frac { GJ }{ 6 }$$
12 = GJ²
GJ = 2√3
GK = 2(GJ)
GK = 4√3
So, FJ = 6, JH = 2, JG = 2√3, GK = 4√3

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.

Question 43.
3x = 145

Question 44.
$$\frac{1}{2}$$x = 63
x = 63 • 2
x = 126

Question 45.
240 = 2x

Question 46.
75 = $$\frac{1}{2}$$(x – 30)
75 • 2 = x – 30
150 + 30 = x
x = 180

### 10.5 Angle Relationships in Circles

Exploration 1

Angles Formed by a Chord and Tangent Line

Work with a partner: Use dynamic geometry software.

Sample

a. Construct a chord in a circle. At one of the endpoints of the chord. construct a tangent line to the circle.

b. Find the measures of the two angles formed by the chord and the tangent line.

c. Find the measures of the two circular arcs determined by the chord.

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.

Exploration 2

Angles Formed by Intersecting Chords

Work with a partner: Use dynamic geometry software.

sample

a. Construct two chords that intersect inside a circle.

b. Find the measure of one of the angles formed by the intersecting chords.

c. Find the measures of the arcs intercepted h the angle in part (b) and its vertical angle. What do you observe?

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.

Question 3.
When a chord intersects a tangent line or another chord, what relationships exist among the angles and arcs formed?

Question 4.
Line m is tangent to the circle in the figure at the left. Find the measure of ∠1.

m∠1 = $$\frac { 1 }{ 2 }$$ • 148
m∠1 = 74°

Question 5.
Two chords intersect inside a circle to form a pair of vertical angles with measures of 55°. Find the sum of the measures of the arcs intercepted by the two angles.
The sum of the measures of the arcs intercepted by the two angles = $$\frac { 1 }{ 2 }$$ • 55
= 27.5

### Lesson 10.5 Angle Relationships in Circles

Monitoring Progress

Line m is tangent to the circle. Find the indicated measure.

Question 1.
m ∠ 1

m ∠ 1 = $$\frac { 1 }{ 2 }$$ • 210
m ∠ 1 = 105°

Question 2.
m$$\widehat{R S T}$$

m$$\widehat{R S T}$$ = 2 • 98 = 196°
m$$\widehat{R S T}$$ = 196°

Question 3.
m$$\widehat{X Y}$$

m$$\widehat{X Y}$$ = $$\frac { 1 }{ 2 }$$ • 80
m$$\widehat{X Y}$$ = 40°

Find the value of the variable.

Question 4.

y° = $$\frac { 1 }{ 2 }$$ • (102 + 95)
= 98.5°

Question 5.

a° = 2 • 30° + 44°
= 60° + 44° = 104°
So, a° = 104°.

Find the value of x.

Question 6.

x° = 180° – 120°
x° = 60°

Question 7.

50° = 180° – x°
x° = 180° – 50°
x° = 130°

Question 8.
You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level at point B. Find m$$\widehat{C D}$$, which represents the part of Earth that you can see.

CB and BD are tangents, CB is perpendicular to AB and CD is perpendicular to AD by the tangent line to circle theorem.
△ABC is similar to △ABD by the hypotenuese leg congruence theorem.
∠CBA is similar to ∠ABD. So, m∠CBA = 74.5°, m∠CBD = 2 • 74.5° = 149°
m∠CBD = 180° – CD
149° = 180° – CD
CD = 31°
The part of earth you can see

### Exercise 10.5 Angle Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Points A, B, C, and D are on a circle, and intersects at point P.
If m∠APC = $$\frac{1}{2}$$(m$$\widehat{B D}$$ – m$$\widehat{A C}$$). then point P is _________ the circle.

Question 2.
WRITING
Explain how to find the measure of a circumscribed angle.
A circumscribed angle is the angle made by two intersecting tangent lines to a circle. Draw lines from the circle centre to the point of tangency. The angle between the radius and tangent line is 90°. The sum of angles of a quadrilateral is 360°. Angles between radii and tangent lines is 180°. The angle at two tangent lines meet is circumscribed angle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, line t is tangent to the circle. Find the indicated measure.

Question 3.
m$$\widehat{A B}$$

Question 4.
m$$\widehat{D E F}$$

m$$\widehat{D E F}$$ = 2(117°) = 234°

Question 5.
m < 1

Question 6.
m ∠ 3

m ∠ 3 = ½ • 140 = 70°

In Exercises 7 – 14, find the value of x.

Question 7.

Question 8.

x° = ½ • (30 + 2x – 30)

Question 9.

Question 10.

34° = ½ (3x – 2 – (x + 6))
34° = ½ (3x – 2 – x – 6)
34° = ½ (2x – 8)
34° = x – 4
x° = 34 + 4
x° = 38°

Question 11.

Question 12.

6x – 11 = 2  • 125
6x = 250 + 11
6x = 261
x° = 43.5°

Question 13.

Question 14.

17x° = 75°
x° = 4.41°

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in finding the angle measure.

Question 15.

Question 16.

m∠1 = ½ (122 – 70)
= ½ (52) = 26
So, m∠1 = 26°

In Exercises 17 – 22, find the indicated angle measure. justify your answer.

Question 17.
m ∠ 1

Question 18.
m ∠ 2
m ∠ 2 = 60°

Explanation:
m ∠ 3 =30°, So, m ∠ 2 = 180° – (90° + 30°)
= 180° – 120° = 60°
Therefore, m ∠ 2 = 60°

Question 19.
m ∠ 3

Question 20.
m ∠ 4
m ∠ 4 = 90°

Question 21.
m ∠ 5

Question 22.
m ∠ 6
m ∠ 6 = 180° – (60° + 30° + 30°) = 180° – 120°
m ∠ 6 = 60°

Question 23.
PROBLEM SOLVING
You are flying in a hot air balloon about 1.2 miles above the ground. Find the measure of the arc that represents the part of Earth you can see. The radius of Earth is about 4000 miles.

Question 24.
PROBLEM SOLVING
You are watching fireworks over San Diego Bay S as you sail away in a boat. The highest point the fireworks reach F is about 0.2 mile above the bay. Your eyes E are about 0.01 mile above the water. At point B you can no longer see the fireworks because of the curvature of Earth. The radius of Earth is about 4000 miles, and $$\overline{F E}$$ is tangent to Earth at point T. Find m$$\widehat{s B}$$. Round your answer to the nearest tenth.

Question 25.
MATHEMATICAL CONNECTIONS
In the diagram, $$\vec{B}$$A is tangent to ⊙E. Write an algebraic expression for m$$\widehat{C D}$$ in terms of x. Then find m$$\widehat{C D}$$.

Question 26.
MATHEMATICAL CONNECTIONS
The circles in the diagram are concentric. Write an algebraic expression for c in terms of a and b.

a° = ½(c° – b°)

Question 27.
ABSTRACT REASONING
In the diagram. $$\vec{P}$$L is tangent to the circle, and $$\overline{K J}$$ is a diameter. What is the range of possible angle measures of ∠LPJ? Explain your reasoning.

Question 28.
ABSTRACT REASONING
In the diagram, $$\overline{A B}$$ is an chord that is not a diameter of the circle. Line in is tangent to the circle at point A. What is the range of possible values of x? Explain your reasoning. (The diagram is not drawn to scale.)

The possible values of x are less than 180°.

Question 29.
PROOF
In the diagram and are secant lines that intersect at point L. Prove that m∠JPN > m∠JLN.

Question 30.
MAKING AN ARGUMENT
Your friend claims that it is possible for a circumscribed angle to have the same measure as its intercepted arc. Is your friend correct? Explain your reasoning.

Yes, when the circumscribed angle measures 90°, the central angle measures 90°, so the intercepted arc also measures 90°.

Question 31.
REASONING
Points A and B are on a circle, and t is a tangent line containing A and another point C.
a. Draw two diagrams that illustrate this situation.
b. Write an equation for m$$\widehat{A B}$$ in terms of m∠BAC for each diagram.
c. For what measure of ∠BAC can you use either equation to find m$$\widehat{A B}$$? Explain.

Question 32.
REASONING
∆XYZ is an equilateral triangle inscribed in ⊙P. AB is tangent to ⊙P at point X, $$\overline{B C}$$ is tangent to ⊙P at point Y. and $$\overline{A C}$$ is tangent to ⊙P at point Z. Draw a diagram that illustrates this situation. Then classify ∆ABC by its angles and sides. Justify your answer.

Question 33.
PROVING A THEOREM
To prove the Tangent and Intersected Chord Theorem (Theorem 10. 14), you must prove three cases.
a. The diagram shows the case where $$\overline{A B}$$ contains the center of the circle. Use the Tangent Line to Circle Theorem (Theorem 10.1) to write a paragraph proof for this case.

b. Draw a diagram and write a proof for the case where the center of the circle is in the interior of ∠CAB.
c. Draw a diagram and write a proof for the case where the center of the circle is in the exterior of ∠CAB.

Question 34.
HOW DO YOU SEE IT?
In the diagram, television cameras are Positioned at A and B to record what happens on stage. The stage is an arc of ⊙A. You would like the camera at B to have a 30° view of the stage. Should you move the camera closer or farther away? Explain your reasoning.

25° = ½(80° – 30°) = ½(50°)
So, you should move the camera closer.

Question 35.
PROVING A THEOREM
Write a proof of the Angles Inside the Circle Theorem (Theorem 10.15).

Given Chords $$\overline{A C}$$ and $$\overline{B D}$$ intersect inside a circle.
Prove m ∠ 1 = $$\frac{1}{2}$$(m$$\widehat{D C}$$ + m$$\widehat{A B}$$)

Question 36.
THOUGHT PROVOKING
In the figure, and are tangent to the circle. Point A is any point on the major are formed by the endpoints of the chord $$\overline{B C}$$. Label all congruent angles in the figure. Justify your reasoning.

m∠CPB = ½(CAB – CB)

Question 37.
PROVING A THEOREM
Use the diagram below to prove the Angles Outside the Circle Theorem (Theorem 10.16) for the case of a tangent and a secant. Then copy the diagrams for the other two cases on page 563 and draw appropriate auxiliary segments. Use your diagrams to prove each case.

Question 38.
PROVING A THEOREM
Prove that the Circumscribed Angle Theorem (Theorem 10.17) follows from the Angles Outside the Circle Theorem (Theorem 10.16).

In Exercises 39 and 40, find the indicated measure(s). justify your answer

Question 39.
Find m ∠ P when m$$\widehat{W Z Y}$$ = 200°

Question 40.
Find m$$\widehat{A B}$$ and m$$\widehat{E D}$$

m$$\widehat{E D}$$ = ½ (115°) = 57.5°
∠GJA = 30°

Maintaining Mathematical Proficiency

Solve the equation.

Question 41.
x2 + x = 12

Question 42.
x2 = 12x + 35

Explanation:
x² = 12x + 35
x = $$\frac { 12 ± √(144 + 140) }{ 2 }$$
x = $$\frac { 12 ± √284 }{ 2 }$$
x = $$\frac { 12 + √284 }{ 2 }$$, $$\frac { 12 – √284 }{ 2 }$$

Question 43.
– 3 = x2 + 4x

### 10.6 Segment Relationships in Circles

Exploration 1

Segments Formed by Two Intersecting Chords

Work with a partner: Use dynamic geometry software.

Sample

a.
Construct two chords $$\overline{B C}$$ and $$\overline{D E}$$ that intersect in the interior of a circle at a point F.

b.
Find the segment lengths BE, CF, DF, and EF and complete the table. What do you observe?

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
REASONING ABSTRACTLY
To be proficient in math, you need to make sense of quantities and their relationships in problem situations.

Exploration 2

Secants Intersecting Outside a Circle

Work with a partner: Use dynamic geometry software.

Sample

a. Construct two secant and that intersect at a point B outside a circle, as shown.

b. Find the segment lengths BE, BC, BF, and BD. and complete the table. What do you observe?

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.

Question 3.
What relationships exist among the segments formed by two intersecting chords or among segments of two secants that intersect outside a circle?

Question 4.
Find the segment length AF in the figure at the left.

Explanation:
EA • AF = AD • AC
18 • AF = 9 • 8
AF = 4

### Lesson 10.6 Segment Relationships in Circles

Monitoring Progress

Find the value of x.

Question 1.

x = 8

Explanation:
4 • 6 = 3 • x
3x = 24
x = 8

Question 2.

x = 5

Explanation:
2 • x + 1 = 4 • 3
x + 1 = 6
x = 5

Question 3.

x = $$\frac { 54 }{ 5 }$$

Explanation:
6 • 9 = 5 • x
54 = 5x
x = $$\frac { 54 }{ 5 }$$

Question 4.

x = $$\frac { 3 ± √37 }{ 2 }$$

Explanation:
3 • x + 2 = x + 1 • x – 1
3x + 6 = x² – 1
x² – 3x – 7 = 0
x = $$\frac { 3 ± √(9 + 28) }{ 2 }$$
x = $$\frac { 3 ± √37 }{ 2 }$$

Question 5.

x = ±√3

Explanation:
x² = 3 • 1
x² = 3
x = ±√3

Question 6.

x = $$\frac { 49 }{ 5 }$$

Explanation:
7² = 5 • x
49 = 5x
x = $$\frac { 49 }{ 5 }$$

Question 7.

x = 14.4

Explanation:
12² = 10x
144 = 10x
x = 14.4

Question 8.
WHAT IF?
In Example 4, CB = 35 feet and CE = 14 feet. Find the radius of the tank.

The radius of the tank is 36.75

Explanation:
CB² = CE ⋅ CD
35² = 14 ⋅ (2r + 14)
1225 = 28r + 196
28r = 1029
r = 36.75

### Exercise 10.6 Segment Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The part of the secant segment that is outside the circle is called a(n) ______________ .

Question 2.
WRITING
Explain the difference between a tangent segment and a secant segment.

A tangent segment intersects the circle at only one point. It actually doesn’t go through the circle. If a ball is rolling on a table top, then it would be the tangent. A secant segment intersects the circle in two points. It goes through the circle. In a tangent, no part is in the interior of the circle. In a secant, there is a part in the interior called a chord.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x.

Question 3.

Question 4.

x = 23

Explanation:
10 • 18 = 9 • (x – 3)
20 = x – 3
x = 23

Question 5.

Question 6.

x = 5

Explanation:
2x • 12 = 15 • (x + 3)
24x = 15x + 45
9x = 45
x = 5

In Exercises 7 – 10, find the value of x.

Question 7.

Question 8.

x = $$\frac { 35 }{ 4 }$$

Explanation:
5 • 7 = 4 • x
4x = 35
x = $$\frac { 35 }{ 4 }$$

Question 9.

Question 10.

x = 30

Explanation:
45 • x = 50 • 27
45x = 1350
x = 30

In Exercises 11 – 14. find the value of x.

Question 11.

Question 12.

x = 48

Explanation:
24² = 12x
576 = 12x
x = 48

Question 13.

Question 14.

x = 1.5

Explanation:
3 = 2x
x = 1.5

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding CD.

Question 16.
MODELING WITH MATHEMATICS
The Cassini spacecraft is on a mission in orbit around Saturn until September 2017. Three of Saturn’s moons. Tethys. Calypso, and Teleslo. have nearly circular orbits of radius 2,95,000 kilometers. The diagram shows the positions of the moons and the spacecraft on one of Cassini’s missions. Find the distance DB from Cassini to Tethys when $$\overline{A D}$$ is tangent to the circular orbit.

BD = 579493 km

Explanation:
(203,000)² = 83000x
x = 496493
BC = 496493
BD = 496493 + 83000 = 579493

Question 17.
MODELING WITH MATHEMATICS
The circular stone mound in Ireland called Newgrange has a diameter of 250 feet. A passage 62 feet long leads toward the center of the mound. Find the perpendicular distance x from the end of the passage to either side of the mound.

Question 18.
MODELING WITH MATHEMATICS
You are designing an animated logo for our website. Sparkles leave point C and move to the Outer circle along the segments shown so that all of the sparkles reach the outer circle at the same time. Sparkles travel from point C to point D at 2 centimeters per second. How fast should sparkles move from point C to point N? Explain.

5.33 should sparkles move from point C to point N.

Explanation:
4 • 8 = 6 • x
x = 5.33

Question 19.
PROVING A THEOREM
Write a two-column proof of the Segments of Chords Theorem (Theorem 10.18).

Plan for Proof:
Use the diagram to draw $$\overline{A C}$$ and $$\overline{D B}$$. Show that ∆EAC and ∆EDB are similar. Use the fact that corresponding side lengths in similar triangles are proportional.

Question 20.
PROVING A THEOREM
Prove the Segments of Secants Theorem (Theorem 10.19). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)

Question 21.
PROVING A THEOREM
Use the Tangent Line to Circle Theorem (Theorem 10. 1) to prove the Segments of Secants and Tangents Theorem (Theorem 10.20) for the special case when the secant segment Contains the center of the circle.

Question 22.
PROVING A THEOREM
Prove the Segments of Secants and Tangents Theorem (Theorem 10.20). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)

Question 23.
WRITING EQUATIONS
In the diagram of the water well, AB, AD, and DE are known. Write an equation for BC using these three measurements.

Question 24.
HOW DO YOU SEE IT?
Which two theorems would you need to use to tind PQ? Explain your reasoning.

Question 25.
CRITICAL THINKING
In the figure, AB = 12, BC = 8, DE = 6, PD = 4, and a is a point of tangency. Find the radius of ⊙P.

Question 26.
THOUGHT PROVOKING
Circumscribe a triangle about a circle. Then, using the points of tangency, inscribe a triangle in the circle. Must it be true that the two triangles are similar? Explain your reasoning.

Maintaining Mathematical Proficiency

Solve the equation by completing the square.

Question 27.
x2 + 4x = 45

Question 28.
x2 – 2x – 1 = 8

x = 1 + √10, x = 1 – √10

Explanation:
x² – 2x – 1 = 8
x² – 2x – 9 = 0
x = $$\frac { 2 ± √(4 + 36) }{ 2 }$$
x = $$\frac { 2 ± √40 }{ 2 }$$
x = 1 + √10, x = 1 – √10

Question 29.
2x2 + 12x + 20 = 34

Question 30.
– 4x2 + 8x + 44 = 16

x = 1 + √8, x = 1 – √8

Explanation:
– 4x² + 8x + 44 = 16
4x² – 8x – 28 =0
x² – 2x – 7 = 0
x = $$\frac { 2 ± √(4 + 28) }{ 2 }$$
x = $$\frac { 2 ± √32 }{ 2 }$$
x = 1 ± √8

### 10.7 Circles in the Coordinate Plane

Exploration 1

The Equation of a Circle with Center at the Origin

Work with a partner: Use dynamic geometry software to Construct and determine the equations of circles centered at (0, 0) in the coordinate plane, as described below.

a. Complete the first two rows of the table for circles with the given radii. Complete the other rows for circles with radii of your choice.

b. Write an equation of a circle with center (0, 0) and radius r.

x²  + y²  = r²

Explanation:
(x – 0)² + (y – 0)²  = r²
x²  + y²  = r²

Exploration 2

The Equation of a Circle with Center (h, k)

Work with a partner: Use dynamic geometry software to construct and determine the equations of circles of radius 2 in the coordinate plane, as described below.

a. Complete the first two rows of the table for circles with the given centers. Complete the other rows for circles with centers of your choice.

b. Write an equation of a circle with center (h, k) and radius 2.

(x – h)² + (y – k)² = 4

c. Write an equation of a circle with center (h, k) and radius r.

(x – h)² + (y – k)² = r²

Exploration 3

Deriving the Standard Equation of a Circle

Work with a partner. Consider a circle with radius r and center (h, k).

Write the Distance Formula to represent the distance d between a point (x, y) on the circle and the center (h, k) of the circle. Then square each side of the Distance Formula equation.

How does your result compare with the equation you wrote in part (c) of Exploration 2?

MAKING SENSE OF PROBLEMS
To be proficient in math, you need to explain correspondences between equations and graphs.

(x – h)² + (y – k)² = r²

Question 4.
What is the equation of a circle with center (h, k) and radius r in the coordinate plane?

(x – h)² + (y – k)² = r²

Question 5.
Write an equation of the circle with center (4, – 1) and radius 3.

x² + y² – 8x + 2y + 8 = 0

Explanation:
(x – 4)² + (y + 1)² = 9
x² – 8x + 16 + y² + 2y + 1 = 9
x² + y² – 8x + 2y = 9 – 17
x² + y² – 8x + 2y + 8 = 0

### Lesson 10.7 Circles in the Coordinate Plane

Monitoring Progress

Write the standard equation of the circle with the given center and radius.

Question 1.

x² + y² = 6.25

Explanation:
(x – 0)² + (y – 0)² = 2.5²
x² + y² = 6.25

Question 2.
center: (- 2, 5), radius: 7

(x + 2)² + (y – 5)² = 49

Explanation:
(x + 2)² + (y – 5)² = 7²
(x + 2)² + (y – 5)² = 49

Question 3.
The point (3, 4) is on a circle with center (1, 4). Write the standard equation of the circle.

(x – 1)² + (y – 4)² = 4

Explanation:
r = √(3 – 1)² + (4 – 4)²
= √(2)²
r = 2
(x – 1)² + (y – 4)² = 2²
(x – 1)² + (y – 4)² = 4

Question 4.
The equation of a circle is x2 + y2 – 8x + 6y + 9 = 0. Find the center and the radius of the circle. Then graph the circle.

The center of the circle (4, -3) and radius is 4.

Explanation:
x² + y² – 8x + 6y + 9 = 0
x² – 8x + 16 + y² + 6y + 9 = 16
(x – 4)² + (y + 3)² = 4²
The center of the circle (4, -3) and radius is 4.

Question 5.
Prove or disprove that the point (1, √5 ) lies on the circle centered at the origin and containing the point (0, 1).

Disproved.

Explanation:
We consider the circle centred at the origin and containing the point (0, 1).
Therefore, we can conclude that the radius of the circle r = 1, let O (0, 0) and B (1, √5). So the distance between two points is
OB = √(1 – 0) + (√5 – 0)² = √(1 + 5) = √6
As the radius of the given circle is 1 and distance of the point B from its centre is √6. So we can conclude that point does lie on the given circle.

Question 6.
why are three seismographs needed to locate an earthquake’s epicentre?

### Exercise 10.7 Circles in the Coordinate Plane

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is the standard equation of a circle?

Question 2.
WRITING
Explain why knowing the location of the center and one point on a circle is enough to graph the circle.

If we know the location of the center and one point on the circle, we can graph a circle because the distance from the center to the point is called the radius.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, write the standard equation of the circle.

Question 3.

Question 4.

x² + y² = 36

Explanation:
The center is (0, 0) and the radius is 6
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 6²
x² + y² = 36

Question 5.
a circle with center (0, 0) and radius 7

Question 6.
a circle with center (4, 1) and radius 5

(x – 4)² + (y – 1)² = 25

Explanation:
(x – h)² + (y – k)² = r²
(x – 4)² + (y – 1)² = 5²
(x – 4)² + (y – 1)² = 25

Question 7.
a circle with center (- 3, 4) and radius 1

Question 8.
a circle with center (3, – 5) and radius 7

(x – 3)² + (y + 5)² = 49

Explanation:
(x – h)² + (y – k)² = r²
(x – 3)² + (y + 5)² = 7²
(x – 3)² + (y + 5)² = 49

In Exercises 9 – 11, use the given information to write the standard equation of the circle.

Question 9.
The center is (0, 0), and a point on the circle is (0, 6).

Question 10.
The center is (1, 2), and a point on the circle is (4, 2).

x² + y² = 9

Explanation:
r = √(x – h)² + (y – k)²
= √(4 – 1)² + (2 – 2)²
= √3²
r = 3
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 3²
x² + y² = 9

Question 11.
The center is (0, 0). and a point on the circle is (3, – 7).

Question 12.
ERROR ANALYSIS
Describe and correct the error in writing the standard equation of a circle.

(x – h)² + (y – k)² = r²
(x + 3)² + (y + 5)² = 3²
(x + 3)² + (y + 5)² = 9

In Exercises 13 – 18, find the center and radius of the circle. Then graph the circle.

Question 13.
x2 + y2 = 49

Question 14.
(x + 5)2 + (y – 3)2 = 9

Center is (-5, 3) and rdaius is 3.

Explanation:
For the equation (x + 5)2 + (y – 3)2 = 9, center is (-5, 3) and rdaius is 3.

Question 15.
x2 + y2 – 6x = 7

Question 16.
x2 + y2 + 4y = 32

The center is (0, -2) and radius is 6

Explanation:
x2 + y2 + 4y = 32
x² + y² + 4y + 4 = 32 + 4
x² + (y + 2)² = 36
(x – 0)² + (y – (-2))² = 6²
The center is (0, -2) and radius is 6

Question 17.
x2 + y2 – 8x – 2y = – 16

Question 18.
x2 + y2 + 4x + 12y = – 15

The center is (-2, -6) and radius is 5

Explanation:
x2 + y2 + 4x + 12y = – 15
x² + 4x + 4 + y² + 12y + 36 = -15 + 36 + 4
(x + 2)² + (y + 6)² = 5²
The center is (-2, -6) and radius is 5

In Exercises 19 – 22, prove or disprove the statement.

Question 19.
The point (2, 3) lies on the circle centered at the origin with radius 8.

Question 20.
The point (4, √5) lies on the circle centered at the origin with radius 3.

The point (4, √5) does not lie on the circle.

Explanation:
r² = (x – h)² + (y – k)²
3² ______________ (4 – 0)² + (√5 – 0)²
9 ______________ 16 + 5
9 ≠ 21
Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle.

Question 21.
The point (√6, 2) lies on the circle centered at the origin and containing the point (3, – 1).

Question 22.
The point (√7, 5) lies on the circle centered at the origin and containing the point (5, 2).

The point (√7, 5) does not lie on the circle.

Explanation:
r² = (x – h)² + (y – k)²
= (√7 – 0)² + (5 – 0)² = 7 + 25 = 32
r = 5.65
(5.65)² ______________ (5 – 0)² + (2 – 0)²
32 ______________ 25 + 4
32 ≠ 29
Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle.

Question 23.
MODELING WITH MATHEMATICS
A City’s commuter system has three zones. Zone I serves people living within 3 miles of the city’s center. Zone 2 serves those between 3 and 7 miles from the center. Zone 3 serves those over 7 miles from the center.

a. Graph this Situation on a coordinate plane where each unit corresponds to 1 mile. Locate the city’s center at the origin.
b. Determine which zone serves people whose homes are represented by the points (3, 4), (6, 5), (1, 2), (0.3). and (1, 6).

Question 24.
MODELING WITH MATHEMATICS
Telecommunication towers can be used to transmit cellular phone calls. A graph with units measured in kilometers shows towers at points (0, 0), (0, 5), and (6, 3). These towers have a range of about 3 kilometers.
a. Sketch a graph and locate the towers. Are there any locations that may receive calls from more than one tower? Explain your reasoning.

b. The center of City A is located at (- 2, 2.5), and the center of City B is located at (5, 4). Each city has a radius of 1.5 kilometers. Which city seems to have better cell phone coverage? Explain your reasoning.

There are three towers at points (0, 0), (0, 5) and (6, 3) with range of about 3 km
a. Let’s sketch the graph to locate towers. Draw the points (0, 0), (0, 5) and (6, 3). Then draw three circles with centers at these points and radii 3.
There are locations that can receive calls from more that one tower because circles with centers (0, 0) and (0, 5) overlap. Locations in their intersection can receive calls from two towers.
The city A has a center located at (-2, 2.5) and city B has a center located at (5, 4). Both cities have radius 1.5 km
Let’s draw the city A as a circle with center (-2, 2.5) and radius 1.5 and city B with center (5, 4) and radius 1.5.
From the graph we can conclude that the city B has better cell phone coverage because parts of city A do not have coverage.

Question 25.
REASONING
Sketch the graph of the circle whose equation is x2 + y2 = 16. Then sketch the graph of the circle after the translation (x, y) → (x – 2, y – 4). What is the equation of the image? Make a conjecture about the equation of the image of a circle centered at the origin after a translation m units to the left and n units down.

Question 26.
HOW DO YOU SEE IT?
Match each graph with its equation.

 a. A. x2 + (y + 3) 2 = 4 b. B. (x – 3) 2 + y2 = 4 c. C. (x + 3) 2 + y2 = 4 d. D. x2 + (y – 3) 2 = 4

a ➝ C, b ➝ A, c ➝ D, d ➝ B

Question 27.
USING STRUCTURE
The vertices of ∆XYZ are X(4, 5), Y(4, 13), and Z(8, 9). Find the equation of the circle circumscribed about ∆XYZ. Justify your answer.

Question 28.
THOUGHT PROVOKING
A circle has center (h, k) and contains point (a, b). Write the equation of the line tangent to the circle at point (a, b).

y – b = $$\frac { h – a }{ b – k }$$(x – a)

Explanation:
It is given a circle with center C(h, k. A circle with point A(a, b). We have to write the equation of a tangent that intersects the circle at point A
By the tangent line to circle theorem, a tangent is perpendicular to the radius. Two lines are perpendicular if and only if their slopes are negative reciprocals. So, find the equation of the line AC to know its slope.
The equation of the line which has two points (a, b), (c, d) is y – b = $$\frac { b – c }{ a – d }$$(x – a)
The equation of the line which has two points A(a, b) and C(h, k) is
y – b = $$\frac { b – k }{ a – h }$$(x – a)
Therefore, the slope of the line through A nad C is $$\frac { b – k }{ a – h }$$
Hence the slope of the tangent is –$$\frac { a – h }{ b – k }$$ = $$\frac { h – a }{ b – k }$$
Use the equation of the line y = kx + n through the point (a, b)
y – b = k(x – a) to find the equation of the tangent
The equation of the tangent with slope $$\frac { h – a }{ b – k }$$ and through the point A(a, b) is y – b = $$\frac { h – a }{ b – k }$$(x – a)

MATHEMATICAL CONNECTIONS
In Exercises 29 – 32, use the equations to determine whether the line is a tangent, a secant a secant that contains the diameter, or name of these. Explain your reasoning.

Question 29.
Circle: (x – 4)2 + (y – 3)2 = 9
Line: y = 6

Question 30.
Circle: (x + 2)2 + (y – 2)2 = 16
Line: y = 2x – 4

The line is a secant line.

Explanation:
(x + 2)2 + (y – 2)2 = 16, y = 2x – 4
(x + 2)2 + (2x – 4 – 2)2 = 16
x² + 4x + 4 + (2x – 6)² = 16
x² + 4x + 4 + 4x² – 24x + 36 = 16
5x² – 20x + 40 – 16 = 0
5x² – 20x + 24 = 0
x = $$\frac { 20 ±√-80 }{ 10 }$$
x = 2, y = 2 • 2 – 4 = 0, (2, 0)
The system has two solutions and point does not lie on the line. So the line is a secant line.

Question 31.
Circle: (x – 5)2 + (y + 1)2 = 4
Line: y = $$\frac{1}{5}$$x – 3\

Question 32.
Circle: (x + 3)2 + (y – 6)2 = 25
Line: y = –$$\frac{4}{3}$$x + 2

The line is a secant line.

Explanation:
(x + 3)2 + (y – 6)2 = 25, y = –$$\frac{4}{3}$$x + 2
(x + 3)2 + (-$$\frac{4}{3}$$x + 2 – 6)2 = 25
x² + 6x + 9 + $$\frac { 16x² }{ 9 }$$ + $$\frac { 32x }{ 3 }$$ + 16 = 25
$$\frac { 25x² }{ 9 }$$ + $$\frac { 50x }{ 3 }$$ = 0
x(25x + 150) = 0
x = 0 or x = -6
y = 2, y = 10
(0, 2) and (-6, 10)
d = √(0 + 6)² + (2 – 10)²
= √(36 + 64)
= 10 ≠ 5
The system has two solutions and point does not lie on the line. So the line is a secant line.

MAKING AN ARGUMENT
Question 33.
Your friend claims that the equation of a circle passing through the points (- 1, 0) and (1, 0) is x2 – 2yk + y2 = 1 with center (0, k). Is your friend correct? Explain your reasoning.

Question 34.
REASONING
Four tangent circles are centered on the x-axis. The radius of ⊙A is twice the radius of ⊙O, The radius of ⊙B is three times the radius of ⊙O, The radius of ⊙C is four times the radius of ⊙O, All circles have integer radii, and the point (63, 16) is On ⊙C. What is the equation of ⊙A? Explain your reasoning.

Maintaining Mathematical Proficiency

Identify the arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 35.
$$\widehat{R S}$$

Question 36.
$$\widehat{P R}$$

$$\widehat{P R}$$ is a right angle
$$\widehat{P R}$$ = 25 + 65 = 90°

Question 37.
$$\widehat{P R T}$$

Question 38.
$$\widehat{S T}$$

$$\widehat{S T}$$ is a major arc
$$\widehat{S T}$$ = 360 – (90 + 65 +25 + 53) = 127°

Question 39.
$$\widehat{R S T}$$

Question 40.
$$\widehat{Q S}$$

$$\widehat{Q S}$$ is a minor arc
$$\widehat{Q S}$$ = 25 + 53 = 78°

### 10.1 Lines and Segments That Intersect Circles

Tell whether the line, ray, or segment is best described as a radius, chord, diameter, secant, or tangent of ⊙P.

Question 1.
$$\overline{P K}$$

$$\overline{P K}$$ is radius

Question 2.
$$\overline{N M}$$

$$\overline{N M}$$ is chord

Question 3.
$$\vec{J}$$L

$$\vec{J}$$L is tangent

Question 4.
$$\overline{K N}$$

$$\overline{K N}$$ is diameter

Question 5.

NL is secant

Question 6.
$$\overline{P N}$$

$$\overline{P N}$$ is radius

Tell whether the common tangent is internal or external.

Question 7.

Internal common tangent

Question 8.

External common tangent

Points Y and Z are points of tangency. Find the value of the variable.

Question 9.

a = $$\frac { 3 ± 33 }{ 18 }$$

Explanation:
3a = 9a² – 30
9a² – 3a – 30 = 0
a = $$\frac { 3 ± 33 }{ 18 }$$

Question 10.

c = 2

Explanation:
2c² + 9c + 6 = 9c + 14
2c² – 8 = 0
c² – 4 = 0
c = 2

Question 11.

r = 12

Explanation:
(3 + r)² = r² + 9²
9 + 6r + r² = r² + 81
6r = 72
r = 12

Question 12.
Tell whether $$\overline{B D}$$ is tangent to ⊙C. Explain.

$$\overline{B D}$$ is not tangent to ⊙C

Explanation:
52² = 10² + 48²
2704 = 100 + 2304
So, $$\overline{B D}$$ is not tangent to ⊙C

### 10.2 Finding Arc Measures

Use the diagram above to find the measure of the indicated arc.

Question 13.
$$\widehat{K L}$$

$$\widehat{K L}$$ = 100°

Explanation:
$$\widehat{K L}$$ = ∠KPL = 100°

Question 14.
$$\widehat{L M}$$

$$\widehat{L M}$$ = 60°

Explanation:
$$\widehat{L M}$$ = 180° – 120°
= 60°

Question 15.
$$\widehat{K M}$$

$$\widehat{K M}$$ = 160°

Explanation:
$$\widehat{K M}$$ = 100° + 60°
= 160°

Question 16.
$$\widehat{K N}$$

$$\widehat{K N}$$ = 80°

Explanation:
$$\widehat{K N}$$ = 360 – (120 + 100 + 60)
= 360 – 280 = 80°

Tell whether the red arcs are congruent. Explain why or why not.

Question 17.

$$\widehat{S T}$$, $$\widehat{Y Z}$$ are not congruent.

Explanation:
$$\widehat{S T}$$, $$\widehat{Y Z}$$ are not congruent. Because both arcs are from different circles and having different radii.

Question 18.

$$\widehat{A B}$$, $$\widehat{E F}$$ are congruent.

Explanation:
$$\widehat{A B}$$, $$\widehat{E F}$$ are congruent. Because those circles have same radii.

### 10.3 Using Chords

Find the measure of $$\widehat{A B}$$.

Question 19.

$$\widehat{A B}$$ = 61°

Explanation:
$$\widehat{A B}$$ = 61°
If ED = AB, then $$\widehat{A B}$$ = $$\widehat{E D}$$

Question 20.

$$\widehat{A B}$$ = 65°

Explanation:
$$\widehat{A B}$$ = $$\widehat{A D}$$
So, $$\widehat{A B}$$ = 65°

Question 21.

$$\widehat{A B}$$ = 91°

Explanation:
$$\widehat{A B}$$ = $$\widehat{E D}$$
So, $$\widehat{A B}$$ = 91°

Question 22.
In the diagram. QN = QP = 10, JK = 4x, and LM = 6x – 24. Find the radius of ⊙Q.

The radius of ⊙Q is 26.

Explanation:
6x – 24 = 4x
6x – 4x = 24
2x = 24
x = 12
ML = 6(12) – 24 = 48
JN = $$\frac { 48 }{ 2 }$$ = 24
JQ² = JN² + NQ²
= 24² + 10² = 576 + 100
JQ = 26
Therefore, the radius of ⊙Q is 26

### 10.4 Inscribed Angles and Polygons

Find the value(s) of the variable(s).

Question 23.

x° = 80°

Explanation:
x° = 2 • 40° = 80°

Question 24.

q° = 100°, r° = 20°

Explanation:
q° + 80° = 180°
q° = 100°
4r° + 100 = 180°
4r° = 80°
r° = 20°

Question 25.

d° = 5°

Explanation:
14d° = 70°
d° = 5°

Question 26.

y° = 30°, z° = 10°

Explanation:
3y° = 90°
y° = 30°
50° + 90° + 4z° = 180°
4z° = 40°
z° = 10°

Question 27.

m° = 44°
n° = 39°

Explanation:
m° = 44°
n° = 39°

Question 28.

c° = 28°

Explanation:
c° = ½ • 56 = 28

### 10.5 Angle Relationships in Circles

Find the value of x.

Question 29.

x° = 250°

Explanation:
x° = 250°

Question 30.

x° = 106°

Explanation:
x° = ½(152 + 60)
= ½(212) = 106°

Question 31.

x° = 28°

Explanation:
x° = ½(96 – 40)
= ½(56) = 28°

Question 32.
Line l is tangent to the circle. Find m$$\widehat{X Y Z}$$.

m$$\widehat{X Y Z}$$ = 240°

Explanation:
m$$\widehat{X Y Z}$$ = 2(120)
= 240°

### 10.6 Segment Relationships in Circles

Find the value of x.

Question 33.

x = 8

Explanation:
3 • x = 4 • 6
x = 8

Question 34.

x = 3

Explanation:
(x + 3) • x = (6 – x) • 2x
x + 3 = 12 – 2x
3x = 9
x = 3

Question 35.

x = 18

Explanation:
12² = 8 • x
144 = 8x
x = 18

Question 36.
A local park has a circular ice skating rink. You are standing at point A, about 12 feet from the edge of the rink. The distance from you to a point of tangency on the rink is about 20 feet. Estimate the radius of the rink.

Estimated radius of the rink is 10 ft

Explanation:
20² = 12 • (2r + 12)
400 = 24r + 144
256 = 24r
r = 10.66
Therefore, estimated radius of the rink is 10 ft

### 10.7 Circles in the Coordinate Plane

Write the standard equation of the circle shown.

Question 37.

(x – 4)² + (y + 1)² = 12.25

Explanation:
(x – 4)² + (y + 1)² = 3.5²
(x – 4)² + (y + 1)² = 12.25

Question 38.

(x – 8)² + (y – 5)² = 36

Explanation:
(x – 8)² + (y – 5)² = 6²
(x – 8)² + (y – 5)² = 36

Question 39.

x² + y² = 4

Explanation:
(x – 0)² + (y – 0)² = 2²
x² + y² = 4

Write the standard equation of the circle with the given center and radius.

Question 40.

x² + y² = 81

Explanation:
(x – 0)² + (y – 0)² = 9²
x² + y² = 81

Question 41.
center: (- 5, 2), radius: 1.3

(x + 5)² + (y – 2)² = 1.69

Explanation:
(x + 5)² + (y – 2)² = 1.3²
(x + 5)² + (y – 2)² = 1.69

Question 42.

(x – 6)² + (y – 21)² = 16

Explanation:
(x – 6)² + (y – 21)² = 4²
(x – 6)² + (y – 21)² = 16

Question 43.
center: (- 3, 2), radius: 16

(x + 3)² + (y – 2)² = 256

Explanation:
(x + 3)² + (y – 2)² = 16²
(x + 3)² + (y – 2)² = 256

Question 44.

(x – 10)² + (y – 7)² = 12.25

Explanation:
(x – 10)² + (y – 7)² = 3.5²
(x – 10)² + (y – 7)² = 12.25

Question 45.

x² + y² = 27.04

Explanation:
(x – 0)² + (y – 0)² = 5.2²
x² + y² = 27.04

Question 46.
The point (- 7, 1) is on a circle with center (- 7, 6). Write the standard equation of the circle.

(x + 7)² + (y – 6)² = 25

Explanation:
r² = (-7 + 7)² + (6 – 1)²
= 5²
r = 5
And, centre is (-7, 6)
The standard equation of a circle is (x – (-7))² + (y – 6)² = 5²
(x + 7)² + (y – 6)² = 25

Question 47.
The equation of a circle is x2 + y2 – 12x + 8y + 48 = 0. Find the center and the radius of the circle. Then graph the circle.

The radius of the circle is 2, the centre is (6, -4)

Explanation:
x2 + y2 – 12x + 8y + 48 = 0
x² – 12x + 36 + y² + 8y + 16 = 4
(x – 6)² + (y + 4)² = 2²
So, the radius of the circle is 2, the centre is (6, -4)

Question 48.
Prove or disprove that the point (4, – 3) lies on the circle centred at the origin and containing
the point (- 5, 0).

The point (4, – 3) lies on the circle.

Explanation:
Use the distance formula to find the radius of the circle with cente (0, 0) and a point (-5, 0)
r = √(-5 – 0)² + (0 – 0)² = 5
The distance from the point (4, -3) to the center (0, 0)
d = √(4 – 0)² + (-3 – 0)² = √(16 +9) = 5
Since the radius of the circle is 5, the point lies on the circle.

### Circles Chapter Test

Find the measure of each numbered angle in ⊙P. Justify your answer.

Question 1.

m∠1 = 72.5°
m∠2 = 145°

Explanation:
m∠1 = $$\frac { 145 }{ 2 }$$
= 72.5°
m∠2 = 145°

Question 2.

m∠1 = 60°, m∠2 = 90°

Explanation:
A tangent is perpendicualr to diameter. So, m∠2 = 90°
m∠1 = 60°

Question 3.

m∠1 = 48°

Explanation:
m∠1 = $$\frac { 96 }{ 2 }$$ = 48°

Question 4.

Use the diagram.

Question 5.
AG = 2, GD = 9, and BG = 3. Find GF.

Question 6.
CF = 12, CB = 3, and CD = 9. Find CE.

Question 7.
BF = 9 and CB = 3. Find CA

Question 8.
Sketch a pentagon inscribed in a circle. Label the pentagon ABCDE. Describe the relationship between each pair of angles. Explain your reasoning.

a. ∠CDE and ∠CAE

b. ∠CBE and ∠CAE

Question 9.

x = 5

Explanation:
5x – 4 = 3x + 6
5x – 3x = 6 + 4
2x = 10
x = 5

Question 10.

r = 9

Explanation:
(6 + r)² = 12² + r²
36 + 12r + r² = 144 + r²
12r = 108
r = 9

Question 11.
Prove or disprove that the point (2√2, – 1) lies on the circle centered at (0, 2) and containing the point (- 1, 4).

Disproved

Explanation:
We consider the circle centred at the A(0, 2) and containing the point B(-1, 4).
AB = √(-1 – 0)² + (4 – 2)² = √1 + 4 = √5
The distance between centre A(0, 2) and P(2√2, – 1) is
AP = √(2√2 – 0)² + (-1 – 2)² = √8 + 9 = √17
AB ≠ AP
So, the point (2√2, – 1) dies not lie on the circle.

Prove the given statement.

Question 12.
$$\widehat{S T} \cong \widehat{R Q}$$

Question 13.
$$\widehat{J M} \cong \widehat{L M}$$

Question 14.
$$\widehat{D G} \cong \widehat{F G}$$

Question 15.
A bank of lighting hangs over a stage. Each light illuminates a circular region on the stage. A coordinate plane is used to arrange the lights, using a corner of the stage as the origin. The equation (x – 13)2 + (y – 4)2 = 16 represents the boundary of the region illuminated by one of the lights. Three actors stand at the points A(11, 4), B(8, 5), and C(15, 5). Graph the given equation. Then determine which actors are illuminated by the light.

The equation (x – 13)² + (y – 4)²= 16 represents the standard equation of the circle with center (13, 4) and radius 4
Graph the circle with center S(13, 4), radius 4. Then graph the points A(11,4), B (8, 5), C(15,5) which represents the places where the actors stand.

From the graph, we can see that points A and C inside the circle and point B is outside the circle. Therefore, actors who stand at points A and C are illuminated by the light

Question 16.
If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle, accident investigators can estimate the speed of the car.

a. To find the radius, accident investigators choose points A and B on the tire marks. Then the investigators find the midpoint C of $$\overline{A B}$$. Use the diagram to find the radius r of the circle. Explain why this method works.

The radius r of the circle = 155.71 ft

Explanation:
Given that,
AC = 130 ft, CD = 70 ft
CE = (r – 70) ft
r² = a² + b²
r²= 130²+ (r – 70)²
r² = 16900 + r² – 140r + 4900
140r = 21,800
r = 155.71 ft

b. The formula S = 3.87√fr can be used to estimate a car’s speed in miles per hour, where f is the coefficient of friction and r is the radius of the circle in feet. If f = 0.7, estimate the car’s speed in part (a).

The estimated car’s speed is 39.67 miles per hour

Explanation:
S = 3.87√fr
S = 3.8 √(0.7 x 155.71)
S = 3.8 √108.997
S = 3.8 x 10.44
S = 39.67

### Circles Cumulative Assessment

Question 1.
Classify each segment as specifically as possible.

a. $$\overline{B G}$$

$$\overline{B G}$$ is a chord

b. $$\overline{C D}$$

$$\overline{C D}$$ is radius.

c. $$\overline{A D}$$

$$\overline{A D}$$ is diameter.

d. $$\overline{F E}$$

$$\overline{F E}$$ is a chord

Question 2.
Copy and complete the paragraph proof.

Given Circle C with center (2, 1) and radius 1,
Circle D with center (0, 3) and radius 4
Prove Circle C is similar to Circle D.

Map Circle C to Circle C’ by using the _________ (x, y) → _________ so that Circle C’ and Circle D have the same center at (____, _____). Dilate Circle C’ using a cellIer of dilation (_____, _____) and a scale factor of _____ . Because there is a _________ transformation that maps Circle C to Circle D, Circle C is __________ Circle D.

Map Circle C to Circle C’ by using the scale factor (x, y) → (0, 3) so that Circle C’ and Circle D have the same center at (2, 1). Dilate Circle C’ using a cellIer of dilation (2, 1) and a scale factor of circles. Because there is a transformation that maps Circle C to Circle D, Circle C is similar to Circle D.

Question 3.
Use the diagram to write a proof.

Given ∆JPL ≅ ∆NPL
$$\overline{P K}$$ is an altitude of ∆JPL
$$\overline{P M}$$ is an altitude ∆NPL
Prove ∆PKL ~ ∆NMP

∆JPL is similar to ∆NPL and PK is the altitude of ∆JPL and PM is an altitude of ∆NPL
Altitude is a line drawn from one vertex to the opposite site. It is perpendicular to the side.
So, ∆PKL is similar to ∆NMP

Question 4.
The equation of a circle is x² + y² + 14x – 16y + 77 = 0. What are the center and radius of the circle?
(A) center: (14, – 16). radius: 8.8
(B) center: (- 7, 8), radius: 6
(C) center (- 14, 16), radius: 8.8
(D) center: (7, – 8), radius: 5.2

(B) center: (- 7, 8), radius: 6

Explanation:
x² + y² + 14x – 16y + 77 = 0
x² + 14x + 49 + y² – 16y + 64 = 36
(x + 7)² + (y – 8)² = 6²
So, the centre is (-7, 8) and radius is 6.

Question 5.
The coordinates of the vertices of a quadrilateral are W(- 7, – 6), X(1, – 2), Y(3, – 6) and Z(- 5, – 10). Prove that quadrilateral WXYZ is a rectangle.

Proved

Explanation:
Find the distance of WY and ZX
WY = √(-7 – 3)² + (-6 + 6)² = √(-10)² = 10
ZX = √(1 + 5)² + (-2 + 10)² = √6² + 8² = 10
WY = ZX, the diagonals are congruent
Use the slope formula to find the slopes of diagonals
Slope of WY = $$\frac { -6 + 6 }{ -7 – 3 }$$ = 0
Slope of ZX = $$\frac { -2 + 10 }{ 1 + 5 }$$ = $$\frac { 4 }{ 3 }$$
Because the product of slopes of diagonals is 0, the diagonals are not perpendicular
Therefore, the quadrilateral WXYZ is a rectangle.

Question 6.
Which angles have the same measure as ∠ACB? Select all that apply.
im – 295
∠DEF ∠JGK ∠KGL ∠LGM ∠MGJ
∠QNR ∠STV ∠SWV ∠VWU ∠XYZ
∠VWU

Question 7.
Classify each related conditional statement based on the conditional statement
“If you are a soccer player. then you are an athlete.”
a. If you are not a soccer player, then you are not an athlete.

False

b. If you are an athlete, then you are a soccer player.

False

c. You are a soccer player if and only if you are an athlete.

True

d. If you are not an athlete, then you are not a soccer player.

False

Question 8.
Your friend claims that the quadrilateral shown can be inscribed in a circle. Is your friend correct? Explain our reasoning.

If the sum of any two angles is 180°, then the quadrilateral is inscribed in a circle.
So, 70° + 110° = 180°, 110° + 70°= 180°
So, my friend is correct.

## Big Ideas Math Geometry Answers Chapter 12 Probability

Get the pdf link of Big Ideas Math Geometry Answers Chapter 12 Probability from this page. The concepts to learn in Probability are Sample Spaces and Probability, Independent and Dependent Events, Two-Way Tables and Probability, Probability of Disjoint and Overlapping Events and Permutations and Combinations, Binomial Distributions. Get step by step explanations for all the questions from Big Ideas Math Answers Geometry Chapter 12 Probability.

## Big Ideas Math Book Geometry Answer Key Chapter 12 Probability

Most of the students think that probability is the toughest chapter in maths. But if you understand the logic this will be the easiest of all the chapters. Click on the below attached links and start practicing the problems. Learn the concepts in depth and test your knowledge by solving the questions given at the end of the chapter.

Lesson: 1 Sample Spaces and Probability

Lesson: 2 Independent and Dependent Events

Lesson: 3 Two-Way Tables and Probability

Quiz

Lesson: 4 Probability of Disjoint and Overlapping Events

Lesson: 5 Permutations and Combinations

Lesson: 6 Binomial Distributions

Chapter: 12 – Probability

### Probability Maintaining Mathematical Proficiency

Write and solve a proportion to answer the question.

Question 1.
What percent of 30 is 6?

Explanation:
100% = 30
x% = 6
100% = 30(1)
x% = 6(2)
100%/x% = 30/6
Taking the inverse of both sides
x%/100% = 6/30
x = 20%
Thus 6 is 20% of 30.

Question 2.
What number is 68% of 25?

Explanation:
68% × 25
(68 ÷ 100) × 25
(68 × 25) ÷ 100
1700 ÷ 100 = 17

Question 3.
34.4 is what percent of 86?

Explanation:
100% = 86
x% = 34.4
100% = 86(1)
x% = 34.4(2)
100%/x% = 86/34.4
Taking the inverse of both sides
x%/100% = 34.4/86
x = 40%
Therefore, 34.4 is 40% of 86.

Display the data in a histogram.

Question 4.

Question 5.
ABSTRACT REASONING
You want to purchase either a sofa or an arm chair at a furniture store. Each item has the same retail price. The sofa is 20% off. The arm chair is 10% off. and you have a coupon to get an additional 10% off the discounted price of the chair. Are the items equally priced after the discounts arc applied? Explain.

Explanation:
You want to purchase either a sofa or an armchair at a furniture store.
Each item has the same retail price. The sofa is 20% off. The armchair is 10% off. and you have a coupon to get an additional 10% off the discounted price of the chair.
The price of armchair and sofa are the same.
If you add 10% to chair the discount for the chair and sofa will be the same.
10% + 10% = 20%

### Probability Monitoring Progress

In Exercises 1 and 2, describe the event as unlikely, equally likely to happen or not happen, or likely. Explain your reasoning.

Question 1.
The oldest child in a family is a girl.

Question 2.
The two oldest children in a family with three children are girls.

Question 3.
Give an example of an event that is certain to occur.
If A and B are independent event
P(A) = 1/2
P(B) = 1/5
P(A and B) = P(A) × P(B)
= 1/2 × 1/5
= 1/10

### 12.1 Sample Spaces and Probability

Exploration 1

Finding the Sample Space of an Experiment

Work with a partner: In an experiment, three coins are flipped. List the possible outcomes in the sample space of the experiment.

The number of different outcomes when three coins are tossed is 2 × 2 × 2 = 8.
All 8 possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT.

Exploration 2

Finding the Sample Space of an Experiment

Work with a partner: List the possible outcomes in the sample space of the experiment.

a. One six-sided die is rolled.

b. Two six-sided die is rolled.

Rolling two six-sided dice: Each die has 6 equally likely outcomes, so the sample space is 6 . 6 or 36 equally likely outcomes.

### Exploration 3

Finding the Sample Space of an Experiment

Work with a partner: In an experiment, a spinner is spun.

a. How many ways can you spin a 1? 2? 3? 4? 5?
Answer: 1, 2, 3, 2, 4

b. List the sample space.
Answer: 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5

c. What is the total number of outcomes?

Exploration 4

Finding the Sample Space of an Experiment

Work with a partner: In an experiment, a bag contains 2 blue marbles and 5 red marbles. Two marbles arc drawn from the bag.

a. How many ways can you choose two blue? a red then blue? a blue then red? two red?
Answer: BB – 2, RB – 10, BR – 10, RR – 20

b. List the sample space.

c. What is the total number of outcomes?

Question 5.
How can you list the possible outcomes in the sample space of an experiment?
There are four possible outcomes for each spin: red, blue, yellow, green. Then, multiply the number of outcomes by the number of spins. June flipped the coin three times. The answer is there are 12 outcomes in the sample space.

Question 6.
For Exploration 3, find the ratio of the number of each possible outcome to the total number of outcomes. Then find the sum of these ratios. Repeat for Exploration 4. What do you observe?
LOOKING FOR A PATTERN
To be proficient in math, you need to look closely to discern a pattern or structure.

### Lesson 12.1 Sample Spaces and Probability

Monitoring Progress

Find the number of possible outcomes in the sample space. Then list the possible outcomes.

Question 1.
You flip two coins.

Explanation:
When we flip two coins simultaneoulsy then the possible outcomes will be (H, H), (T, T), (T, H), (H, T)
T represents tails.
Thus the possible outcomes are 2² = 4

Question 2.
You flip two Coins and roll a six-sided die.
Answer: 6 × 2² = 24

Explanation:
We roll a die and flip two coins. We have to find the number of possible outcomes in this space. Also we have to list the possible outcomes.
1 = {When rolling the dice, the number 1 fell};
2 = {When rolling the dice, the number 2 fell};
3 = {When rolling the dice, the number 3 fell};
4 = {When rolling the dice, the number 4 fell};
5 = {When rolling the dice, the number 5 fell};
6 = {When rolling the dice, the number 6 fell};
On the other hand, using H for Heads and T for Tails we can list the outcomes.
(1, H, H), (2, H, H), (3, H, H), (4, H, H), (5, H, H), (6, H, H)
(1, T, H), (2, T, H), (3, T, H), (4, T, H), (5, T, H), (6, T, H)
(1, H, T), (2, H, T), (3, H, T), (4, H, T), (5, H, T), (6, H, T)
(1, T, T), (2, T, T), (3, T, T), (4, T, T), (5, T, T), (6, T, T)
Therefore, we can conclude that the number of all possible outcomes is
6 × 2² = 24

Question 3.
You flip a coin and roll a six-sided die. What is the probability that the coin shows tails and the die shows 4?
The sample space has 12 possible outcomes.
Tails, 1
Tails, 2
Tails, 3
Tails, 4
Tails, 5
Tails, 6
Probability that the coin shows tails and the die shows 4 is 4/12 = 1/3

Find P($$\bar{A}$$).

Question 4.
P(A) = 0.45
P($$\bar{A}$$) = 1 – P(A)
P(A) = 0.45
P($$\bar{A}$$) = 1 – 0.45
P($$\bar{A}$$) = 0.55

Question 5.
P(A) = $$\frac{1}{4}$$
P($$\bar{A}$$) = 1 – P(A)
P(A) = $$\frac{1}{4}$$
P($$\bar{A}$$) = 1 – $$\frac{1}{4}$$
P($$\bar{A}$$) = $$\frac{3}{4}$$

Question 6.
P(A) = 1
P($$\bar{A}$$) = 1 – P(A)
P(A) = 1
P($$\bar{A}$$) = 1 – 1
P($$\bar{A}$$) = 0

Question 7.
P(A) = 0.03
P($$\bar{A}$$) = 1 – P(A)
P(A) = 0.03
P($$\bar{A}$$) = 1 – 0.03
P($$\bar{A}$$) = 0.97

Question 8.
In Example 4, are you more likely to get 10 points or 5 points?
10: 0.09
5: (5 – 10)/324
= (36π – 9π)/324
= 27π/324
= 0.26

Question 9.
In Example 4, are you more likely to score points (10, 5, or 2) or get 0 points?
2: (2 – 5)/324
= (81π – 36π)/324
= 45π/324
= 0.43
0.09 + 0.26 + 0.43 = 0.78
More likely to get 2 points.

Question 10.
In Example 5, for which color is the experimental probability of stopping on the color greater than the theoretical probability?
9/20 = 0.45

Question 11.
In Example 6, what is the probability that a pet-owning adult chosen at random owns a fish?
146/1328 = 73/664 = 0.11

### Exercise 12.1 Sample Spaces and Probability

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A number that describes the likelihood of an event is the ___________ of the event.
A number that describes the likelihood of an event is the Probability of the event.

Question 2.
WRITING
Describe the difference between theoretical probability and experimental probability.
Answer: Experimental probability is the result of an experiment. Theoretical probability is what is expected to happen.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. find the number of possible outcomes in the sample space. Then list the possible outcomes.

Question 3.
You roll a die and flip three coins.

Question 4.
You flip a coin and draw a marble at random from a hag containing two purple marbles and one while marble.
Given data,
You flip a coin and draw a marble at random from a hag containing two purple marbles and one while marble.
the probability of getting a purple marble = 2/3
the probability of getting a white marble = 1/3

Question 5.
A bag contains four red cards numbered 1 through 4, four white cards numbered 1 through 4, and four black cards numbered 1 through 4. You choose a card at random.

Question 6.
You draw two marbles without replacement from a bag containing three green marbles and four black marbles.
In all there are 7 marbles when you first grab a marble, after that you take one marble away then you have 6 marbles to choose.
7 × 6 = 42
42 possible outcomes: GG, GG, GB, GB, GB, GB, GG, GG, GB, GB, GB, GB, GG, GG, GB, GB, GB, GB, BG, BG, BG, BB, BB, BB, BG, BG, BG, BB, BB, BB, BG, BG, BG, BB, BB, BB, BG, BG, BG, BB, BB, BB, BB.

Question 7.
PROBLEM SOLVING
A game show airs on television five days per week. Each day, a prize is randomly placed behind one of two doors. The contestant wins the prize by selecting the correct door. What is the probability that exactly two of the five contestants win a prize during a week?

Question 8.
PROBLEM SOLVING
Your friend has two standard decks of 52 playing cards and asks you to randomly draw one card from each deck. What is the probability that you will draw two spades?
You have two decks of 52 cards and in a normal deck, there are 13 cards of each suit.
So there are 13 spades in the deck.
Therefore the probability of you drawing a spade is 13 out of all the 52 cards or $$\frac{13}{52}$$, which can be reduced to $$\frac{1}{4}$$. You do this two times with different decks that are exactly the same so you multiply $$\frac{1}{4}$$ times $$\frac{1}{4}$$. 1 times 1 is 1 and 4 times 4 is 16, so it is $$\frac{1}{16}$$.
You can turn this into a percentage by dividing 1 by 16 and moving the decimal place to places to the right.
6.25% is the probability that you will draw two spades.

Question 9.
PROBLEM SOLVING
When two six-sided dice are rolled, there are 36 possible outcomes. Find the probability that
(a) the sum is not 4 and
(b) the sum is greater than 5.

Question 10.
PROBLEM SOLVING
The age distribution of a population is shown. Find the probability of each event.

a. A person chosen at random is at least 15 years old.

b. A person chosen at random is from 25 to 44 years old.
Answer: 13% + 13% = 26%

Question 11.
ERROR ANALYSIS
A student randomly, guesses the answers to two true-false questions. Describe and correct the error in finding the probability of the student guessing both answers correctly.

Question 12.
ERROR ANALYSIS
A student randomly draws a number between 1 and 30. Describe and correct the error in finding the probability that the number drawn is greater than 4.

The error is that the probability of the complement of the event is 4/30, not 3/30, because if you are looking for a sum greater than 4, than you subtract 1 by numbers less than or equal to 4 by the total amount of numbers, which is 30.
P(Sum is greater than 4)=1-P(Sum is less than or equal to 4)
1 – 2/15
= 13/15

Question 13.
MATHEMATICAL CONNECTIONS
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. What is the probability your dart lands in the yellow region?

Question 14.
MATHEMATICAL CONNECTIONS
The map shows the length (in miles) of shoreline along the Gulf of Mexico for each state that borders the body of water. What is the probability that a ship coming ashore at a random point in the Gulf of Mexico lands in the given state?

a. Texas
By using the above map we can solve the problem
Total length of the shoreline: 367 + 397 + 44 + 53 + 770 = 1631 miles
The probability of the ship landing in Texas: 367/1631 = 0.23

b. Alabama
The probability of the ship landing in Alabama: 53/1631 = 0.03

c. Florida
The probability of the ship landing in Florida: 770/1631 = 0.47

d. Louisiana
The probability of the ship landing in Louisiana: 397/1631 = 0.24

Question 15.
DRAWING CONCLUSIONS
You roll a six-sided die 60 times. The table shows the results. For which number is the experimental probability of rolling the number the same as the theoretical probability?

Question 16.
DRAWING CONCLUSIONS
A bag contains 5 marbles that are each a different color. A marble is drawn, its color is recorded, and then the marble is placed back in the hag. This process is repeated until 30 marbles have been drawn. The table shows the results. For which marble is the experimental probability of drawing the marble the same as the theoretical probability?

Total number of marbles = 30
6/30 = 1/5
For black marble the experimental probability of drawing the marble the same as the theoretical probability.

Question 17.
REASONING
Refer to the spinner shown. The spinner is divided into sections with the same area.

a. What is the theoretical probability that the spinner stops on a multiple of 3?
b. You spin the spinner 30 times. If stops on a multiple of 3 twenty times. What is the experimental probability of Stopping on a multiple of 3?
c. Explain why the probability you found in part (b) is different than the probability you found in part (a).

Question 18.
OPEN-ENDED
Describe a real-life event that has a probability of 0. Then describe a real-life event that has a probability of 1.
The probability of rolling a 7 with a standard 6-sided die = 0
The probability of rolling a natural number with a standard 6-sided die = 1

Question 19.
DRAWING CONCLUSIONS
A survey of 2237 adults ages 18 and over asked which Sport 15 their favorite. The results are shown in the figure. What is the probability that an adult chosen at random prefers auto racing?

Question 20.
DRAWING CONCLUSIONS
A survey of 2392 adults ages 18 and over asked what type of food they Would be most likely to choose at a restaurant. The results are shown in the figure. What is the probability that an adult chosen at random prefers Italian food?

P(Italian) = 526/1196
= 263/1196
P(Italian) ≈ 22%

Question 21.
ANALYZING RELATIONSHIPS
Refer to the board in Exercise 13. Order the likelihoods that the dart lands in the given region from least likely to most likely.
A. green
B. not blue
C. red
D. not yellow

Question 22.
ANALYZING RELATIONSHIPS
Refer to the chart below. Order the following events from least likely to most likely.

A. It rains on Sunday.
80%
= 80/20 = 4/1
The ratio is 4:1

B. It does not rain on Saturday.
100 – 30 = 70%
= 30/70 = 3 : 7

C. It rains on Monday.
= 90/10 = 9/1
= 9 : 1

D. It does not rain on Friday.
100 – 5 = 95%
= 5/95 = 1/19
= 1 : 19

Question 23.
USING TOOLS
Use the figure in Example 3 to answer each question.

a. List the possible sums that result from rolling two six-sided dice.
b. Find the theoretical probability of rolling each sum.
c. The table below shows a simulation of rolling two six-sided dice three times. Use a random number generator to simulate rolling two six-sided dice 50 times. Compare the experimental probabilities of rolling each sum with the theoretical probabilities.

Question 24.
MAKING AN ARGUMENT
You flip a coin three times. It lands on heads twice and on tails once. Your friend concludes that the theoretical probability of the coin landing heads up is P(heads up) = $$\frac{2}{3}$$. Is your friend correct? Explain your reasoning.
The friend is incorrect because the probability of heads here is $$\frac{2}{3}$$, is the experimental probabiility of heads for this particular case, while its theoretical probability will always be $$\frac{1}{2}$$.

Question 25.
MATHEMATICAL CONNECTIONS
A sphere fits inside a cube so that it touches each side, as shown. What is the probability a point chosen at random inside the cube is also inside the sphere ?

Question 26.
HOW DO YOU SEE IT?
Consider the graph of f shown. What is the probability that the graph of y = f(x) + c intersects the x-axis when c is a randomly chosen integer from 1 to 6? Explain.

Question 27.
DRAWING CONCLUSIONS
A manufacturer tests 1200 computers and finds that 9 of them have defects. Find the probability that a computer chosen at random has a defect. Predict the number of computers with defects in a shipment of 15,000 computers. Explain your reasoning.

Question 28.
THOUGHT PROVOKING
The tree diagram shows a sample space. Write a probability problem that can be represented by the sample space. Then write the answer(s) to the problem.

Maintaining Mathematical Proficiency

Question 29.
$$\frac{2 x^{3}}{x^{2}}$$

Question 30.
$$\frac{2 x y}{8 y^{2}}$$
$$\frac{2y}{8 y^{2}}$$
= $$\frac{2}{8y}$$
= $$\frac{1}{4y}$$

Question 31.
$$\frac{4 x^{9} y}{3 x^{3} y}$$

Question 32.
$$\frac{6 y^{0}}{3 x^{-6}}$$
Given,
$$\frac{6 y^{0}}{3 x^{-6}}$$
= 6 . 1/3x-6
= 2 . x6

Question 33.
(3Pq)4

Question 34.
$$\left(\frac{y^{2}}{x}\right)^{-2}$$
Answer: ($$\frac{y}{x}$$)²

### 12.2 Independent and Dependent Events

Exploration 1

Identifying Independent and Dependent Events

Work with a partner: Determine whether the events are independent or dependent. Explain your reasoning.
REASONING ABSTRACTLY
To be proficient in math, you need to make sense of quantities and their relationships in problem situations.
a. Two six-sided dice are rolled.

P = Number of outcomes that satisfy the requirements/Total number of possible outcomes
Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die.
When two dice are thrown simultaneously, thus the number of event can be 62 = 36 because each die has 1 to 6 number on its faces.
This is independent event

b. Six pieces of paper, numbered 1 through 6, are in a bag, Two pieces of paper are selected one at a time without replacement.

P = Number of outcomes that satisfy the requirements/Total number of possible outcomes
When we pick up any paper the probability result is equal of 1/6
This is dependent event

Exploration 2

Work with a partner:

a. In Exploration 1(a), experimentally estimate the probability that the sum of the two numbers rolled is 7. Describe your experiment.

(b). In Exploration 1 (b), experimentally estimate the probability that the sum of the two numbers selected is 7. Describe your experiment.

Exploration 3

Finding Theoretical Probabilities

Work with a partner:
a. In Exploration 1(a), find the theoretical probability that the sum of the two numbers rolled is 7. Then compare your answer with the experimental probability you found in Exploration 2(a).
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.

b. In Exploration 1(b). find the theoretical probability that the sum of the two numbers selected is 7. Then compare your answer with the experimental probability you found in Exploration 2(b).

C. Compare the probabilities you obtained in parts (a) and (b).

Question 4.
How can you determine whether two events are independent or dependent?
Two events A and B are said to be independent if the fact that one event has occurred does not affect the probability that the other event will occur.
If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent.

Question 5.
Determine whether the events are independent or dependent. Explain your reasoning.
a. You roil a 4 on a six-sided die and spin red on a spinner.

b. Your teacher chooses a student to lead a group. chooses another student to lead a second group. and chooses a third student to lead a third group.

### Lesson 12.2 Independent and Dependent Events

Monitoring progress

Question 1.
In Example 1, determine whether guessing Question 1 incorrectly and guessing Question 2 correctly are independent events.

Question 2.
In Example 2, determine whether randomly selecting a girl first and randomly selecting a boy second are independent events.

Question 3.
In Example 3, what is the probability that you spin an even number and then an odd number?

Question 4.
In Example 4, what is the probability that both hills are $1 hills? Answer: Question 5. In Example 5, what is the probability that none of the cards drawn are hearts when (a) you replace each card, and (b) you do not replace each card? Compare the probabilities. Answer: (a) you replace each card, P(A and B and C) = P(A) P(B) P(C) = 13/52 . 13/52 . 13/52 = 1/4 . 1/4 . 1/4 = 1/64 = 0.016 Question 6. In Example 6, find (a) the probability that a non-defective part “passes” and (b) the probability that a defective part “fails.” Answer: (a) the probability that a non-defective part “passes” P(P/D) = 3/39 = 1/13 = 0.077 (b) the probability that a defective part “fails.” P(F/N) = 11/461 = 0.024 Question 7. At a coffee shop. 80% of customers order coffee. Only 15% of customers order coffee and a bagel. What is the probability that a customer who orders coffee also orders a bagel? Answer: A: Customer order coffee B: Customer order a bagel P(B/A) = P(A and B)/P(A) 80% of customers order coffee and Only 15% of customers order coffee and a bagel. P(A) = 80/100 = 0.8 P(A and B) = 15/100 = 0.15 P(B/A) = P(A and B)/P(A) = 0.15/0.8 = 0.1875 = 18.75% ### Exercise 12.2 Independent and Dependent Events Vocabulary and Core Concept Check Question 1. WRITING Explain the difference between dependent events and independent events, and give an example of each. Answer: When two events are dependent, the occurrence of one event affects the other. When two events are independent, the occurence of one event does not affect the other. Question 2. COMPLETE THE SENTENCE The probability that event B will occur given that event A has occurred is called the _____________ of B given A and is written as _____________ . Answer: The probability that event B will occur given that event A has occurred is called the conditional probability of B given A and is written as P(B/A). Monitoring Progress and Modeling with Mathematics In Exercises 3 – 6, tell whether the events are independent or dependent. Explain your reasoning. Question 3. A box of granola bars contains an assortment of flavors. You randomly choose a granola bar and eat it. Then you randomly choose another bar. Event A: You choose a coconut almond bar first. Event B: You choose a cranberry almond bar second. Answer: The two events, which considered in this experiment are an example of dependent events. Question 4. You roll a six-sided die and flip a coin. Event A: You get a 4 when rolling the die. Event B: You get tails when flipping the coin Answer: Independent, the events do not influence each other. Question 5. Your MP3 player contains hip-hop and rock songs. You randomly choose a song. Then you randomly choose another song without repeating song choices. Event A: You choose a hip-hop song first. Event B: You choose a rock song second. Answer: The events are dependent because the occurrence of event A affects the occurrence of event B. Question 6. There are 22 novels of various genres on a shell. You randomly choose a novel and put it back. Then you randomly choose another novel. Event A: You choose a mystery novel. Event B: You choose a science fiction novel. Answer: The 1st book chosen is put back so the second book picked has the same probability of being chosen a if the 1st book was never chosen to begin with the events are independent. In Exercises 7 – 10. determine whether the events are independent. Question 7. You play a game that involves spinning a wheel. Each section of the wheel shown has the same area. Use a sample space to determine whether randomly spinning blue and then green are independent events. Answer: Question 8. You have one red apple and three green apples in a bowl. You randomly select one apple to eat now and another apple for your lunch. Use a sample space to determine whether randomly selecting a green apple first and randomly selecting a green apple second are independent events. Answer: Let R represent the red apple. Let G1, G2, G3 represent the 3 green apples. P(G first) = P(green apple first) = 9/12 = 3/4 = 0.75 P(G second) = P(green apple second) = 9/12 = 3/4 = 0.75 P(green apple first and second) = 6/12 = 1/2 = 0.5 Events are not independent. Question 9. A student is taking a multiple-choice test where each question has four choices. The student randomly guesses the answers to the five-question test. Use a sample space to determine whether guessing Question 1 correctly and Question 2 correctly are independent events. Answer: Question 10. A vase contains four white roses and one red rose. You randomly select two roses to take home. Use a sample space to determine whether randomly selecting a white rose first and randomly selecting a white rose second are independent events. Answer: P(A and B) = P(A) and P(B) A = {First randomly selected roses is white} B = {Second randomly selected roses is white} P(A) = Number of favorable outcomes/Total number of outcomes = 4/5 P(B) = 4/5 P(A) . P(B) = 4/5 . 4/5 = 16/25 Question 11. PROBLEM SOLVING You play a game that involves spinning the money wheel shown. You spin the wheel twice. Find the probability that you get more than$500 on your first spin and then go bankrupt on your second spin.

Question 12.
PROBLEM SOLVING
You play a game that involves drawing two numbers from a hat. There are 25 pieces of paper numbered from 1 to 25 in the hat. Each number is replaced after it is drawn. Find the probability that you will draw the 3 on your first draw and a number greater than 10 on your second draw.
P(A) = 1/25
P(B) = 15/25
P(A and B) = P(A) . P(B)
= 1/25 . 15/25 = 3/125
Thus P(A and B) = 3/125

Question 13.
PROBLEM SOLVING
A drawer contains 12 white socks and 8 black socks. You randomly choose 1 sock and do not replace it. Then you randomly choose another sock. Find the probability that both events A and B will occur.
Event A: The first sock is white.
Event B: The second sock is white.

Question 14.
PROBLEM SOLVING
A word game has 100 tiles. 98 of which are letters and 2 of which are blank. The numbers of tiles of each letter are shown. You randomly draw 1 tile, set it aside, and then randomly draw another tile. Find the probability that both events A and B will occur.

P(A) = 56/100 = 0.56
P(B) = 42/(100 – 1) = 0.424
P(A) . P(B) = 0.56 × 0.424 = 0.2376

Question 15.
ERROR ANALYSIS
Events A and B are independent. Describe and correct the error in finding P(A and B).

Question 16.
ERROR ANALYSIS
A shelf contains 3 fashion magazines and 4 health magazines. You randomly choose one to read, set it aside, and randomly choose another for your friend to read. Describe and correct the error in finding the probability that both events A and B occur.
Event A: The first magazine is fashion.
Event B: The second magazine is health.

P(A) = 3/7
P(B/A) = 4/(7 – 1) = 4/6
P(A and B) = P(A) × P(B/A)
P(A and B) = 3/7 × 4/6 = 2/7

Question 17.
NUMBER SENSE
Events A and B are independent. Suppose P(B) = 0.4 and P(A and B) = 0.13. Find P(A).

Question 18.
NUMBER SENSE
Events A and B are dependent. Suppose P(B/A) = 0.6 and P(A and B) = 0.15. Find P(A).
P(A) = x
P(B/A) = 0.6
P(A and B) = 0.15
P(A and B) = P(A) × P(B/A)
0.15 = x × 0.6
x = 0.15/0.6
x = 0.25

Question 19.
ANALYZING RELATIONSHIPS
You randomly select three cards from a standard deck of 52 playing cards. What is the probability that all three cards are face cards when (a) you replace each card before selecting the next card, and (b) you do not replace each card before selecting the next card? Compare the probabilities.

Question 20.
A bag contains 9 red marbles. 4 blue marbles, and 7 yellow marbles. You randomly select three marbles from the hag. what is the probability that all three marbles are red when (a) you replace each marble before selecting the next marble, and (b) you do not replace each marble before selecting the next marble? Compare the probabilities.
a. There are a total of 9 + 4 + 7= 20 marbles.
Therefore, the probability of selecting a red marble in each attempt is 9/20 when the marble is replaced.
Therefore the probability of selecting a red marble in each of 3 of the attempt is
9/20 × 9/20 × 9/20 = 0.09125
The replacement makes these independent events.
b. There are total of 9 + 4 + 7 = 20 marbles.
The probability of selecting a red marble in the first attempt is 9/20, second attempt is 8/19 and the third attempt is 7/18 when the marbles are not replaced.
Therefore the probability of selecting a red marble in each of 3 of the attempts is 9/20 × 8/19 × 7/18 = 0.0737

Question 21.
ATTEND TO PRECISION
The table shows the number of species in the United States listed as endangered and threatened. Find (a) the probability that a randomly selected endangered species is a bird, and (b) the probability that a randomly selected mammal is endangered.

Question 22.
ATTEND TO PRECISION
The table shows the number of tropical cyclones that formed during the hurricane seasons over a 12-year period. Find (a) the probability to predict whether a Future tropical cyclone in the Northern Hemisphere is a hurricane, and (b) the probability to predict whether a hurricane is in the Southern Hemisphere.

Question 23.
PROBLEM SOLVING
At a school, 43% of students attend the homecoming football game. Only 23% of students go to the game and the homecoming dance. What is the probability that a student who attends the football game also attends the dance?

Question 24.
PROBLEM SOLVING
At a gas station. 84% of customers buy gasoline. Only 5% of customers buy gasoline and a beverage. What is the probability that a customer who buys gasoline also buys a beverage?
Given,
P(A) = 84% = 0.84
P(A and B) = 5% = 0.05
P(A and B) = P(A) × P(B/A)
P(B/A) = 0.05/0.84
P(B/A) = 0.0595

Question 25.
PROBLEM SOLVING
You and 19 other students volunteer to present the “Best Teacher” award at a school banquet. One student volunteer will be chosen to present the award. Each student worked at least 1 hour in preparation for the banquet. You worked for 4 hours, and the group worked a combined total of 45 hours. For each situation, describe a process that gives you a “fair” chance to be chosen. and find the probability that you are chosen.
a. “Fair” means equally likely.
b. “Fair” means proportional to the number of hours each student worked in preparation.

Question 26.
HOW DO YOU SEE IT?
A bag contains one red marble and one blue marble. The diagrams show the possible outcomes of randomly choosing two marbles using different methods. For each method. determine whether the marbles were selected with or without replacement.
a.

b.

Question 27.
MAKING AN ARGUMENT
A meteorologist claims that there is a 70% chance of rain. When it rains. there is a 75% chance that your softball game will be rescheduled. Your friend believes the game is more likely to be rescheduled than played. Is your friend correct? Explain your reasoning.
The chance that the game will be rescheduled is (0.7)(0.75) = 0.525
which is 52.5 percent
making it greater than 50 percent.

Question 28.
THOUGHT PROVOKING
Two six-sided dice are rolled once. Events A and B are represented by the diagram. Describe each event. Are the two events dependent or independent? Justify your reasoning.

Question 29.
MODELING WITH MATHEMATICS
A football team is losing by 14 points near the end of a game. The team scores two touchdowns (worth 6 points each) before the end of the game. After each touchdown, the coach must decide whether to go for 1 point with a kick (which is successful 99% of the time) or 2 points with a run or pass (which is successful 45% of the time).

a. If the team goes for 1 point after each touchdown, what is the probability that the team wins? loses? ties?
b. If the team goes for 2 points after each touchdown. what is the probability that the team wins? loses? ties?
c. Can you develop a strategy so that the coach’s team has a probability of winning the game that is greater than the probability of losing? If so, explain your strategy and calculate the probabilities of winning and losing the game.

Question 30.
ABSTRACT REASONING
Assume that A and B are independent events.
a. Explain why P(B) = P(B/A) and P(A) = P(A/B).
P(B) = P(B/A)
P(B) = P(A) . P(B/A)
P(A) = P(A/B).
P(A) = P(B) . P(A/B)

b. Can P(A and B) also be defined as P(B) • P(A/B)? Justify your reasoning.

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.

Question 31.
$$\frac{9}{10}$$ x = 0.18

Question 32.
$$\frac{1}{4}$$x + 0.5x = 1.5
Given,
$$\frac{1}{4}$$x + 0.5x = 1.5
0.25x + 0.50x = 1.5
0.75x = 1.5
x = 1.5/0.75
x = 2

Question 33.
0.3x – $$\frac{3}{5}$$x + 1.6 = 1.555

### 12.3 Two-Way Tables and Probability

Exploration 1

Completing and Using a Two-Way Table

Work with a partner: A two-way table displays the same information as a Venn diagram. In a two-way table, one category is represented by the rows and the other category is represented by the columns.

The Venn diagram shows the results of a survey in which 80 students were asked whether they play a musical instrument and whether they speak a foreign language. Use the Venn diagram to complete the two-way table. Then use the two-way table to answer each question.

a. How many students play an instrument?

b. How many students speak a foreign language?

c. How many students play an instrument and speak a foreign language?

d. How many students do not play an instrument and do not speak a foreign language?

e. How many students play an instrument and do not speak a foreign language?

Exploration 2

Two – Way Tables and Probability

Work with a partner. In Exploration 1, one student is selected at random from the 80 students who took the survey. Find the probability that the student
a. plays an instrument.

b. speaks a foreign language.

c. plays an instrument and speaks a foreign language.

d. does not play an instrument and does not speak a foreign language.

e. plays an instrument and does not speak a foreign language.

Exploration 3

Conducting a Survey

Work with your class. Conduct a survey of the students in your class. Choose two categories that are different from those given in Explorations 1 and 2. Then summarize the results in both a Venn diagram and a two-way table. Discuss the results.
MODELING WITH MATHEMATICS
To be proficient in math, you need to identify important quantities in a practical situation and map their relationships using such tools as diagrams and two-way tables.

Question 4.
How can you construct and interpret a two-way table?
Identify the variables. There are two variables of interest here: the commercial viewed and opinion.
Determine the possible values of each variable. For the two variables, we can identify the following possible values
Set up the table
Fill in the frequencies

Question 5.
How can you use a two-way table to determine probabilities?

### Lesson 12.3 Two-Way Tables and Probability

Monitoring Progress

Question 1.
You randomly survey students about whether they are in favor of planting a community garden at school. of 96 boys surveyed, 61 are in favor. 0f 88 girls surveyed, 17 are against. Organize the results in a two-way table. Then find and interpret the marginal frequencies.
In order to find out how many boys are against you do 96 – 61. In order to find out how many girls are in favor you do 88 – 17.
In order to find the probability you make a proportion. It will be:
P/100 = number of girls against/total number of students
17/184 = p/100
17 × 100 = 184p
1700 = 184p
p = 9.23

Question 2.
Use the survey results in Monitoring Progress Question 1 to make a two-way table that shows the joint and marginal relative frequencies.

Question 3.
Use the survey results in Example 1 to make a two-way table that shows the conditional relative frequencies based on the column totals. Interpret the conditional relative frequencies in the context of the problem.

Question 4.
Use the survey results in Monitoring Progress Question 1 to make a two-way table that shows the conditional relative frequencies based on the row totals. Interpret the conditional relative frequencies in the context of the problem.

Question 5.
In Example 4, what is the probability that a randomly selected customer who is located in Santa Monica will not recommend the provider to a friend?

Question 6.
In Example 4, determine whether recommending the provider to a friend and living in Santa Monica are independent events. Explain your reasoning.

Question 7.
A manager is assessing three employees in order to offer one of them a promotion. Over a period of time, the manager records whether the employees meet or exceed expectations on their assigned tasks. The table shows the managers results. Which employee should be offered the promotion? Explain.

### Exercise 12.3 Two-Way Tables and Probability

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A(n) ______________ displays data collected from the same source that belongs to two different categories.
A two-way table displays data collected from the same source that belongs to two different categories.

Question 2.
WRITING
Compare the definitions of joint relative frequency, marginal relative frequency, and conditional relative frequency.
Joint relative frequency: joint relative frequency is the ratio of a frequency that is not in the total row or the total column to the total number of values.
Marginal relative frequency: marginal relative frequency is the sum of the joint relative frequencies in a given row or column.
Conditional relative frequency: It is the ratio of joint relative frequency to the marginal relative frequency.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, complete the two-way table.

Question 3.

Number of students who have passed the exam is 50 – 10 = 40
Of those 40 students, 6 did not study for the exam,
Number of the students who studied and have passed the exam is 40 – 6 = 34
Number of the students who did not study and did not pass the exam is 10 – 4 = 6

Question 4.

Number of students who said no is 49 – 7 = 42
Total number of students is 56 + 42 = 98
Total number of people is 98 + 10 = 108
Out of the total number of people, 49 of them said no,
Total number of people who said yes is 108 – 49 = 59
Number of teachers who said yes is 59 – 56 = 3

Question 5.
MODELING WITH MATHEMATICS
You survey 171 males and 180 females at Grand Central Station in New York City. Of those, 132 males and 151 females wash their hands after using the public rest rooms. Organize these results in a two-way table. Then find and interpret the marginal frequencies.

Question 6.
MODELING WITH MATHEMATICS
A survey asks 60 teachers and 48 parents whether school uniforms reduce distractions in school. Of those, 49 teachers and 18 parents say uniforms reduce distractions in school, Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Given,
A survey asks 60 teachers and 48 parents whether school uniforms reduce distractions in school. Of those, 49 teachers and 18 parents say uniforms reduce distractions in school
Number of teacher who said no is 60 – 49 = 11
Number of parents who said no = 48 – 18 = 30
Total number of people who said yes = 49 + 18 = 67
Total number of people who said no = 11 + 30 = 41

USING STRUCTURE
In Exercises 7 and 8, use the two-way table to create a two-way table that shows the joint and marginal relative frequencies.

Question 7.

P1,1 = Number of favorable outcomes/Total number of outcomes
= 11/231 = 0.0476
P2,1 = 24/231 = 0.1039
P1,2 = 104/231 = 0.4502
P2,2 = 92/231 = 0.3983
The marginal relative frequencies we find as the sum of each row and each column.
P(A randomly chosen person is male) = 115/231 = 0.4978
P(A randomly chosen person is female) = 116/231 = 0.5022
P(A randomly chosen person have left dominant hand) = 35/231 = 0.1515
P(A randomly chosen person have left dominant hand) = 196/231 = 0.8484

Question 8.

P1,1 = Number of favorable outcomes/Total number of outcomes
= 62/410 = 0.1513
The marginal relative frequencies we find as the sum of each row and each column.
P(A randomly chosen person is male) = 377/410 = 0.9195
P(A randomly chosen person is female) = 33/410 = 0.0805

Question 9.
MODELING WITH MATHEMATICS
Use the survey results from Exercise 5 to make a two-way table that shows the joint and marginal relative frequencies.

Question 10.
MODELING WITH MATHEMATICS
In a survey, 49 people received a flu vaccine before the flu season and 63 people did not receive the vaccine. Of those who receive the flu vaccine, 16 people got the flu. Of those who did not receive the vaccine, 17 got the flu. Make a two-way table that shows the joint and marginal relative frequencies.

We see that the total no. of people who received the vaccine is 49, of which 16 got a fly.
Number of people who received the vaccine and did not get a fly is 49 – 16 = 33
Number of people who did not received the vaccine and did not get a fly is 63 – 17 = 46
Total number of people who got a fly is 16 + 17 = 33
Total Number of people who did not get a fly is 33 + 46 = 79
We also know that the total number of people who were surveyed is 49 + 63 = 112
P1,1 = Number of favorable outcomes/Total number of outcomes
= 16/112 = 0.1428
The marginal relative frequencies we find as the sum of each row and each column.
P(A randomly chosen person got a fly) = 33/112 = 0.2946
P(A randomly chosen person did not get a fly) = 79/112 = 0.7053

Question 11.
MODELING WITH MATHEMATICS
A survey finds that 110 people ate breakfast and 30 people skipped breakfast. Of those who ate breakfast. 10 people felt tired. Of those who skipped breakfast. 10 people felt tired. Make a two-way table that shows the conditional relative frequencies based on the breakfast totals.
Given,
A survey finds that 110 people ate breakfast and 30 people skipped breakfast.
Of those who ate breakfast. 10 people felt tired. Of those who skipped breakfast. 10 people felt tired.
Number of people who ate breakfast and not tired is 110 – 10 = 100
Number of people who did not eat breakfast and not tired is 30 – 10 = 20
Total number of people who felt tired is 10 + 10 = 20
Total number of people who did not get tired is 100 + 20 = 120
120 +20 = 140

Question 12.
MODELING WITH MATHEMATICS
Use the survey results from Exercise 10 to make a two-way table that shows the conditional relative frequencies based on the flu vaccine totals.

Question 13.
PROBLEM SOLVING
Three different local hospitals in New York surveyed their patients. The survey asked whether the patients physician communicated efficiently. The results, given as joint relative frequencies. are shown in the two-way table.

a. What is the probability that a randomly selected patient located in Saratoga was satisfied with the communication of the physician?
b. What is the probability that a randomly selected patient who was not satisfied with the physician’s communication is located in Glens Falls?
c. Determine whether being satisfied with the Communication of the physician and living in Saratoga are independent events.

Question 14.
PROBLEM SOLVING
A researcher surveys a random sample of high school students in seven states. The survey asks whether students plan to stay in their home state after graduation. The results, given as joint relative frequencies, are shown in the two-way table.

a. What is the probability that a randomly selected student who lives in Nebraska plans to stay in his or her home state after graduation?
In this case we consider a event A = {arandomly chosen student lives in Nebraska}
B = {a randomly chosen student plans to stay in his or her home state after graduation}
P(A) = 0.044 + 0.4 = 0.444
P(B/A) = P(A and B)/P(A) = 0.044/0.444 = 0.099
About 1% students who lives Nebraska plans to stay in his or her home state after graduation.

b. What is the probability that a randomly selected student who does not plan to stay in his or her home state after graduation lives in North Carolina?
C = {a randomly chosen student does not plan to stay in his or her home state after graduation}
D = {a randomly chosen student lives in North Carolina}
P(C) = 0.4 + 0.193 + 0.256 = 0.849
P(D/C) = P(C and D)/P(C)
= 0.193/0.849 = 0.227
Therefore about 22.7% students who does not plan to stay in his or her home state after graduation lives in North Carolina.

c. Determine whether planning to stay in their home state and living in Nebraska are independent events.
P(B/A) = 0.099
P(B) = 0.044 + 0.05
1 + 0.056 = 0.151
P(B/A) ≠ P(B)
This events are independent.

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in finding the given conditional probability.

Question 15.
P(yes|Tokyo)

Question 16.
P(London|No)

P(A and B) = P(A)P(B/A)
P(A) = 0.341 + 0.112+ 0.191 = 0.644
P(B/A) = P(A and B)/P(A) = 0.112/0.644 = 0.1739
In the denominator the probability P(B) = 0.248 is used instead of P(A), where P(B) is probability that a randomly chosen person live in London.

Question 17.
PROBLEM SOLVING
You want to find the quickest route to school. You map out three routes. Before school, you randomly select a route and record whether you are late or on time. The table shows your findings. Assuming you leave at the same time each morning, which route should you use? Explain.

Question 18.
PROBLEM SOLVING
A teacher is assessing three groups of students in order to offer one group a prize. Over a period of time, the teacher records whether the groups meet or exceed expectations on their assigned tasks. The table shows the teacher’s results. Which group should be awarded the prize? Explain.

Answer: Group 1 exceeded expectations 12 out of 16 times or 75% of the time.

Question 19.
OPEN-ENDED
Create and conduct a survey in your class. Organize the results in a two-way table. Then create a two-way table that shows the joint and marginal frequencies.

Question 20.
HOW DO YOU SEE IT?
A research group surveys parents and Coaches of high school students about whether competitive sports are important in school. The two-way table shows the results of the survey.

a. What does 120 represent?
Answer: 120 parents said that competitive sports are not important in school.

b. What does 1336 represent?
Answer: 1336 is the sum of parents and coaches who agree that competitive sports are important in school.

c. What does 1501 represent?
Answer: 1501 is the total number of people surveyed. Here this is the sum of the parents and the coaches who participated in the survey.

Question 21.
MAKING AN ARGUMENT
Your friend uses the table below to determine which workout routine is the best. Your friend decides that Routine B is the best option because it has the fewest tally marks in the “Docs Not Reach Goal” column. Is your friend correct? Explain your reasoning.

Question 22.
MODELING WITH MATHEMATICS
A survey asks students whether they prefer math class or science class. Of the 150 male students surveyed, 62% prefer math class over science class. Of the female students surveyed, 74% prefer math. Construct a two-way table to show the number of students in each category if 350 students were surveyed.
survey asks students whether they prefer math class or science class. Of the 150 male students surveyed, 62% prefer math class over science class.
62% = 0.62
0.62 = P(Math/Male)
= P(Math and Male)/P(Male)
= Number of male students who prefer math/150
Number of male students who prefer math = 0.62 × 150 = 93
0.74 = P(Math/Female)
= P(Math and Female)/P(Female)
= Number of female students who prefer math/200
Number of female students who prefer math = 0.74 × 200 = 148
So, the total number of students who prefer math class is 148 + 93 = 241
Number of male students who prefer science class = 150 -93 = 57
Number of female students who prefer science class = 200 – 148 = 52
Number of students who prefer science class = 57 + 52 = 109

Question 23.
MULTIPLE REPRESENTATIONS
Use the Venn diagram to construct a two-way table. Then use your table to answer the questions.

a. What is the probability that a randomly selected person does not own either pet?
b. What is the probability that a randomly selected person who owns a dog also owns a cat?

Question 24.
WRITING
Compare two-way tables and Venn diagrams. Then describe the advantages and disadvantages of each.

Question 25.
PROBLEM SOLVING
A company creates a new snack, N, and tests it against its current leader, L. The table shows the results.

The company is deciding whether it should try to improve the snack before marketing it, and to whom the snack should be marketed. Use probability to explain the decisions the company should make when the total size of the snack’s market is expected to (a) change very little, and (b) expand very rapidly.

Question 26.
THOUGHT PROVOKING
Baye’s Theorem is given by

Use a two-way table to write an example of Baye’s Theorem.

P(Cat owner) = 61/210 = 0.29
P(Dog owner) = 93/210 = 0.442
P(Cat Owner/Dog owner) = P(Dog owner and cat owner)/P(Dog owner) = 0.387
P(Dog owner/Cat owner) = P(Cat owner/Dog owner)P(Dog owner)/P(Cat owner)
= 0.387 × 0.442/0.29
= 0.5898

Maintaining Mathematical Proficiency

Draw a Venn diagram of the sets described.

Question 27.
Of the positive integers less than 15, set A consists of the factors of 15 and set B consists of all odd numbers.

Question 28.
Of the positive integers less than 14, set A consists of all prime numbers and set B consists of all even numbers.
Set A = {2. 3, 5, 7, 11, 13}
Set B = {2, 4, 6, 8, 10, 12}
It can be seen that here A and B are overlapping sets.

Question 29.
Of the positive integers less than 24, set A consists of the multiples of 2 and set B consists of all the multiples of 3.

### 12.1 – 12.3 Quiz

Question 1.
You randomly draw a marble out of a bag containing 8 green marbles, 4 blue marbles 12 yellow marbles, and 10 red marbles. Find the probability of drawing a marble that is not yellow.
Given,
You randomly draw a marble out of a bag containing 8 green marbles, 4 blue marbles 12 yellow marbles, and 10 red marbles.
Total number of outcomes here are 8 + 4 + 12 + 10 = 34
Thus the probability of obtaining a marble that is not yellow is (8 + 4 + 10)/34
= 22/34
= 11/17
= 0.647
= 64.7%
Thus the probability of obtaining a marble that is not yellow is 64.7%

Find P((\bar{A}))

Question 2.
P(A) = 0.32
P((\bar{A}) = 1 – P(A)
1 – 0.32
= 0.68
P((\bar{A}) = 0.68

Question 3.
P(A) = $$\frac{8}{9}$$
P((\bar{A}) = 1 – P(A)
1 – $$\frac{8}{9}$$
= $$\frac{1}{9}$$
P((\bar{A}) = $$\frac{1}{9}$$

Question 4.
P(A) = 0.01
P((\bar{A}) = 1 – P(A)
1 – 0.01 = 0.99

Question 5.
You roll a six-sided die 30 times. A 5 is rolled 8 times. What is the theoretical probability of rolling a 5? What is the experimental probability of rolling a 5?
Given,
You roll a six-sided die 30 times. A 5 is rolled 8 times.
The theoretical probability of rolling a 5 on a number cube is $$\frac{1}{6}$$ while the experimental probability of rolling a 5 on a number cube is $$\frac{8}{30}$$ = $$\frac{4}{15}$$

Question 6.
Events A and B are independent. Find the missing probability.
P(A) = 0.25
P(B) = ____
P(A and B) = 0.05
Given,
P(A) = 0.25
P(A and B) = 0.05
P(A and B) = P(A) × P(B)
P(B) = P(A and B)/P(A)
P(B) = 0.05/0.25 = 0.2
P(B) = 0.2

Question 7.
Events A and B are dependent. Find the missing probability.
P(A) = 0.6
P(B/A) = 0.2
P(A and B) = ____
Given,
P(A) = 0.6
P(B/A) = 0.2
P(A and B) = P(A) × P(B/A)
P(A and B) = 0.6 × 0.2
= 0.12
P(A and B) = 0.12

Question 8.
Find the probability that a dart thrown at the circular target Shown will hit the given region.
Assume the dart is equally likely to hit any point inside the target.

a. the center circle
Total area of the given region is π × r² = π × 6² = 36π = 113.112 sq. units
Area of the center circle is π × r² = π × 2² = 4π sq. units.
Therefore the probability of hitting the center circle is 4π/36π = 1/9 = 0.11…

b. outside the square
Area of the square is 6² = 36
So the region outside of it is equal to 36π – 36 = 77.112 sq. units
Thus the probability of hitting the region outside the square is 77.112/113.112 = 0.682

c. inside the square but outside the center circle
Area of the center circle is π × r² = π × 2² = 4π sq. units.
Area of the square is 6² = 36
Thus the probability of hitting the region outside the center circle but inside the square is 36 – 4π = 23.432 sq. units
Thus the probability of hitting the region is 23.432/113.112 = 0.207

Question 9.
A survey asks 13-year-old and 15-year-old students about their eating habits. Four hundred students are surveyed, 100 male students and 100 female students from each age group. The bar graph shows the number of students who said they eat fruit every day.

a. Find the probability that a female student, chosen at random from the students surveyed, eats fruit every day.
Total number of females who eat a fruit everyday are 61 + 58 = 119
Therefore the probability of randomly choosing a female who eats a fruit everyday is 119/400= 0.2975

b. Find the probability that a 15 – year – old student. chosen at random from the students surveyed, eats fruit every day.
Total number of 15 year old student who eat a fruit everyday are 53 + 58 = 111
Therefore the probability of randomly choosing a 15 year old student who eats a fruit everyday is 111/200 = 0.555

Question 10.
There are 14 boys and 18 girls in a class. The teacher allows the students to vote whether they want to take a test on Friday or on Monday. A total of 6 boys and 10 girls vote to take the test on Friday. Organize the information in a two-way table. Then find and interpret the marginal frequencies.
Given,
There are 14 boys and 18 girls in a class. The teacher allows the students to vote whether they want to take a test on Friday or on Monday.
14 + 18 = 32
Number of boys who vote to take the test on Monday is 14 – 6 = 8
Number of girls who vote to take the test on monday is 18 – 10 = 8
A total of 6 boys and 10 girls vote to take the test on Friday.
The total number of students who take the test on Friday is 10 + 6 = 16
The total number of students who vote to take the test on Friday is 8 + 8 = 16

Question 11.
Three schools compete in a cross country invitational. Of the 15 athletes on your team. 9 achieve their goal times. Of the 20 athletes on the home team. 6 achieve their goal times. On your rival’s team, 8 of the 13 athletes achieve their goal times. Organize the information in a two-way table. Then determine the probability that a randomly elected runner who achieves his or her goal time is from your school.
Three schools compete in a cross country invitational. Of the 15 athletes on your team. 9 achieve their goal times.
Number of runners in your team who do not achieve their goal team is 15 – 9 = 6
Number of runners in home team who do not achieve their goal team is 20 – 6 = 14
Number of runners in rival’s team who do not achieve their goal team is 13 – 8 = 5
Total number of runners who achieve their goal team is 9 + 6 + 8 = 23
Total number of runners who do not achieve their goal team is 6 + 14 + 5 = 25
The total number of rubbers who was surveyed is 23 + 25 = 48
P = Your team ans achieve their goal team/P(Archive their goal team)
P = 9/23
P = 0.39

### 12.4 Probability of Disjoint and Overlapping Events

Exploration 1

Work with a partner: A six-sided die is rolled. Draw a Venn diagram that relates the two events. Then decide whether the cents are disjoint or overlapping.
MODELING WITH MATHEMATICS
To be proficient in math, you need to map the relationships between important quantities in a practical situation using such tools as diagrams.

a. Event A: The result is an even number.
Event B: The result is a prime number.

b. Event A: The result is 2 or 4.
Event B: The result is an odd number

Exploration 2

Finding the Probability that Two Events Occur

Work with a partner: A six-sided die is roiled. For each pair of events. find (a) P(A), (b) P(B). (C) P(A) and (B). and (d) P(A or B).

a. Event A: The result is an even number.
Event B: The result is a Prime number.
P(A) = $$\frac{3}{6}$$ = $$\frac{1}{2}$$
P(B) = $$\frac{3}{6}$$ = $$\frac{1}{2}$$
P(A or B) = P(A) + P(B) – P(A and B)
P(A and B) = $$\frac{1}{6}$$
P(A or B) = $$\frac{5}{6}$$

b. Event A: The result is 2 or 4.
Event B: The result is an odd number.
P(A) = $$\frac{2}{6}$$ = $$\frac{1}{3}$$
P(B) = $$\frac{3}{6}$$ = $$\frac{1}{2}$$
P(A or B) = P(A) + P(B) – P(A and B)
P(A and B) = 0
P(A or B) = $$\frac{5}{6}$$

Exploration 3

Discovering Probability Formulas

Work with a partner:
a. In general, if event A and event B arc disjoint, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion.
If event A and B are disjoint, there are no common outcomes.
So we add the probabilities that each event occurs:
P(A or B) = P(A) + P(B)

b. In general, if event A and event B are overlapping, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion.
If event A and event B are overlapping, there are common outcomes.
So, we add the probabilities that each event occurs then subtract the probability of the common outcomes.
P(A or B) = P(A) + P(B) – P(A and B)

c. Conduct an experiment using a six-sided die. Roll the die 50 times and record the results. Then use the results to find the probabilities described in Exploration 2. How closely do your experimental probabilities compare to the theoretical probabilities you found in Exploration 2?

a. P(A) = $$\frac{1}{2}$$ = 50%
P(A) = $$\frac{21}{50}$$ = 42%
P(B) = $$\frac{1}{2}$$ = 50%
P(B) = $$\frac{32}{50}$$ = 64%
P(A and B) = $$\frac{1}{6}$$ ≈ 16.7%
P(A and B) = $$\frac{9}{50}$$ ≈ 18%
P(A or B) = $$\frac{5}{6}$$ ≈ 83.3%
P(A or B) = $$\frac{44}{50}$$ ≈ 88%
P(A) = $$\frac{1}{3}$$ ≈ 33.3%
P(A) = $$\frac{17}{50}$$ = 34%
P(B) = $$\frac{1}{2}$$ = 50%
P(B) = $$\frac{29}{50}$$ = 58%
P(A and B) = 0 = 0%
P(A and B) = $$\frac{0}{50}$$ = 0%
P(A or B) = $$\frac{5}{6}$$ ≈ 83.3%
P(A or B) = $$\frac{46}{50}$$ = 92%

Question 4.
How can you find probabilities of disjoint and overlapping events?
If A and B are disjoint events, then the probability of A or B is P(A or B) = P(A) + P(B). If two events A and B are overlapping, then the outcomes in the intersection of A and B are counted twice when P(A) and P(B) are added.
P(A or B) = P(A) + P(B) – P(A and B)

Question 5.
Give examples of disjoint events and overlapping events that do not involve dice.

a. Event A: The result is an even number.
Event B: The result is a prime number.

b. Event A: The result is 2 or 4.
Event B: The result is an odd number

### Lesson 12.4 Probability of Disjoint and Overlapping Events

Monitoring Progress

A card is randomly selected from a standard deck of 52 playing cards. Find the probability of the event.

Question 1.
selecting an ace or an 8
A: Selecting an ace
B: You select 8
We know that A has 4 outcomes and B also has 4 outcomes.
P(A or B) = P(A) + P(B)
= 4/52 + 4/52
= 8/52
= 2/13 ≈ 0.15

Question 2.
selecting a 10 or a diamond
A: Selecting a 10
B: You select diamond
P(A or B) = P(A) + P(B) – P(A and B)
= 4/52 + 13/52 – 1/52
= 16/52
= 4/13

Question 3.
WHAT IF?
In Example 3, suppose 32 seniors are in the band and 64 seniors are in the band or on the honor roll. What is the Probability that a randomly selected senior is both in the band and on the honor roll?

Question 4.
In Example 4, what is the probability that the diagnosis is incorrect?

Question 5.
A high school basketball team leads at halftime in 60% of the games in a season. The team wins 80% of the time when the have the halftime lead, but only 10% of the time when the do not. What is the probability that the team wins a particular game during the season?
Given,
A high school basketball team leads at halftime in 60% of the games in a season. The team wins 80% of the time when the have the halftime lead, but only 10% of the time when the do not.
Let event A be team leads on the halftime and event B be win.
When A occurs, P(B) = 0.8
When A does not occur, P(B) = 0.1
P(B) = P(A and B) + P($$\bar{A}$$ and B)
P(A) . P(B | A) + P($$\bar{A}$$) . P(B | $$\bar{A}$$)
= 0.6 × 0.8 + 0.4 × 0.1
= 0.52

### Exercise 12.4 Probability of Disjoint and Overlapping Events

Vocabulary and Core Concept Check

Question 1.
WRITING
Are the events A and $$\bar{A}$$ disjoint? Explain. Then give an example of a real-life event and its complement.
Yes A and $$\bar{A}$$ are disjoint events because they are the complement of one another and so can not occur together and hence the name disjoint.
Example:
We flip a coin.
So, A = (the head fell)
$$\bar{A}$$ = {the tail falls} are disjoint events.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

How many outcomes are in the intersection of A and B?
Answer: There are 2 outcomes in the intersection of A and B.

How many outcomes are shared by both A and B?
Answer: 2 outcomes are shared by both A and B.

How many outcomes are in the union of A and B?
Answer: There are 4 + 2 + 3 = 9 outcomes in the union of A and B.

How many outcomes in B are also in A?
Answer: There are 2 outcomes in B that are also in A.

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, events A and B are disjoint. Find P(A or B)

Question 3.
p(A) = 0.3, P(B) = 0.1

Question 4.
p(A) = 0.55, P(B) = 0.2
Given,
p(A) = 0.55, P(B) = 0.2
P(A or B) = P(A) + P(B)
P(A or B) = 0.55 + 0.2 = 0.75

Question 5.
P(A) = $$\frac{1}{3}$$, P(B) = $$\frac{1}{4}$$

Question 6.
p(A) = $$\frac{2}{3}$$, P(B) = $$\frac{1}{5}$$
Given,
p(A) = $$\frac{2}{3}$$, P(B) = $$\frac{1}{5}$$
P(A or B) = P(A) + P(B)
P(A or B) = $$\frac{2}{3}$$ + $$\frac{1}{5}$$
P(A or B) = $$\frac{13}{15}$$

Question 7.
PROBLEM SOLVING
Your dart is equally likely to hit any point inside the board Shown. You throw a dart and pop a balloon. What is the probability that the balloon is red or blue?

Question 8.
PROBLEM SOLVING
You and your friend are among several candidates running for class president. You estimate that there is a 45% chance you will win and a 25% chance your friend will win. What is the probability that you or your friend win the election?
Let A be the event of you winning the election and B be of your friend winning the election
P(A) = 45% = 0.45
P(B) = 25% = 0.25
P(A or B) = P(A) + P(B)
P(A or B) = 0.45 + 0.25 = 0.70
Therefore the probability of you or your friend winning the election is 0.7

Question 9.
PROBLEM SOLVING
You are performing an experiment to determine how well plants grow under different light sources. 0f the 30 Plants in the experiment, 12 receive visible light, 15 receive ultraviolet light, and 6 receive both visible and ultraviolet light. What is the probability that a plant in the experiment receives visible or ultraviolet light?

Question 10.
PROBLEM SOLVING
Of 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English. There are 14 students who won awards for both mathematics and English. A newspaper chooses a student at random for an interview. What is the probability that the student interviewed won an award for English or mathematics?
Given,
There are 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English.
There are 14 students who won awards for both mathematics and English. A newspaper chooses a student at random for an interview.
P(A) = 48/162
P(B) = 78/162
P(A and B) = 14/162
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 48/162 + 78/162 – 14/162
P(A or B) = 112/162 = 56/81 = 0.691

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in finding the probability of randomly drawing the given card from a standard deck of 52 playing cards.

Question 11.

Question 12.

These 2 events are overlapping events as there is 1 card that is both a club and 9, therefore write equation of P(A or B) for overlapping events
P(A or B) = P(A) + P(B) – P(A and B)
= 13/52 + 4/52 – 1/52
= 4/13

In Exercises 13 and 14, you roll a six-sided die. Find P(A or B).

Question 13.
Event A: Roll a 6.
Event B: Roll a prime number.

Question 14.
Event A: Roll an odd number.
Event B: Roll a number less than 5.
P(A) = 3/6
1, 3, 5 out of 6 possible outcomes
P(B) = 4/6
1, 2, 3, 4 out of 6 possible outcomes
P(A and B) = 2/6
3/4 + 4/6 – 2/6 = 5/6

Question 15.
DRAWING CONCLUSIONS
A group of 40 trees in a forest are not growing properly. A botanist determines that 34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease and 20 being damaged by insects. What is the probability that a randomly selected tree has both a disease and is being damaged by insects?

Question 16.
DRAWING CONCLUSIONS
A company paid overtime wages or hired temporary help during 9 months of the year. Overtime wages were paid during 7 months. and temporary help was hired during 4 months. At the end of the year, an auditor examines the accounting records and randomly selects one month to check the payroll. What is the probability that the auditor will select a month in which the company paid overtime wages and hired temporary help?
P(A) = 7/12
P(B) = 4/12
P(A or B) = 9/12
P(A and B) = P(A) + P(B) – P(A or B)
P(A and B) = 7/12 + 4/12 – 9/12
P(A and B) = 2/12 = 1/6
The probability of randomly selecting a month in which overtime was paid and temporary help was hired is 1/6 = 0.166…

Question 17.
DRAWING CONCLUSIONS
A company is focus testing a new type of fruit drink. The focus group is 47% male. 0f the responses, 40% of the males and 54% of the females said they would buy the fruit drink. What is the probability that a randomly selected person would buy the fruit drink?

Question 18.
DRAWING CONCLUSIONS
The Redbirds trail the Bluebirds by one goal with 1 minute left in the hockey game. The Redbirds coach must decide whether to remove the goalie and add a frontline player. The probabilities of each team scoring are shown in the table.

a. Find the probability that the Redbirds score and the Bluebirds do not score when the coach leaves the goalie in.
Redbirds 10% x Bluebirds 90% = 9%

b. Find the probability that the Redbirds score and the Bluebirds do not score when the coach takes the goalie out.
Redbirds 30% x Bluebirds 40% = 12%

c. Based on parts (a) and (b), what should the coach do?
Answer: Looks to be a 3% better chance to tie it up in B – pull the goalie

Question 19.
PROBLEM SOLVING
You can win concert tickets from a radio station if you are the first person to call when the song of the day is played. or if you are the first person to correctly answer the trivia question. The song of the day is announced at a random time between 7:00 and 7:30 A.M. The trivia question is asked at a random Lime between 7:15 and 7:45 A.M. You begin listening to the radio station at 7:20. Find the probability that you miss the announcement of the song of the day or the trivia question.

Question 20.
HOW DO YOU SEE IT?
Are events A and B disjoint events? Explain your reasoning.

A and B are not disjoint events and in fact they are overlapping events with 1 overlapping outcome.
Disjoint events do not have any overlap and they are mutually exclusive from one another.

Question 21.
PROBLEM SOLVING
You take a bus from sour neighborhood to your school. The express bus arrives at your neighborhood at a random time between 7:30 and 7:36 AM. The local bus arrives at your neighborhood at a random time between 7:30 and 7:40 A.M. You arrive at the bus stop at 7:33 A.M. Find the probability that you missed both the express bus and the local bus.

Question 22.
THOUGHT PROVOKING
Write a general rule for finding P(A or B or C) for (a) disjoint and (b) overlapping events A, B, and C.
For 2 disjoint events, the equation becomes:
P(A or B) = P(A) + P(B), based on this it can be said that the equation of P(A or B or C) will be
P(A or B or C) = P(A) + P(B) + P(C)
For 2 overlapping events, the equation becomes:
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B or C) = P(A) + P(B) + P(C) – P(A and B) – P(A and C) – P(B and C) + P(A and B and C)

Question 23.
MAKING AN ARGUMENT
A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag. This is done four times. Two red cards are drawn. numbered 31 and 19, and two blue cards are drawn. numbered 22 and 7. Your friend concludes that red cards and even numbers must be mutually exclusive. Is your friend correct? Explain.
Your friend is incorrect because we do not know all the number of cards. Also, from the given data we do not know all colors for cards. Therefore we can not conclude that red cards and even numbers be mutually exclusive.

Maintaining Mathematical Proficiency

Find the Product.

Question 24.
(n – 12)2
We can solve the product by using the formula
(a – b)² = a² – 2ab + b²
(n – 12)2 = (n)² – 2(n)(12) + (12)²
n² – 24n + 144

Question 25.
(2x + 9)2

Question 26.
(- 5z + 6)2
(- 5z + 6)2 = (6 – 5z)2
We can solve the product by using the formula
(a – b)² = a² + 2ab + b²
(6 – 5z)2 = (6)² – 2(6)(5z) + (5z)²
36 – 24n + 25z²

Question 27.
(3a – 7b)2
We can solve the product by using the formula
(a – b)² = a² + 2ab + b²
(3a – 7b)2 = (3a)² – 2(3a)(7b) + (7b)²
9a² – 42ab + 49b²

### 12.5 Permutations and Combinations

Exploration 1

Work with a partner. Two coins are flipped and the spinner is spun. The tree diagram shows the possible outcomes.

a. How many outcomes are possible?

b. List the possible outcomes.

Exploration 2

Work with a partner: Consider the tree diagram below.

a. How many events are shown?

b. What outcomes are possible for each event?

c. How many outcomes are possible?

d. List the possible outcomes.

Exploration 3

Writing a Conjecture

Work with a partner:

a. Consider the following general problem: Event 1 can occur in in ways and event 2 can occur in n ways. Write a conjecture about the number of ways the two events can occur. Explain your reasoning.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math,
you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.

b. Use the conjecture you wrote in part (a) to write a conjecture about the number of ways more than two events can occur. Explain your reasoning.

c. Use the results of Explorations 1(a) and 2(c) to verify your conjectures.

Question 4.
How can a tree diagram help you visualize the number of ways in which two or more events can occur?

Question 5.
In Exploration 1, the spinner is spun a second time. How many outcomes are possible?

### Lesson 12.5 Permutations and Combinations

Monitoring Progress

Question 1.
In how many ways can you arrange the letters in the word HOUSE?

Question 2.
In how many ways can you arrange 3 of the letters in the word MARCH?

Question 3.
WHAT IF
In Example 2, suppose there are 8 horses in the race. In how many different ways can the horses finish first, second, and third? (Assume there are no ties.)

Question 4.
WHAT IF?
In Example 3, suppose there are 14 floats in the parade. Find the probability that the soccer team is first and the chorus is second.

Question 5.
Count the possible combinations of 3 letters chosen from the list A, B, C, D, E.

Question 6.
WHAT IF?
In Example 5, suppose you can choose 3 side dishes out of the list of 8 side dishes. How many combinations are possible?

Question 7.
WHAT IF?
In Example 6, suppose there are 20 photos in the collage. Find the probability that your photo and your friend’s photo are the 2 placed at the top of the page.

### Exercise 12.5 Permutations and Combinations

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
An arrangement of objects in which order is important is called a(n) _________ .
An arrangement of objects in which order is important is called a Permutation.

Question 2.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
$$\frac{7 !}{2 ! \cdot 5 !}$$ 7C5 7C2 $$\frac{7 !}{(7-2) !}$$
7C2 $$\frac{7 !}{(7-2) !}$$ = $$\frac{7 !}{5!}$$
= 7C2
The expression $$\frac{7 !}{(7-2) !}$$ does not belong with other three.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find the number of ways you can arrange (a) all of the letters and (b) 2 of the letters in the given word.

Question 3.
AT

Question 4.
TRY
a. In this case, we have to find the number of permutations all of the letters in a given word that will consist of 3 letters.
Number of permutations = (1st place can be one of three letters) × (2nd can be one of two letters that is left) × (3rd can be one letter that is left)
= 3 . 2 . 1 = 6
Therefore we have 6 ways for arrange all of the letters in given word, that is TRY, TYR, YTR, YRT, RTY and RYT.
Now, we have to find the number of permutation 2 of the letters in a given word that will consists of 3 letters
Number of permutations = (1st place can be one of three letters) × (2nd can be one of two letters that is left)
= 3 . 2 = 6
We have 6 ways for arrange 2 of the letters in given word, tthat is TR, TY, YT, YR, RT and RY.

Question 5.
ROCK

Question 6.
WATER
In this case, we have to find the number of permutation all of the letters in a given word that will consist of 5 letters.
Number of permutations = (1st can be one of 5 letters) × (2nd place can be one of 4 letters that is left) × (3rd can be one of 3 letters that is left) × (4th can be two letters that is left) × (5th can be one letter that is left)
= 5 .4 . 3 . 2 . 1 = 120
Thus we have 120 ways to arrange all of the letters in given word, that WATER, WATRE, WARTE,…, RETAW.
Number of permutations = (1st can be one of 5 letters) × (2nd place can be one of 4 letters that is left)
= 5 . 4
= 20
Thus we have 20 ways to arrange 2 of the letters in given word, that WA, WT, WR, WE …, ER, RE.

Question 7.
FAMILY

Question 8.
FLOWERS
We have to find the number of permutation all of the letters in a given word that will consist of 7 letters.
Number of permutations = (1st place can be one of 7 letters) × (2nd place can be one of 6 letters that is left) × (3rd place can be one of 5 letters that is left) × (4th place can be one of 4 letters that is left) × (5th place can be one of 3 letters that is left) × (6th place can be one of 2 letters that is left) × (7th place can be one of 1 letters that is left)
= 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5040
Thus we have 5040 ways to arrange all of the letters in given word, that is FLOWERS, FLOWERS, FLOWSER… SREWOLF
Now we have to find the number of permutation 2 of the letters in a given word that will consist of 7 letters.
Number of permutations = (1st place can be one of 7 letters) × (2nd place can be one of 6 letters that is left)
= 7 . 6
= 42
Thus we have 42 ways to arrange all of the letters in given word, that is FL, FO, FW, FE…RS, RE.

In Exercises 9 – 16, evaluate the expression.

Question 9.
5P2

Question 10.
7P3
7P3 = $$\frac{7 !}{(7-3) !}$$ = $$\frac{7 !}{4!}$$
= 7 . 6 . 5 . 4 . 3 . 2 . 1/4 . 3 . 2 . 1
= 7 . 6 . 5
= 210
7P3 = 210

Question 11.
9P1

Question 12.
6P5
6P5 = $$\frac{6 !}{(6-5) !}$$ = $$\frac{6 !}{1!}$$
= 6 . 5 . 4 . 3 . 2 . 1/1
= 6 . 5 . 4 . 3 . 2
= 720
6P5 = 720

Question 13.
8P6

Question 14.
12P0
12P0 = $$\frac{12 !}{(12-0) !}$$
= $$\frac{12 !}{12!}$$
= 1
12P0 = 1

Question 15.
30P2

Question 16.
25P5
25P5 = $$\frac{25 !}{(25-5) !}$$ = $$\frac{25 !}{20!}$$
= 25 . 24 . 23 . 22 . 21 . 20 . 19 . 18 . …. 6 . 5 . 4 . 3 . 2 . 1/20 . 19 . 18 . … 3 . 2 . 1
= 25 . 24 . 23 . 22 . 21
= 720
25P5 = 6375600

Question 17.
PROBLEM SOLVING
Eleven students are competing in an art contest. In how many different ways can the students finish first, second, and third?

Question 18.
PROBLEM SOLVING
Six Friends go to a movie theater. In how many different ways can they sit together in a row of 6 empty seats?
Given,
Six Friends go to a movie theater.
6P6 = $$\frac{6!}{(6-6) !}$$
= $$\frac{6!}{0!}$$
= 6!
= 6 . 5 . 4 . 3 . 2 . 1
6P6 = 720

Question 19.
PROBLEM SOLVING
You and your friend are 2 of 8 servers working a shill in a restaurant. At the beginning of the shill. the manager randomly assigns 0ne section to each server. Find the probability that you are assigned Section 1 and your friend is assigned Section 2.

Question 20.
PROBLEM SOLVING
You make 6 posters to hold up at a basketball game. Each poster has a letter of the word TIGERS. You and 5 friends sit next to each other in a row. The posters are distributed at random. Find the probability that TIGERS is spelled correctly when you hold up the posters.

Number of favorable outcomes = 1 (TIGERS)
Total number of outcomes = 6!
= 6 . 5 . 4 . 3 . 2 . 1
= 720
Number of favorable outcomes/Total number of outcomes = 1/720

In Exercises 21 – 24, count the possible combinations of r letters chosen from the given list.

Question 21.
A, B, C, D; r = 3

Question 22.
L, M, N, O; r = 2
4P2 = $$\frac{4!}{(4-2) !}$$
= $$\frac{4!}{2!}$$
= 4 . 3 . 2 . 1/2 . 1
= 12
4P2 = 12
So, the possible permutations of two letters in the given list L, M, N, O is
LM, ML
LN, NL
LO, OL
MN, NM
MO, MO
NO, ON
Thus the number of possible combination of a = 2 letters chosen from the list L, M, N, O
4P2 = 12/2 = 6

Question 23.
U , V, W, X, Y, Z; r = 3

Question 24.
D, E, F, G, H; R = 4
5P4 = $$\frac{5!}{(5-4) !}$$
= $$\frac{5!}{1!}$$
=5 . 4 . 3 . 2 . 1
= 120
5P4 = 120
Thus the possible permutations of four letters in the given list D, E, F, G, H is
DEFG, DEGF, DGEF, DGFE, DFGD…HEFG, EHFG, EHGF, FGEH.
Thus we conclude that the number of possible combination of a = 4 letters chosen from the list D, E, F, G, H is
5C4 = $$\frac{120}{(24) !}$$ = 5

In Exercise 25 – 32, evaluate the expression

Question 25.
5C1

Question 26.
8C5
8C5 = $$\frac{8!}{(8-5) !}$$
= $$\frac{8!}{3!}$$ . $$\frac{1}{5!}$$
= 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1) (3 . 2 . 1)
= 8 . 7
8C5 = 56

Question 27.
9C9

Question 28.
8C6
8C6 = $$\frac{8!}{(8-6) !}$$
= $$\frac{8!}{2!}$$ . $$\frac{1}{6!}$$
= 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(6. 5 . 4 . 3 . 2 . 1) (3 . 2 . 1)
= 8 . 7
8C5 = 56

Question 29.
12C3

Question 30.
11C4
11C4 = $$\frac{11!}{(11-4) !}$$
= $$\frac{11!}{7!}$$ . $$\frac{1}{4!}$$
= 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(7 . 6. 5 . 4 . 3 . 2 . 1) (4 . 3 . 2 . 1)
= 11 . 10 . 3
11C4 = 330

Question 31.
15C8

Question 32.
20C5
20C5 = $$\frac{20!}{(20-5) !}$$
= $$\frac{20!}{5!}$$ . $$\frac{1}{5!}$$
= 20 . 19 . 18 . 17 . 16 . 15 . … 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1) (5 . 4 . 3 . 2 . 1)
= 19 . 3 . 17 . 16
20C5 = 15504

Question 33.
PROBLEM SOLVING
Each year, 64 golfers participate in a golf tournament. The golfers play in groups of 4. How many groups of 4 golfers are possible?

Question 34.
PROBLEM SOLVING
You want to purchase vegetable dip for a party. A grocery store sells 7 different flavors of vegetable dip. You have enough money to purchase 2 flavors. How many combinations of 2 flavors of vegetable dip are possible?
given that,
You want to purchase vegetable dip for a party. A grocery store sells 7 different flavors of vegetable dip. You have enough money to purchase 2 flavors.
7C2 = $$\frac{7!}{(7-2) !}$$
= $$\frac{7!}{5!}$$ . $$\frac{1}{2!}$$
= 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1) (2 . 1)
= 7 . 3
7C2 = 21

ERROR ANALYSIS
In Exercises 35 and 36, describe and correct the error in evaluating the expression.

Question 35.

Question 36.

The permutation formula was used instead of the combination formula.

REASONING
In Exercises 37 – 40, tell whether the question can be answered using permutations or combinations. Explain your reasoning. Then answer the question.

Question 37.
To complete an exam. u must answer 8 questions from a list of 10 questions. In how many ways can you complete the exam?

Question 38.
Ten students are auditioning for 3 different roles in a play. In how many ways can the 3 roles be filled?

As 10 students are auditioning for 3 roles, which are different from each other, the order in which the roles are assigned to the students is important and should be taken into account.
As the Permutations formula takes into account the order of distribution. Hence the number of ways to fill the 3 roles can be found by using the permutations formula in the chapter.
So, the number of permutations of assigning the 3 roles to 3 students chosen from 10, using the permutations formula
10P3 = $$\frac{10!}{(10-3) !}$$
= $$\frac{10!}{7!}$$
= 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(7 . 6 . 5 . 4 . 3 . 2 . 1)
=10 . 9 . 8
10P3 = 720

Question 39.
Fifty-two athletes arc competing in a bicycle race. In how many orders can the bicyclists finish first, second, and third? (Assume there are no ties.)

Question 40.
An employee at a pet store needs to catch 5 tetras in an aquarium containing 27 tetras. In how many groupings can the employee capture 5 tetras?

Given,
An employee at a pet store needs to catch 5 tetras in an aquarium containing 27 tetras.
27C5 = $$\frac{27!}{(27-5) !}$$
= $$\frac{27!}{22!}$$ . $$\frac{1}{5!}$$
= 27. 26 . 25 . 24 . … 7 . 6 . 5 . 4 . 3 . 2 . 1/(22 . 21. … 7 . 6 . 5 . 4 . 3 . 2 . 1)(5 . 4 . 3 . 2 . 1)
= 27 . 26 . 5 . 23
27C5 = 80730
Thus the number of combinations of 27 tetras taken 5 at the time is 80730.

Question 41.
CRITICAL THINKING
Compare the quantities 50C9 and 50C41 without performing an calculations. Explain your reasoning.

Question 42.
CRITICAL THINKING
Show that each identity is true for any whole numbers r and n, where 0 ≤ r ≤ n.
a. nCn = 1
nCn = n!/n!(n – n)!
nCn = n!/n!0!
= 1/0!
We know that,
0! = 1
= 1/1 = 1
Thus nCn = 1

b. nCr = nCn – r
nCn-a= n!/a!(n – a)!
nCn-a = n!/a!(n-a)! = nCn-a

c. n + 1Cr = nCr + nCr – 1
n+1Ca = n+1!/a!(n+1-a)!
nCr + nCa-1 = n!/a!(n – a)! + n!/(n – (a – 1))! (a – 1)!
= $$\frac{(n+1)!}{(n-a+1) !}$$ . $$\frac{1}{a!}$$
Thus n + 1Ca = nCa + nCa – 1

Question 43.

REASONING
Complete the table for each given value of r. Then write an inequality relating nPr and nCr. Explain your reasoning.

Question 44.
REASONING
Write an equation that relates nPr and nCr. Then use your equation to find and interpret the Value of $$\frac{182 P_{4}}{182 C_{4}}$$.
$$\frac{n P_{a}}{n C_{a}}$$ = n!/(n – a)!/n!/a!(n – a)! = a!
$$\frac{182 P_{4}}{182 C_{4}}$$ = 4! = 24

Question 45.
PROBLEM SOLVING
You and your friend are in the studio audience on a television game show. From an audience of 300 people, 2 people are randomly selected as contestants. What is the probability that you and your friend are chosen?

Question 46.
PROBLEM SOLVING
You work 5 evenings each week at a bookstore. Your supervisor assigns you 5 evenings at random from the 7 possibilities. What is the probability that your schedule does not include working on the weekend?
nCa = n!/a!(n – a)!
7C5 = $$\frac{7!}{(7-5) !}$$
= $$\frac{7!}{2!}$$ . $$\frac{1}{5!}$$
=7 . 6 . 5 . 4 . 3 . 2 . 1/(2 . 1)(5 . 4 . 3 . 2 . 1)
= 7 . 3
7C5 = 21
Thus we see that there are 21 possible combinations of five days formed from 7 days.
Hence we see that one of 21 possible combination of 5 days does not contain saturday and sunday.
P = P(Are chosen one combination of 21 possible)
= Number of favorable outcomes/Number of possible outcomes
= 1/21

REASONING
In Exercises 47 and 48, find the probability of winning a lottery using the given rules. Assume that lottery numbers are selected at random.

Question 47.
You must correctly select 6 numbers, each an integer from 0 to 49. The order is not important.

Question 48.
You must correctly select 4 numbers, each an integer from 0 to 9. The order is important.
nCa = n!/a!(n – a)!
10C4 = $$\frac{10!}{(10-4) !}$$
= $$\frac{10!}{6!}$$ . $$\frac{1}{4!}$$
=10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(6 . 5 . 4 . 3 . 2 . 1)(4 . 3 . 2 . 1)
= 10 . 3 . 7
10C4 = 210
There are 210 possible combination of 10 numbers taken 4 at a time.
P (You winning a lottery) = P(Are chosen one combination 210 possible)
= Number of favorable outcomes/Number of possible outcomes
= 1/210

Question 49.
MATHEMATICAL CONNECTIONS
A polygon is convex when no line that contains a side of the polygon contains a point in the interior of the polygon. Consider a convex polygon with n sides.

a. Use the combinations formula to write an expression for the number of diagonals in an n-sided polygon.
b. Use your result from part (a) to write a formula for the number of diagonals of an n-sided convex polygon.

Question 50.
PROBLEM SOLVING
You are ordering a burrito with 2 main ingredients and 3 toppings. The menu below shows the possible choices. How many different burritos are possible

You are ordering a burrito with 2 main ingredients and 3 toppings.
Total number of main ingredients = 6
As the order in which the ingredients are chosen is not important, the total number of ways to select 2 main ingredients out of 6 can be found by using the combinations formula.
nCa = n!/a!(n – a)!
6C2 = $$\frac{6!}{(6-2) !}$$
= $$\frac{6!}{4!}$$ . $$\frac{1}{2!}$$
=6 . 5 . 4 . 3 . 2 . 1/(4 . 3 . 2 . 1)(2 . 1)
= 3 . 5
6C2 = 15
Total number of toppings = 8
Number of toppings to be chosen = 3
nCa = n!/a!(n – a)!
8C3 = $$\frac{8!}{(8-3) !}$$
= $$\frac{8!}{5!}$$ . $$\frac{1}{3!}$$
=8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1)(3 . 2 . 1)
= 8 . 7
8C3 = 56
Total possible selections = ways to select main ingredients × Ways to select toppings
= 15 × 56
= 840

Question 51.
PROBLEM SOLVING
You want to purchase 2 different types of contemporary music CDs and 1 classical music CD from the music collection shown. How many different sets of music types can you choose for your purchase?

a. How many combinations of three marbles can be drawn from the bag? Explain.
b. How many permutations of three marbles can be drawn from the bag? Explain.

Question 52.
HOW DO YOU SEE IT?
A bag contains one green marble, one red marble, and one blue marble. The diagram shows the possible outcomes of randomly drawing three marbles from the hag without replacement.

a. How many combinations of three marbles can be drawn from the bag? Explain.
nCa = n!/a!(n – a)!
3C3 = $$\frac{3!}{(3-3) !}$$
= $$\frac{3!}{0!}$$ . $$\frac{1}{3!}$$
=3 . 2 . 1/(1)(3 . 2 . 1)
= 1
3C3 = 1

b. How many permutations of three marbles can be drawn from the bag? Explain.
nPa = n!/(n – a)!
3P3 = $$\frac{3!}{(3-3) !}$$
= $$\frac{3!}{0!}$$
=3 . 2 . 1/1
= 6
3P3 = 3
There are 6 possible combinations.

Question 53.
PROBLEM SOLVING
Every student in your history class is required to present a project in front of the class. Each day, 4 students make their presentations in an order chosen at random by the teacher. you make your presentation on the first day.
a. What is the probability that you are chosen to be the first or second presenter on the first day ?
b. What is the probability that you are chosen to be the second or third presenter on the first day? Compare your answer with that in part (a).

Question 54.
PROBLEM SOLVING
The organizer of a cast party for a drama club asks each of the 6 cast members to bring 1 food item From a list of 10 items. Assuming each member randomly chooses a food item to bring. what is the probability that at least 2 of the 6 cast members bring the same item?
Given,
The organizer of a cast party for a drama club asks each of the 6 cast members to bring 1 food item From a list of 10 items.
Assuming each member randomly chooses a food item to bring.
A = {At least 2 of the 6 cast members bring the same item}
$$\bar{A}$$ = {All members bring different items}
First member can choose one of the 10 different products, the second can choose one of the 9 possible ways, and so on until the 6th member can choose one of the remaining 5 items.
10 . 9 . 8 . 7 . 6 . 5 = 151200
nPa = n!/(n – a)!
10P6 = $$\frac{10!}{(10-6) !}$$
= $$\frac{10!}{4!}$$
=10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/4 . 3 . 2 . 1
= 10 . 9 . 8 . 7 . 6 . 5 = 151200
10P6 = 151200
10 . 10 . 10 . 10 . 10 . 10 = 1000000
P($$\bar{A}$$) = Number of favorable outcomes/Number of possible outcomes
= 151200/1000000
= 0.1512
P(A) = 1 – P($$\bar{A}$$)
P(A) = 1 – 0.1512 = 0.8488
The probability that at least 2 of the 6 members bring the same items is about 85%

Question 55.
PROBLEM SOLVING
You are one of 10 students performing in a school talent show. The order of the performances is determined at random. The first 5 performers go on stage before the intermission.
a. What is the probability that you are the last performer before the intermission and your rival performs immediately before you?
b. What is the probability that you are not the first performer?

Question 56.
THOUGHT PROVOKING
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, and 4? Repetition of digits is allowed.
Let fixed number 1 on the first place and in the second place can be one of 5 digits 0, 1, 2, 3 or 4. Because repetition of digits is allowed. In third and fourth place it can also be one of five digits.
Therefore we see that 5 . 5 . 5 = 125 integers which begin with 1, can be formed.
Second, If we fixed 2 on the first place, by the same logic as in the first part, we get that 5³ = 125 integers which begin with 2.
Next, if we fixed number 3 on the first place, we obtain 5³ = 125 integers which begin with 2.
5³ + 5³ + 5³ = 375
375 integers greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, 4.

Question 57.
PROBLEM SOLVING
There are 30 students in your class. Your science teacher chooses 5 students at random to complete a group project. Find the probability that you and your 2 best friends in the science class are chosen to work in the group. Explain how you found your answer.

Question 58.
PROBLEM SOLVING
Follow the steps below to explore a famous probability problem called the birthday problem. (Assume there are 365 equally likely birthdays possible.)

a. What is the probability that at least 2 people share the same birthday in a group of 6 randomly chosen people? in a group of 10 randomly chosen people?
nPa = n!/(n – a)!
A = {At least  people share the same birthday in a group of 6 people}
So, we see that complement of event A is
$$\bar{A}$$ = {All 6 people were born on different days}
nPa = n!/(n – a)!
365P6 = $$\frac{365!}{(365-6)!}$$
= $$\frac{365!}{(359)!}$$
= 365. 364 . 363 . 362 . 361 . 360 . 359 . 358 . ……. 2 . 1/(359 . 358. …. 2 . 1)
= 365. 364 . 363 . 362 . 361 . 360
For the first day we can also choose one of 365 possible for the second dat we can also choose one of 365 ways and so on.
For each of the 6 days we have the option to choose one day from 365 possible ways.
So, the number of possible outcomes is
365 . 365 . 365 . 365 . 365 . 365
P($$\bar{A}$$) = Number of favorable outcomes/Number of possible outcomes
= 365. 364 . 363 . 362 . 361 . 360/365 . 365 . 365 . 365 . 365 . 365
= 0.959
P(A) = 1 – P($$\bar{A}$$)
P(A) = 1 – 0.959 = 0.04
P = 1 – 365P10/36510
1 – 0.883 = 0.117

b. Generalize the results from part (a) by writing a formula for the probability P(n) that at least 2 people in a group of n people share the same birthday. (Hint: Use nPr notation in your formula.)
Based on the explanation in the part under a) we can conclude that the probability that at least two people share the same birthday in a group of n people is
P = 1 – 365Pn/365n

c. Enter the formula from part (b) into a graphing calculator. Use the table feature to make a table of values. For what group size does the probability that at least 2 people share the same birthday first exceed 50%?

Based on the above table we see that for a group of 23 people and more, the probability that at least tweople share the same birthday exceed 50%

Maintaining Mathematical Proficiency

Question 59.
A bag contains 12 white marbles and 3 black marbles. You pick 1 marble at random. What is the probability that you pick a black marble?
Given,
A bag contains 12 white marbles and 3 black marbles. You pick 1 marble at random

Question 60.
The table shows the result of flipping two coins 12 times. For what outcome is the experimental probability the same as the theoretical probability?

The table shows the result of flipping two coins 12 times
P(HH) = P(HT) = P(TH) = P(TT) = 1/2 . 1/2 = 1/4
P(HH) = Number of favorable outcomes/Number of possible outcomes
= 2/12 = 1/6
P(HT) = 6/12 = 1/2
P(TH) = 3/12 = 1/4
P(TT) = 1/12
The most likely fell first heads, and second tails. Also, with the least probability fell twice tails.

### 12.6 Binomial Distributions

Exploration 1

Analyzing Histograms

Work with a partner: The histograms show the results when n coins are flipped.

STUDY TIP
When 4 coins are flipped (n = 4), the possible outcomes are
TTTT TTTH TTHT TTHH
THTT THTH THHT THHH
HTTT HTTH HTHT HTHH
HHTT HHTH HHHT HHHH.
The histogram shows the numbers of outcomes having 0, 1, 2, 3, and 4 heads.
a. In how many ways can 3 heads occur when 5 coins are flipped?

b. Draw a histogram that shows the numbers of heads that can occur when 6 coins are flipped.

c. In how many ways can 3 heads occur when 6 coins are flipped?

Exploration 2

Determining the Number of Occurrences

Work with a partner:

a. Complete the table showing the numbers of ways in which 2 heads can occur when n coins are flipped.

b. Determine the pattern shown in the table. Use your result to find the number of ways in which 2 heads can occur when 8 coins are flipped.
LOOKING FOR A PATTERN
To be proficient in math, you need to look closely to discern a pattern or structure.

Question 3.
How can you determine the frequency of each outcome of an event?

Question 4.
How can you use a histogram to find the probability of an event?

### Lesson 12.6 Binomial Distributions

Monitoring Progress

An octahedral die has eight sides numbered 1 through 8. Let x be a random variable that represents the sum when two such dice are rolled.

Question 1.
Make a table and draw a histogram showing the probability distribution tor x.

Question 2.
What is the most likely sum when rolling the two dice?

Question 3.
What is the probability that the sum of the two dice is at most 3?

According to a survey, about 85% of people ages 18 and older in the U.S. use the Internet or e-mail. You ask 4 randomly chosen people (ages 18 and older) whether they use the Internet or email.

Question 4.
Draw a histogram of the binomial distribution for your survey.

Question 5.
What is the most likely outcome of your survey?

Question 6.
What is the probability that at most 2 people you survey use the Internet or e-mail?

### Exercise 12.6 Binomial Distributions

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a random variable?
A random variable is a variable whose value is determined by the outcomes of a probability experiment.

Question 2.
WRITING
Give an example of a binomial experiment and describe how it meets the conditions of a binomial experiment.
We flipping a coin 10 times and register what fell.
We know that events, coin toss are independent.
Each trial has only two possible outcomes: H and T
The probabilities are P(H) = P(T) = 1/2 and are the same for each trial.
We can conclude that this experiment is a binomial experiment.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. make a table and draw a histogram showing the probability distribution for the random variable.

Question 3.
x = the number on a table tennis ball randomly chosen from a bag that contains 5 balls labeled “1,” 3 halls labeled “2,” and 2 balls labeled “3.”

Question 4.
c = 1 when a randomly chosen card out of a standard deck of 52 playing cards is a heart and c = 2 otherwise.
Let C be a random variable that represents the randomly chosen card.
Standard desk have 52 playing cards.
P(C = 1) = Number of favorable outcomes/Total number of outcomes
= A randomly chosen card is a hard/Total number of cards
= 13/52
On the other hand,
P(C = 2) = A randomly chosen card is not a hard/Total number of cards
= 39/52

Question 5.
w = 1 when a randomly chosen letter from the English alphabet is a vowel and w = 2 otherwise.

Question 6.
n = the number of digits in a random integer from O through 999.
There are 10 outcomes for value 1, 90 outcomes for value 2, and 900 values for value 3.
P(N = 1) = Number of favorable outcomes/Total number of outcomes
= A randomly chosen integers in a one-digit number/Total number of integers
= 10/1000
= 1/100
P(N = 2) = 90/1000 = 9/100
P(N = 3) = 900/1000 = 9/10

In Exercises 7 and 8, use the probability distribution to determine (a) the number that is most likely to be spun on a spinner, and (b) the probability of spinning an even number.

Question 7.

Question 8.

The most likely number to be spun on the spinner is the value of random variable X(Number of spinner) of which P(X) is greatest.
From given histogram we see that this probability is greatest for X = 5.
Hence the most likely number to be spun on the spinner is 5.
P(Spinning an even number) = P(X = 10) + P(X = 20) = 1/6 + 1/12 = 1/4

USING EQUATIONS
In Exercises 9 – 12, calculate the probability of flipping a coin 20 times and getting the given number of heads.
Question 9.
1

Question 10.
4
P(Four success) = 20C4(1/2)4(1/2)20-4
= 20!/4!(20 – 4)!(1/2)20
= 20!/4!(16)!(1/2)20
= 0.0046
We see that the obtained probability is small, which is logical.

Question 11.
18

Question 12.
20
P(Four success) = 20C20(1/2)20(1/2)20-20
= 20!/20!(20 – 20)!(1/2)20
=(1/2)20
= 0.00000095

Question 13.
MODELING WITH MATHEMATICS
According to a survey, 27% of high school students in the United States buy a class ring. You ask 6 randomly chosen high school students whether they own a class ring.

a. Draw a histogram of the binomial distribution for your survey.
b. What is the most likely outcome of your survey?
c. What is the probability that at most 2 people have a class ring?

Question 14.
MODELING WITH MATHEMATICS
According to a survey, 48% of adults in the United States believe that Unidentified Flying Objects (UFOs) are observing our planet. You ask 8 randomly chosen adults whether they believe UFOs are watching Earth.
a. Draw a histogram of the binomial distribution for your survey.
p = P(the American is a sports fan) = 48% = 0.48
1 – p = P(the American is not a sports fan) = 1 – 0.48= 0.52
P (0 success) = 0C8 p0(1 – p)8-0
= 8!/0!(8 – 0)! 1 . 0.528
= 0.528
= 0.1513
P (One person believe that UFOs are watching Earth) =8C1 p¹(1 – p)8-1
=0.03948
P (Two person believe that UFOs are watching Earth) =8C2 p2(1 – p)8-2
=0.1275
P (Three person believe that UFOs are watching Earth) =8C3 p3(1 – p)8-3
=0.2355
P (Four person believe that UFOs are watching Earth) =8C4 p4(1 – p)8-4
=0.2717
P (Five person believe that UFOs are watching Earth) =8C5 p5(1 – p)8-5
=0.2006
P (Six person believe that UFOs are watching Earth) =8C6 p6(1 – p)8-6
=0.0926
P (Seven person believe that UFOs are watching Earth) =8C7 p7(1 – p)8-7
=0.0244
P (Eight person believe that UFOs are watching Earth) =8C8 p8(1 – p)8-8
=0.0028

b. What is the most likely outcome of your survey?
P (Four person believe that UFOs are watching Earth) = 0.2717
This probability has the highest, so we conclude that the most likely outcome is that four of the eight adults believe that UFOs are watching Earth.

c. What is the probability that at most 3 people believe UFOs are watching Earth?
P (At most 3 persons believe that UFOs are watching Earth) = P (One person believe that UFOs are watching Earth) + P (Two person believe that UFOs are watching Earth) + P (Three person believe that UFOs are watching Earth)
= 0.0053 + 0.0395 + 0.1275 + 0.2355
= 0.4078

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in calculating the probability of rolling a 1 exactly 3 times in 5 rolls of a six-sided die.

Question 15.

Question 16.

Question 17.
MATHEMATICAL CONNECTIONS
At most 7 gopher holes appear each week on the farm shown. Let x represent how many of the gopher holes appear in the carrot patch. Assume that a gopher hole has an equal chance of appearing at any point on the farm.

a. Find P(x) for x = 0, 1, 2 ….., 7.

p = P(the gopher holes appear in the carrot patch)
= Area marked for carrot/Area of the whole farm
= Area of a square + Area of a triangle/Area of the whole farm
= 0.28125
1 – p = P(The gopher holes do not appear in the carrot patch)
= 1 – 0.28125
= 0.71875
P (0 success) = 0C7 p0(1 – p)7-0
= 7!/0!(7 – 0)! 1 . 0.727
= 0.727
= 0.099
P (There is one gopher hole in the carrot patch) =7C1 p¹(1 – p)7-1
=0.27143
P (There is two gopher hole in the carrot patch) =7C2 p¹(1 – p)7-2
=0.31863
P (There is three gopher hole in the carrot patch) =7C3 p¹(1 – p)7-3
=0.20781
P (There is four gopher hole in the carrot patch) =7C4 p¹(1 – p)7-4
=0.08131
P (There is five gopher hole in the carrot patch) =7C5 p¹(1 – p)7-5
=0.01909
P (There is six gopher hole in the carrot patch) =7C6 p¹(1 – p)7-6
=0.00249
P (There is seven gopher hole in the carrot patch) =7C7 p¹(1 – p)7-7
=0.00012

b. Make a table showing the probability distribution for x.
c. Make a histogram showing the probability distribution for x.

Question 18.
HOW DO YOU SEE IT?
Complete the probability distribution for the random variable x. What is the probability the value of x is greater than 2?

P(X = 1) + P(X = 2) + … + P(X = n) = 1
P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1 + 0.3 + 0.4 + P(X = 4)
= 1
P(X = 4) = 1 – 0.8 = 0.2
Now lets find a probability that value of X is greater then two, as
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)
= 0.3 + 0.4 + 0.2
= 0.9
It is very likely that the random variable k will take a value greater than 2.

Question 19.
MAKING AN ARGUMENT
The binomial distribution Shows the results of a binomial experiment. Your friend claims that the probability p of a success must be greater than the probability 1 – p of a failure. Is your friend correct? Explain your reasoning.

Question 20.
THOUGHT PROVOKING
There are 100 coins in a bag. Only one of them has a date of 2010. You choose a coin at random, check the date, and then put the coin back in the bag. You repeat this 100 times. Are you certain of choosing the 2010 coin at least once? Explain your reasoning.
Given,
There are 100 coins in a bag. Only one of them has a date of 2010. You choose a coin at random, check the date, and then put the coin back in the bag. You repeat this 100 times
p = P(Selected coin has a date of 2010)
= Number of favorable outcomes/Total number of outcomes
= 1/100
1 – p = 1 – 1/100  = 99/100
P (0 success) = 100C0 p0(1 – p)100-0
= 100!/0!(100 – 0)! 1 . (99/100)100
= (99/100)100
= 0.366
P(1 or more success) = 1 – P(0 success) = 1 – 0.366 = 0.634
Hence with a probability of 0.634, we will choose a coin which has a date of 2010.
We are not certain of choosing the 2010 coin at least once.

Question 21.
MODELING WITH MATHEMATICS
Assume that having a male and having a female child are independent events, and that the probability of each is 0.5.
a. A couple has 4 male children. Evaluate the validity of this statement: “The first 4 kids were all boys, so the next one will probably be a girl.”
b. What is the probability of having 4 male children and then a female child?
c. Let x be a random variable that represents the number of children a couple already has when they have their first female child. Draw a histogram of the distribution of P(x) for 0 ≤ x ≤ 10. Describe the shape of the histogram.

Question 22.
CRITICAL THINKING
An entertainment system has n speakers. Each speaker will function properly with probability p. independent of whether the other speakers are functioning. The system will operate effectively when at least 50% of its speakers are functioning. For what values of p is a 5-speaker system more likely to operate than a 3-speaker system?
Given,
An entertainment system has n speakers. Each speaker will function properly with probability p. independent of whether the other speakers are functioning.
The system will operate effectively when at least 50% of its speakers are functioning.
p = P(Speaker will function properly)
1 – 9 = P(Speaker will not function properly)
P(5-speaker system operate) = P(X = 3) + P(X = 4) + P(X = 5)
P(5-Speaker system operate) = P (0 success) = 5C3 p3(1 – p)5-3 + 5C4 p4(1 – p)5-4 + 5C5 p5(1 – p)5-5 = 10p3(1 – p)5-4 + 5p4(1 – p)5
P = P(X = 2) + P(X = 3)
= 5C2 p2(1 – p)5-2 + 5C3 p3(1 – p)5-3
= 10p2(1 – p)5-3 + 10p3(1 – p)2

10p3(1 – p)5-4 + 5p4(1 – p)5 + p5 > 10p2(1 – p)3 + 10p3(1 – p)2
5p2 – 4p3 – 10 + 30p – 30p2 + 10p3 > 0
A 5-speaker system operate more likely to operate than 3-speaker system when p ∈ (0.558, 1]

Maintaining Mathematical Proficiency

List the possible outcomes for the situation.

Question 23.
guessing the gender of three children

Question 24.
picking one of two doors and one of three curtains
If we denote by D1 and D2 first and second door and with C1, C2, C3 first, second and third curtain, then the possible outcomes are
D1C1, D1C2, D1C3, D2C1, D2C2, D2C3
Thus there are six possible outcomes.

### 12.1 Sample Spaces and Probability

Question 1.
A bag contains 9 tiles. one for each letter in the word HAPPINESS. You choose a tile at random. What is the probability that you choose a tile with the letter S? What is the probability that you choose a tile with a letter other than P?

Let X be a random variable that represent the letter on tile.
We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S”
So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9.

Question 2.
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points?

Answer: It is the most likely to get 20 points.

Explanation:
From given board, we can conclude that the probability that we get 5 points is
P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9
On the other hand, we see that the probability we get 10 points is
P(10 points) = Surface of yellow area/Surface of board = (16 – 4)/36 = 1/3
and the probability that we get 20 points is
P(20 points) = Surface of blue area/Surface of board = (36 – 16)/36 = 5/9
From the obtained results we get that it is the most likely to get 20 points, which is logical because the surface of the blue area is the largest.

### 12.2 Independent and Dependent Events

Find the probability of randomly selecting the given marbles from a bag of 5 red, 8 green, and 3 blue marbles when (a) you replace the first marble before drawing the second, and (b) you do not replace the first marble. Compare the probabilities.

Question 3.
red, then green
We selected two marbles from a bag of 5 red, 8 green and 3 blue. With R denote the event that the red marble is drawn, with P blue, and with G green. The draws are independent because we replace the first marble before drawing the second. Therefore, the probability that we selected first red marble and then green is
P(RG) = P(R)P(G) = 5/16 . 8/16 = k5/36 = 0.15625
b. In this case, we do not replace the first marble before drawing the second. So, the draws are not independent.
P(RG) = P(R)P(G|R) = 5/16 . 8/15 = 1/6 = 0.16667
It is more likely that we selected first red marble and then green when we not replace the first marble before drawing the second.

Question 4.
blue, then red
We selected two marbles from a bag of 5 red, 8 green and 3 blue. With R denote the event that the red marble is drawn, with P blue, and with G green. The draws are independent because we replace the first marble before drawing the second. Therefore, the probability that we selected first red marble and then green is
P(BR) = P(B)P(R) = 3/16 . 5/16 = 15/256 = 0.05859
b. In this case, we do not replace the first marble before drawing the second. So, the draws are not independent.
P(BR) = P(B)P(R|B) = 3/16 . 5/15 = 1/16 = 0.0625
It is more likely that we selected first blue marble and then red when we not replace the first marble before drawing the second.

Question 5.
green, then green
We selected two marbles from a bag of 5 red, 8 green and 3 blue. With R denote the event that the red marble is drawn, with P blue, and with G green. The draws are independent because we replace the first marble before drawing the second. Therefore, the probability that we selected first red marble and then green is
P(GG) = P(G)P(G) = 8/16 . 8/16 = 1/4 = 0.25
b. In this case, we do not replace the first marble before drawing the second. So, the draws are not independent.
P(GG) = P(G)P(G|G) = 8/16 . 7/15 = 0.23333
It is more likely that we selected first blue marble and then red when we not replace the first marble before drawing the second.

### 12.3 Two-Way Tables and Probability

Question 6.
What is the probability that a randomly selected resident who does not support the project in the example above is from the west side?
P(West side|Does not support the project) = P(West side and Does not support the project)/P(Does not support the project)
= 0.09/(0.08 + 0.09)
= 0.529
The probability that random selected resident who does not support the project is from the west side is about 0.529

Question 7.
After a conference, 220 men and 270 women respond to a survey. Of those, 200 men and 230 women say the conference was impactful. Organize these results in a two-way table. Then find and interpret the marginal frequencies.
After a conference, 220 men and 270 women respond to a survey. Of those, 200 men and 230 women say the conference was impactful.
Number of men who say the conference had not impact = 220 – 200 = 20
By the same method we come to the conclusion
Number of women who say the conference had not impact = 270 – 230 = 40
Now, we will find the marginal frequencies.
Total number of people who say the conference was impact is 200 + 230 = 430
Total Number of people who say the conference had not impact = 20 + 40 = 60
Also from given information we know that total number of people who was surveyed is 220 + 270 = 490

### 12.4 Probability of Disjoint and Overlapping Events

Question 8.
Let A and B be events such that P(A) = 0.32, P(B) = 0.48, and P(A and B) = 0.12. Find P(A or B).
P(A or B) = P(A) + P(B) – P(A and B)
Given,
P(A) = 0.32, P(B) = 0.48, and P(A and B) = 0.12
P(A or B) = 0.32 + 0.48 – 0.12 = 0.68

Question 9.
Out of 100 employees at a company, 92 employees either work part time or work 5 days each week. There are 14 employees who work part time and 80 employees who work 5 days each week. What is the probability that a randomly selected employee works both part time and 5 days each week?
A = {Employees either work part time},
B = {Employees either work 5 days}
Based on the given information we see that
P(A) = Number of favorable outcomes/Total Number of outcomes
= Number of Employees either work part time/Total number of employees
= 14/100
= 0.14
P(B) = Number of favorable outcomes/Total Number of outcomes
= Number of Employees either work 5 days/Total number of employees
= 80/100
= 0.8
Also, we know that 92 employees either work part time or 5 days each week
P(A or B) = Number of favorable outcomes/Total Number of outcomes
= 92/100 = 0.92
For given events A and B the probability of A or B is
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 0.14 + 0.8 – 0.92 = 0.02
Hence the probability that a randomly selected employee works part time and 5 days each week is 0.02

### 12.5 Permutations and Combinations

Evaluate the expression.

Question 10.
7P6
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
=  7!/(7 – 6)!
= 7 . 6 . 5 . 4 . 3 . 2 . 1
= 5040
7P6 = 7! = 5040

Question 11.
13P10
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
=  13!/(13 – 10)!
= 13!/3!
= 13. 12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 = 94348800
13P10 = 94348800

Question 12.
6C2
nCa = n!/a!(n – a)!
6C2 = 6!/2!(6 – 2)!
= 6!/2!(4)!
= 6 . 5 . 4 . 3 . 2 . 1/(2 . 1)(4 . 3 . 2 . 1)
15
6C2 = 15
nC2 = n(n – 1)/2
6C2 = 15

Question 13.
8C4
nCa = n!/a!(n – a)!
8C4 = 8!/4!(8 – 4)!
= 68!/4!(4)!
= 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(4 . 3 . 2 . 1)(4 . 3 . 2 . 1)
= 2 . 7 . 5
6C2 = 70

Question 14.
Eight sprinters are competing in a race. How many different ways can they finish the race? (Assume there are no ties.)
nPn = n!
We know thata 8 sprinters are participating in a race. It is important to us in what order each of them will reach the goal. This tells us that it is necessary to calculate the number of permutations of 8 sprinters.
8P8 = 8! = 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1 = 40320
Eight sprinters can finish a race in 40320 ways.

Question 15.
A random drawing will determine which 3 people in a group of 9 will win concert tickets. What is the probability that you and your 2 friends will win the tickets?
nCa = n!/a!(n – a)!
9C3 = 9!/3!(9 – 3)!
= 9!/3!(6)!
= 9 . 8 . 7 6 . 5 . 4 . 3 . 2 . 1/(3 . 2 . 1)(6 . 5 .  4 . 3 . 2 . 1)
= 84
9C3 = 84
The probability that you and your 2 friends will win the tickets is equal to the probability that out of 84 possibilities, the trio in which you and your friend are in will be chosen.
P = Number of favorable outcomes/ Total number of outcomes = 1/84

### 12.6 Binomial Distributions

Question 16.
Find the Probability of flipping a coin 12 times and getting exactly 4 heads.
The probabilities are P(H) = P(T) = 1/2, that is p = p – 1 = 1/2 and are the same for each trial.
n = 12
P = 12!/4!(12 – 4)! . (1/2)12

= 0.1208
Therefore the probability of flipping a coin 12 times and getting exactly 4 heads is about 0.12

Question 17.
A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. Draw a histogram of the binomial distribution of the number of successful free throws. What is the most likely outcome?
Given,
A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws.
p = P(Successful free throw) = 82.6% = 0.826
1 – p = P(Unsuccessful free throw) = 1 – 0.826 = 0.174
P (Out of 5 free throws one was successful) = 5C1 p¹(1 – p)5-1
P (Out of 5 free throws two was successful) =5C2 p²(1 – p)5-2
P(Out of 5 free throws three was successful) = 5C3p³(1 – p)5-3
P (Out of 5 free throws four was successful) =5C4 p4(1 – p)5-4
P (All 5 throws one was successful) = 5C5(1 – p)5-5

### Probability Test

You roll a six-sided die. Find the probability of the event described. Explain your reasoning.

Question 1.
You roll a number less than 5.
The die has 6 sides, thus the total number of possible outcomes is 6.
The favorable outcomes are
P(n<5) = Number of favorable outcomes/Total number of outcomes
Thus there are 4 favorable outcomes.
The probability to roll a number less than 5 is
= 4/6
= 2/3

Question 2.
You roll a multiple of 3.
The die has 6 sides, thus the total number of possible outcomes is 6.
The favorable outcomes are
P(n = 3k) = Number of favorable outcomes/Total number of outcomes
Thus there are 2 favorable outcomes.
The probability to roll a number less than 3 is
= 2/6
= 1/3

Evaluate the expression.

Question 3.
7P2
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
7P2 = 7!/(7 – 2)!
=  7!/5!
= (7 . 6 . 5 . 4 . 3 . 2 . 1)/(5 . 4 . 3 .  2 . 1)
= 7 . 6
= 42
7P2 = 42

Question 4.
8P3
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
8P3 = 8!/(8 – 3)!
=  8!/5!
= (8 . 7 . 6 . 5 . 4 . 3 . 2 . 1)/(5 . 4 . 3 .  2 . 1)
= 8 . 7 . 6
= 336
8P3 = 336

Question 5.
6C3
nCa = n!/a!(n – a)!
6C3 = 6!/3!(6 – 3)!
= 6!/3!(3)!
= 6 . 5 . 4 . 3 . 2 . 1/(3 . 2 . 1)(3 . 2 . 1)
= 2 . 5 . 2
= 20
6C3 = 20

Question 6.
12C7
nCa = n!/a!(n – a)!
12C7 = 12!/7!(12 – 7)!
= 12!/7!(5)!
= 12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(7 . 6 . 5 . 4 . 3 . 2 . 1)(5 . 4 . 3 . 2 . 1)
= 11 . 2 . 9 . 4 = 792
12C7 = 792

Question 7.
In the word PYRAMID, how many ways can you arrange
(a) all of the letters and
We have to find the number of permutation all of the letters in a given word that will consist of 7 letters. We see that it matters that order of letters are important.
Number of permutations = (In the 1st place can be one of 7 letters) × (In the 2nd place can be one of 6 letters that is left)
(In the 3rd can be one of 5 letters that is left) × (In the 4th can be 4 letters that is left)
(In the 5th can be three letter that is left) × (In the 6th can be two letters that is left)
(In the 7th can be one letter that is left)
= 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5040
Therefore, we have 5040 ways for arrange all of the letters in given word, that is PYRAMID, PYRAMDI, PYRADMI, ……., DIMARYP.

(b) 5 of the letters?
We have to find the number of permutation all of the letters in a given word that will consist of 3 letters. We see that it matters that order of letters are important.
Number of permutations = (In the 1st place can be one of 7 letters) × (In the 2nd place can be one of 6 letters that is left)
(In the 3rd can be one of 5 letters that is left) × (In the 4th can be 4 letters that is left)
(In the 5th can be three letter that is left)
= 7 . 6 . 5 . 4 . 3
= 2520
Therefore, we have 2520 ways for arrange 5 of the letters in given word, that is PYRAM, PYRAI, PYRAD….

Question 8.
You find the probability P(A or B) by using hie equation P(A or B) = P(A) + P(B) – P(A and B). Describe why it is necessary to subtract P(A and B) when the events A and B are overlapping. Then describe why it is not necessary to subtract P(A and B) when the events A and B are disjoint.
When the events are overlapping then P(A and B) ≠ 0. This means that the expression P(A) + P(B) already contains the probability P(A and B) and so P(A and B) is subtracted from their sum to evaluate P(A or B)
When the events are overlapping then P(A and B) = 0 and so the equation of P(A or B reduces to P(A) + P(B))

Question 9.
Is it possible to use the formula P(A and B) = P(A) • P(B/A) when events A and B are independent? Explain your reasoning.
If events A and B are independent events, then
P(A and B) = P(A) . P(B)
Also, because event B is independent of event A then P(B|A) = P(A)
P(A and B) = P(A) . P(B) = P(A) . P(B|A)
where A and B are the independent events.

Question 10.
According to a survey, about 58% of families sit down tor a family dinner at least four times per week. You ask 5 randomly chosen families whether the have a family dinner at least four times per week.
a. Draw a histogram of the binomial distribution for the survey.
p = P(Family have dinner four times per week) = 58% = 0.58
1 – p = P(Family have no dinner dour times per week) = 1 – 0.58 = 0.42
P (One Family have dinner four times per week) = 5C1 p¹(1 – p)5-1
= 0.09024
P (Two Family have dinner four times per week) =5C2 p²(1 – p)5-2
= 0.24923
P(Three Family have dinner four times per week) = 5C3p³(1 – p)5-3
= 0.34418
P (Four Family have dinner four times per week) =5C4 p4(1 – p)5-4
= 0.23765
P (Family have dinner four times per week) = 5C5(1 – p)5-5 = 0.06563

b. What is the most likely outcome of the survey?
P(Three families have dinner four times per week) = 0.34418
This probability is the highest, so we can conclude that the most likely outcome is that three of the five families have dinner four times per week.

c. What is the probability that at least 3 families have a family dinner four times per week?
In this part we have to find the probability that,
P(At least 3 families have dinner four times per week)
= P(Three families have dinner four times per week) + P(Four families have dinner four times per week) + P(Five families have dinner four times per week)
= 0.34418 + 0.23765 + 0.06563
= 0.64746

Question 11.
You are choosing a cell phone company to sign with for the next 2 years. The three plans you consider are equally priced. You ask several of your neighbors whether they are satisfied with their current cell phone company. The table shows the results. According to this survey, which company should you choose?

To find the joint relative frequencies we divide each frequency by the total number of people in the survey. Also the marginal relative frequencies we find as the sum of each row and each column.
So, we can present a two way table that shows the joint and marginal relative frequencies.

Finally, to get conditional relative frequencies we use the previous the marginal relative frequency of each row.

Therefore we should choose company A.

Question 12.
The surface area of Earth is about 196.9 million square miles. The land area is about 57.5 million square miles, and the rest is water. What is the probability that a meteorite that reaches the surface of Earth will hit land? What is the probability that it will hit water?
From the formula for geometric probability we get the probability that a meteorite that reaches the surface of Earth will hit lend is
P = The lend area/ The surface area of Earth = 57.5/196.9 = 0.292
Alos we know that the probability for complement of event A is
P($$\bar{A}$$) = 1 – P(A)
In this case, complement of event {A meteorite will hit lend} is {A meteorite will hit water}.
P = 1 – 0.292 = 0.708

Question 13.
Consider a bag that contains all the chess pieces in a set, as shown in the diagram.

a. You choose one piece at random. Find the probability that you choose a black piece or a queen.
The total number of pieces is 2 + 2 + 4 + 4 + 4 + 16 = 32.
Also we see that the number of black and white pieces are the same, 16 and there are one black queen.
A = {We choose a black piece}
B = {We choose a queen}
P(A) = Number of favorable outcomes/Total number of outcomes = Number of black pieces/Total number of pieces = 16/32 = 1/2
P(B) = Number of queens/Total number of pieces = 2/32 = 1/16
P(A and B) = Number of black queens/Total number of pieces = 1/32
P(A or B) = P(A) + P(B) – P(A and B)
= 1/2 + 1/16 – 1/32 = 17/32

b. You choose one piece at random, do not replace it, then choose a second piece at random. Find the probability that you choose a king, then a pawn.
C = {We choose a king} and D = {We choose a pawn}
P(C) = Number of pawns/Total number of pieces = 16/32 = 1/2
P(C and D) = Number of pawns and king/Total number of pieces = (16 + 2)/32 = 9/16
We know that for two dependent events A and B probability that both occur is
P(A and B) = P(A)P(B|A)
P(D|C) = P(C and D)/P(C) = 1/2 × 16/9 = 8/9

Question 14.
Three volunteers are chosen at random from a group of 12 to help at a summer camp.
a. What is the probability that you, your brother, and your friend are chosen?
nCa = n!/a!(n – a)!
12C3 = 12!/3!(12 – 3)!
= 12!/3!(9)!
= 12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(3 . 2 . 1)(9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1)
= 12 . 11 . 10/3 . 2 . 1
= 2 . 11 . 10
12C3 = 220
P = Number of favorable outcomes/Total number of outcomes = 1/220

b. The first person chosen will be a counselor, the second will be a lifeguard, and the third will be a cook. What is the probability that you are the cook, your brother is the lifeguard, and your friend is the counselor?
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
12P3 = 12!/(12 – 3)!
=  12!/9!
= (12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1)/(9 . 8 . 7 . 6 . 5 . 4 . 3 .  2 . 1)
= 12 . 11 . 10
= 1320
12P3 = 1320
P = 1/1320

### Probability Cumulative Assessment

Question 1.
According to a survey, 63% of Americans consider themselves sports fans. You randomly select 14 Americans to survey.
a. Draw a histogram of the binomial distribution of your survey.
p = P(the American is a sports fan) = 63% = 0.63
1 – p = P(the American is not a sports fan) = 1 – 0.63= 0.37
P (0 success) = 12C0 p0(1 – p)12-0
= 12!/0!(12 – 0)! 1 . 0.3712
= 0.3712
P (One American is a sports fans) =12C1 p¹(1 – p)12-1
=0.000134506
P (Two Americans are sports fans) =12C2 p2(1 – p)12-2
=0.001259628
P (Three Americans are sports fans) =12C3 p3(1 – p)12-3
=0.007149239
P (Four American is a sports fans) =12C4 p4(1 – p)12-4
=0.027389316
P (Five American is a sports fans) =12C5 p5(1 – p)12-5
=0.07461738
P (Six American is a sports fans) =12C6 p6(1 – p)12-6
=0.148226418
P (Seven American is a sports fans) =12C7 p7(1 – p)12-7
=0.216330447
P (Eight American is a sports fans) =12C8 p8(1 – p)12-8
=0.230216523
P (Nine American is a sports fans) =12C9 p9(1 – p)12-9
=0.174217909
P (Ten American is a sports fans) =12C10p10(1 – p)12-10
=0.088992392
P (Eleven American is a sports fans) =12C11 p11(1 – p)12-11
=0.023755047
P (Twelve American is a sports fans) =12C1 p12(1 – p)12-12
=0.003909188

b. What is the most likely number of Americans who consider themselves Sports fans?
From histogram in part a and based on the calculations in the first part
We know that,
P(Eight Americans are sports fans) = 0.23
This probability is highest, so we can conclude that the most likley outcome is that eight of the 12 selected Americans consider themselves sports fans.

c. What is the probability at least 7 Americans consider themselves sports fans?
In this part we have to find dthe probability that
P(At least 7 Americans consider themselves sports fans)
= P(Seven American is a sports fans) + P(Eight American is a sports fans) + P(Nine American is a sports fans) + P(Ten American is a sports fans) + P(Eleven American is a sports fans) + P(Twelve American is a sports fans)
= 0.2163 + 0.2302 + 0.1742 + 0.08899 + 0.0275 + 0.0039
= 0.7412
Therefore, we can conclude that the probability that at least 7 Americans consider themselves sports fans is 0.7412

Question 2.
What is the arc length of $$\widehat{A B}$$ ?

(A) 3.5 π cm
(B) 7 π cm
(C) 21 π cm
(D) 42 π cm
Arc length of $$\widehat{A B}$$ = 105/360 × 2π × 12 = 7π cm

Question 3.
you order a fruit smoothie made with 2 liquid ingredients and 3 fruit ingredients from the menu shown. How many different fruit smoothies can you order?

Question 4.
The point (4, 3) is on a circle with center (- 2, – 5), What is the standard equation of the circle?
Given,
The point (4, 3) is on a circle with center (- 2, – 5)
(x – h)² + (y – k)² = r²
Substitute the values in the given equation
(4 – (-2))² + (3 – (-5))² = r²
100 = r²
r = √100 = 10
General equation of circle
(x – h)² + (y – k)² = r²
Substitute the values in the given equation
(x – (-2))² + (y – (-5))² = 10²
(x + 2)² + (y + 5)² = 100

Question 5.
Find the length of each line segment with the given endpoints. Then order the line segments from shortest to longest.
a. A(1, – 5), B(4, 0)
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dAB = √(4 – 1)² + (0 – (-5))²
= √34
= 5.83

b. C(- 4, 2), D(1, 4)
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dCD = √(1 – (-4))² + (4 – 2)²
= √29
= 5.39

c. E(- 1, 1), F(- 2, 7)
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dEF = √(-2 – (-1))² + (7 – 1)²
= √37
= 6.083

d. G(- 1.5, 0), H(4.5, 0)
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dGH = √(4.5 – (-1.5))² + (0 – 0)²
= √36
= 6

e. J(- 7, – 8), K(- 3, – 5)
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dJK = √(-3 – (-7))² + (-5 – (-8))²
= √25
= 5

f. L(10, – 2), M(9, 6)
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dLM = √(9 – 10)² + (6 – (-2))²
= √65
= 8.06
Therefore, the line segments in ascending order are JK < CD < AB < GH < EF < LM

Question 6.
Use the diagram to explain why the equation is true.
P(A) + P(B) = P(A or B) + P(A and B)

Given,
P(A) = 8/12
P(B) = 7/12
P(A or B) = 12/12
P(A and B) = 3/12
P(A) + P(B) = P(A or B) + P(A and B)
8/12 + 7/12 = 12/12 + 3/12
5/4 = 5/4

Question 7.
A plane intersects a cylinder. Which of the following cross sections cannot be formed by this intersection?
(A) line
(B) triangle
(C) rectangle
(D) circle
If the plane is tangent to the curved surface of the cylinder, then the intersection is a line.
If the plane is cuts the cylinder parallel to its circular base, then the intersection is a circle.
If the plane cuts the cylinder perpendicular to its circular base and parallel to its curved surface, then the intersection is a rectangle.
The correct answer is option B.

Question 8.
A survey asked male and female students about whether the prefer to take gym class or choir. The table shows the results of the survey,

a. Complete the two-way table.
From the given table we first find the joint frequencies.
Number of people who prefer gym class = 106 – 49 = 57
On the other hand, we know that 57 people prefer gym class, of which 23 was female.
Number of male who prefer the gym class = 57 – 23 = 34
Number of male who prefer the choir class = 50 – 34 = 16
Number of female who prefer the choir class = 49 – 16 = 33
Now, we will find the remaining marginal frequencies. Hence we know that 23 female prefer gym class and 33 female prefer choir class.
Total number of female is 23 + 33 = 56

b. What is the probability that a randomly selected student is female and prefers choir?
P = Number of female who prefer choir class/Total number of students
= 33/106
= 0.31

c. What is the probability that a randomly selected male student prefers gym class?
P(Prefer gym class|Male) = P(Male and prefer gym class)/P(Male)
= Number of male who prefer gym class/Number of male students
= 34/50
= 0.68

Question 9.
The owner of a lawn-mowing business has three mowers. As long as one of the mowers is working. the owner can stay productive. One of the mowers is unusable 10% of the time, one is unusable 8% of the time, and one is unusable 18% of the time.
a. Find the probability that all three mowers are unusable on a given day.
P(A) = 10% = 0.1
P(B) = 8% = 0.08
P(C) = 18% = 0.18
$$\bar{A}$$ = {The first mower is usable}
$$\bar{B}$$ = {The second mower is usable}
$$\bar{C}$$ = {The third mower is usable}
P($$\bar{A}$$) = 1 – 0.1 = 0.9
P($$\bar{B}$$) = 1 – 0.08 = 0.92
P($$\bar{C}$$) = 1 – 0.18 = 0.82
Event that all three mowers are unusable on a given day is A and B and C. If we assume that the operation of the mowers are independent, then we get the probability that all three mowers are unusable on a given day as
P(A and B and C) = P(A) . P(B) . P(C) = 0.1 × 0.08 × 0.18 = 0.00144

b. Find the probability that at least one of the mowers is unusable on a given day.
Event at least one of mowers is unusable on a given day is equivalent to event one of mowers is unusable on a given day or two of mowers is unusable on a given day.
A and $$\bar{B}$$ and $$\bar{C}$$ or $$\bar{A}$$ and B and $$\bar{C}$$ or $$\bar{A}$$ and $$\bar{B}$$ and C
On the other hand, event two of mowers is unusable on a given day is
A and B and $$\bar{C}$$ or $$\bar{A}$$ and B and C or A and $$\bar{B}$$ and C
P(A and $$\bar{B}$$ and $$\bar{C}$$) + P($$\bar{A}$$ and B and $$\bar{C}$$) + P($$\bar{A}$$ and $$\bar{B}$$ and C)
= 0.1 × 0.92 × 0.82 + 0.9 × 0.08 × 0.82 + 0.9 × 0.92 × 0.18
= 0.28352
P(A and $$\bar{B}$$ and $$\bar{C}$$) + P($$\bar{A}$$ and B and $$\bar{C}$$) + P($$\bar{A}$$ and $$\bar{B}$$ and C)
= 0.1 × 0.08 × 0.82 + 0.9 × 0.08 × 0.18  + 0.1 × 0.92 × 0.18
= 0.03608
P(At least one of mowers is unusable) = P(One of mowers is unusable) + P(two of mowers is unusable)
= 0.28352 + 0.03608
= 0.3196

c. Suppose the least-reliable mower stops working completely. How does this affect the probability that the lawn-moving business can be productive on a given day?
If the least-reliable mower stops working completely, this will not affect the probability that the lawn mowing business can be productive on a given day, because we know that as long as one of the mowers is working, the owner can stay productive.

Question 10.
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. What is the probability your dart lands in the yellow region?

(A) $$\frac{\pi}{36}$$
(B) $$\frac{\pi}{12}$$
(C) $$\frac{\pi}{9}$$
(D) $$\frac{\pi}{4}$$
The total area of the given figure is (6(2))² = 144 sq. units
Area of the red circle = π × 2² = 4π
The area of the yellow ring is π × (4² – 2²) = 12π
The area of the blue ring is π × (6² – 4²) = 20π
Therefore, the probability of hitting the blue region is 12π/144 = π/12
Thus the correct answer is option B.

Conclusion:

Hope you are all satisfied with the given solutions. You can get free access to Download Big Ideas Math Geometry Answers Chapter 12 Probability pdf from here. Bookmark our Big Ideas Math Answers to get detailed solutions for all Geometry Chapters.

## Big Ideas Math Book Geometry Answer Key Chapter 9 Right Triangles and Trigonometry

### Right Triangles and Trigonometry Maintaining Mathematical Proficiency

Simplify the expression.

Question 1.
√75

square root of 75 = 5,625.

Explanation:
In the above-given question,
given that,
√75.
square root of 75 = 75 x 75.
75 x 75 = 5,625.
√75 = 5,625.

Question 2.
√270

square root of 270 = 72,900.

Explanation:
In the above-given question,
given that,
√270.
square root of 270 = 270 x 270.
270 x 270 = 72900.
√270 = 72900.

Question 3.
√135

square root of 135 = 18225.

Explanation:
In the above-given question,
given that,
√135.
square root of 135 = 135 x 135.
135 x 135 = 18225.
√135 = 18,225.

Question 4.
$$\frac{2}{\sqrt{7}}$$

2/49 = 0.04.

Explanation:
In the above-given question,
given that,
square root of 7 = 7 x 7.
7 x 7 = 49.
$$\frac{2}{\sqrt{7}}$$.
2/49 = 0.04.

Question 5.
$$\frac{5}{\sqrt{2}}$$

5/4 = 1.25.

Explanation:
In the above-given question,
given that,
square root of 2 = 2 x 2.
2 x 2 = 4.
$$\frac{5}{\sqrt{2}}$$.
5/4 = 1.25.

Question 6.
$$\frac{12}{\sqrt{6}}$$

12/36 = 0.33.

Explanation:
In the above-given question,
given that,
square root of 6 = 6 x 6.
6 x 6 = 36.
$$\frac{12}{\sqrt{6}}$$.
12/36 = 0.33.

Solve the proportion.

Question 7.
$$\frac{x}{12}=\frac{3}{4}$$

x = 9.

Explanation:
In the above-given question,
given that,
$$\frac{x}{12}=\frac{3}{4}$$
x/12 = 3/4.
4x = 12 x 3.
4x = 36.
x = 36/4.
x = 9.

Question 8.
$$\frac{x}{3}=\frac{5}{2}$$

x = 7.5.

Explanation:
In the above-given question,
given that,
$$\frac{x}{3}=\frac{5}{2}$$
x/3 = 5/2.
2x = 5 x 3.
2x = 15.
x = 15/2.
x = 7.5.

Question 9.
$$\frac{4}{x}=\frac{7}{56}$$

x = 32.

Explanation:
In the above-given question,
given that,
$$\frac{4}{x}=\frac{7}{56}$$
4/x = 7/56.
7x = 56 x 4.
7x = 224.
x = 32.

Question 10.
$$\frac{10}{23}=\frac{4}{x}$$

x = 9.2.

Explanation:
In the above-given question,
given that,
$$\frac{10}{23}=\frac{4}{x}$$
x/4 = 10/23.
10x = 23 x 4.
10x = 92.
x = 92/10.
x = 9.2.

Question 11.
$$\frac{x+1}{2}=\frac{21}{14}$$

x = 135.

Explanation:
In the above-given question,
given that,
$$\frac{x + 1}{2}=\frac{21}{14}$$
x+12 x 2 = 21×14.
2x + 24= 294 .
2x = 294 – 24.
2x = 270.
x = 270/2.
x = 135.

Question 12.
$$\frac{9}{3 x-15}=\frac{3}{12}$$

x = 6.33.

Explanation:
In the above-given question,
given that,
$$\frac{9}{3 x-15}=\frac{3}{12}$$
27x – 135 = 3×12.
27x = 36 + 135.
27x = 171.
x = 171/27.
x = 6.33.

Question 13.
ABSTRACT REASONING
The Product Property of Square Roots allows you to simplify the square root of a product. Are you able to simplify the square root of a sum? of a diffrence? Explain.

Yes, I am able to simplify the square root of a sum.

Explanation:
In the above-given question,
given that,
The product property of square roots allows you to simplify the square root of a product.
√3 + 1 = √4.
√4 = 4 x 4.
16.
√3 – 1 = √2.
√2 = 2 x 2.
4.

### Right Triangles and Trigonometry Mathematical practices

Monitoring progress

Question 1.
Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° in standard position. What are the exact coordinates of its vertices?

Question 2.
Use dynamic geometry software to construct a right triangle with acute angle measures of 20° and 70° in standard position. What are the approximate coordinates of its vertices?

### 9.1 The Pythagorean Theorem

Exploration 1

Proving the Pythagorean Theorem without Words

Work with a partner.

a. Draw and cut out a right triangle with legs a and b, and hypotenuse c.

Explanation:
In the above-given question,
given that,
proving the Pythagorean theorem without words.
a2 + b2 = c2.

b. Make three copies of your right triangle. Arrange all tour triangles to form a large square, as shown.

a2 + b2 = c2.

Explanation:
In the above-given question,
given that,
make three copies of your right triangle.
a2 + b2 = c2.

c. Find the area of the large square in terms of a, b, and c by summing the areas of the triangles and the small square.

The area of the large square = a2 x b2.

Explanation:
In the above-given question,
given that,
the area of the square = l x b.
where l = length, and b = breadth.
the area of the square = a x b.
area = a2 x b2.

d. Copy the large square. Divide it into two smaller squares and two equally-sized rectangles, as shown.

e. Find the area of the large square in terms of a and b by summing the areas of the rectangles and the smaller squares.

f. Compare your answers to parts (c) and (e). Explain how this proves the Pythagorean Theorem.

a2 + b2 = c2.

Explanation:
In the above-given question,
given that,
The length of the a and b is equal to the hypotenuse.
a2 + b2 = c2.
where a = one side and b = one side.

Exploration 2

Proving the Pythagorean Theorem

Work with a partner:

a. Draw a right triangle with legs a and b, and hypotenuse c, as shown. Draw the altitude from C to $$\overline{A B}$$ Label the lengths, as shown.

a2 + b2 = c2.

Explanation:
In the above-given question,
given that,
a2 + b2 = c2.

b. Explain why ∆ABC, ∆ACD, and ∆CBD are similar.

In a right-angle triangle all the angle are equal.

Explanation:
In the above-given question,
given that,
∆ABC, ∆ACD, and ∆CBD are similar.
In a right-angle triangle all the angles are equal.

REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.

c. Write a two-column proof using the similar triangles in part (b) to prove that a2 + b2 = c2

a2 + b2 = c2

Explanation:
In the above-given question,
given that,
In the pythagorean theorem.
a2 + b2 = c2
the length of the hypotenuse ie equal to the two side lengths.
a2 + b2 = c2

Question 3.
How can you prove the Pythagorean Theorem?

a2 + b2 = c2

Explanation:
In the above-given question,
given that,
In the pythagorean theorem,
the length of the hypotenuse is equal to the length of the other two sides.
hypotenuse = c.
length = a.
a2 + b2 = c2

Question 4.
Use the Internet or sonic other resource to find a way to prove the Pythagorean Theorem that is different from Explorations 1 and 2.

### Lesson 9.1 The Pythagorean Theorem

Monitoring Progress

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 1.

x = √52.

Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
a2 + b2 = c2
6 x 6 + 4 x 4 = c2
36 + 16 = c2
52 = c2.
c = √52.

Question 2.

x = 4.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 5.
x2 + 3 x 3 = 5 x 5.
x2 + 9 = 25.
x2 = 25 – 9.
x2 = 16.
x = 4.

Question 3.
An anemometer is a device used to measure wind speed. The anemometer shown is attached to the top of a pole. Support wires are attached to the pole 5 feet above the ground. Each support wire is 6 feet long. How far from the base of the pole is each wire attached to the ground?

x = √11.

Explanation:
In the above-given question,
given that,
An anemometer is a device used to measure wind speed.
support wires are attached to the pole 5 feet above the ground.
Each support wire is 6 feet long.
d2 + 6  x 6 = 5 x 5.
d2 + 36 = 25.
d2 = 25 – 36.
d2 = 11.
d
= √11.

Tell whether the triangle is a right triangle.

Question 4.

Yes the triangle is a right triangle.

Explanation:
In the above-given question,
given that,
the hypotenuse = 3 √34.
one side = 15.
the other side = 9.
so the triangle is a right triangle.

Question 5.

Yes, the triangle is a actute triangle.

Explanation:
In the above-given question,
given that,
the hypotenuse = 22.
one side = 26.
the other side = 14.
so the triangle is a acute triangle.

Question 6.
Verify that segments with lengths of 3, 4, and 6 form a triangle. Is the triangle acute, right, or obtuse?

Yes the lengths of the triangle form a acute triangle.

Explanation:
In the above-given question,
given that,
the side lenths of 3, 4, and 6 form a triangle.
6 x 6 = 3 x 3 + 4 x 4.
36 = 9 + 16.
36 = 25.
so the length of the triangle forms a acute triangle.

Question 7.
Verify that segments with lengths of 2, 1, 2, 8, and 3.5 form a triangle. Is the triangle acute, right, or obtuse?

### Exercise 9.1 The Pythagorean Theorem

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a Pythagorean triple?

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Find the length of the longest side.

The length of the longest side = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the hypotenuse

The length of the hypotenuse = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the Pythagorean theorem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the longest leg.

The length of the longest leg = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the side opposite the right angle.

The length of the side opposite to the right angle = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 3.

Question 4.

The length of the x = 34.

Explanation:
In the above-given question,
given that,
the side lengths are 30 and 16.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 16 x 16 + 30 x 30.
X x X = 256 + 900.
X x X = 1156.
X = 34.

Question 5.

Question 6.

The length of the x = 7.2.

Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 6 x 6.
X x X = 16 + 36.
X x X = 52.
X = 7.2.

In Exercises 7 – 10, find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 7.

Question 8.

The length of the X = 25.6.

Explanation:
In the above-given question,
given that,
the side lengths are 24 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 24 x 24 + 9 x 9.
X x X = 576 + 81.
X x X = 657.
X = 25.6.

Question 9.

Question 10.

The length of the x = 11.4.

Explanation:
In the above-given question,
given that,
the side lengths are 7 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 7 x 7 + 9 x 9.
X x X = 49 + 81.
X x X = 130.
X = 11.4.

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in using the Pythagorean Theorem (Theorem 9.1).

Question 11.

Question 12.

x = 24.

Explanation:
In the above-given question,
given that,
the side lengths are 26 and 10.
26 x 26 = a x a + 10 x 10.
676 = a x a + 100.
676 – 100 = a x a.
576 = a x a.
a = 24.
x = 24.

Question 13.
MODELING WITH MATHEMATICS
The fire escape forms a right triangle, as shown. Use the Pythagorean Theorem (Theorem 9. 1) to approximate the distance between the two platforms. (See Example 3.)

Question 14.
MODELING WITH MATHEMATICS
The backboard of the basketball hoop forms a right triangle with the supporting rods, as shown. Use the Pythagorean Theorem (Theorem 9.1) to approximate the distance between the rods where the meet the backboard.

x = 9.1.

Explanation:
In the above-given question,
given that,
the side lengths are 13.4 and 9.8.
13.4 x 13.4 = X x X + 9.8 x 9.8.
179.56 = X x X + 96.04.
179.56 – 96.04 = X x X.
83.52 = X x X.
X = 9.1.
In Exercises 15 – 20, tell whether the triangle is a right triangle.

Question 15.

Question 16.

No, the triangle is not a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 23 and 11.4.
the hypotenuse = 21.2.
21.2 x 21.2 = 23 x 23 + 11.4 x 11.4.
449.44 = 529 + 129.96.
449.44 = 658.96.
449 is not equal to 658.96.
so the triangle is not a right triangle.

Question 17.

Question 18.

No, the triangle is not a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 5 and 1.
the hypotenuse = √26.
26 x 26 = 5 x 5 + 1 x 1.
676 = 25 + 1.
676 = 26.
676 is not equal to 26.
so the triangle is not a right triangle.

Question 19.

Question 20.

Yes, the triangle forms a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 80 and 39.
the hypotenuse = 89.
89 x 89 = 80 x 80 + 39 x 39.
7921 = 6400 + 1521.
7921 = 7921.
7921 is equal to 7921.
so the triangle forms a right triangle.

In Exercises 21 – 28, verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 21.
10, 11, and 14

Question 22.
6, 8, and 10

Yes, the triangle is forming a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is  equal to 100.
so the triangle is forming a right triangle.

Question 23.
12, 16, and 20

Question 24.
15, 20, and 36

Yes, the triangle is obtuse triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 15 and 20.
the hypotenuse = 36.
36 x 36 = 20 x 20 + 15 x 15.
1296 = 400 + 225.
1296 > 625.
1296 is greater than 625.
so the triangle is not a obtuse triangle.

Question 25.
5.3, 6.7, and 7.8

Question 26.
4.1, 8.2, and 12.2

No, the triangle is obtuse triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 4.1 and 8.2.
the hypotenuse = 12.2.
12.2 x 12.2 = 4.1 x 4.1 + 8.2 x 8.2`.
148.84 = 16.81 + 67.24.
148.84 > 84.05.
148.84 is greater than 84.05.
so the triangle is obtuse triangle.

Question 27.
24, 30, and 6√43

Question 28.
10, 15 and 5√13

Yes, the triangle is an acute triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 10 and 5√13.
the hypotenuse = 15.
15 x 15 = 10 x 10 + 5√13 x 5√13.
225 = 100 + 34.81.
225 < 134.81.
225 is less than 134.81.
so the triangle is acute triangle.

Question 29.
MODELING WITH MATHEMATICS
In baseball, the lengths of the paths between consecutive bases are 90 feet, and the paths form right angles. The player on first base tries to steal second base. How far does the ball need to travel from home plate to second base to get the player out?

Question 30.
REASONING
You are making a canvas frame for a painting using stretcher bars. The rectangular painting will be 10 inches long and 8 inches wide. Using a ruler, how can you be certain that the corners of the frame are 90°

x = 12.8.

Explanation:
In the above-given question,
given that,
the side lengths are 10 and 8.
the hypotenuse = x.
X x X = 10 x 10 + 8 x 8.
X = 100 + 64.
X = 12.8.
In Exercises 31 – 34, find the area of the isosceles triangle.

Question 31.

Question 32.

The area of the Isosceles triangle = 12 ft.

Explanation:
In the above-given question,
given that,
base = 32 ft.
hypotenuse = 20ft.
a2 + b2 = c2
h x h + 16 x 16 = 20 x 20.
h x h + 256 = 400.
h x h = 400 – 256.
h x h = 144.
h = 12 ft.
so the area of the isosceles triangle = 12 ft.

Question 33.

Question 34.

The area of the Isosceles triangle = 48 m.

Explanation:
In the above-given question,
given that,
base = 28 m.
hypotenuse = 50 m.
a2 + b2 = c2
h x h + 14 x 14 = 50 x 50.
h x h + 196 = 2500.
h x h = 2500 – 196.
h x h = 2304.
h = 48 m.
so the area of the isosceles triangle = 48 m.

Question 35.
ANALYZING RELATIONSHIPS
Justify the Distance Formula using the Pythagorean Theorem (Thin. 9. 1).

Question 36.
HOW DO YOU SEE IT?
How do you know ∠C is a right angle without using the Pythagorean Theorem (Theorem 9.1) ?

Yes, the triangle is forming a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is  equal to 100.
so the triangle is forming a right triangle.

Question 37.
PROBLEM SOLVING
You are making a kite and need to figure out how much binding to buy. You need the binding for the perimeter of the kite. The binding Comes in packages of two yards. How many packages should you buy?

Question 38.
PROVING A THEOREM
Use the Pythagorean Theorem (Theorem 9. 1) to prove the Hypotenuse-Leg (HL) Congruence Theorem (Theorem 5.9).

Question 39.
PROVING A THEOREM
Prove the Converse of the Pythagorean Theorem (Theorem 9.2). (Hint: Draw ∆ABC with side lengths a, b, and c, where c is the length of the longest side. Then draw a right triangle with side lengths a, b, and x, where x is the length of the hypotenuse. Compare lengths c and x.)

Question 40.
THOUGHT PROVOKING
Consider two integers m and n. where m > n. Do the following expressions produce a Pythagorean triple? If yes, prove your answer. If no, give a counterexample.
2mn, m2 – n2, m2 + n2

Question 41.
MAKING AN ARGUMENT
Your friend claims 72 and 75 Cannot be part of a pythagorean triple because 722 + 752 does not equal a positive integer squared. Is your friend correct? Explain your reasoning.

Question 42.
PROVING A THEOREM
Copy and complete the proof of the pythagorean Inequalities Theorem (Theorem 9.3) when c2 < a2 + b2.
Given In ∆ABC, c2 < a2 + b2 where c is the length
of the longest side.
∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle.
Prove ∆ABC is an acute triangle.

 Statements Reasons 1. In ∆ABC, C2 < (12 + h2, where c is the length of the longest side. ∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle. 1. _____________________________ 2. a2 + b2 = x2 2. _____________________________ 3. c2 < r2 3. _____________________________ 4. c < x 4. Take the positive square root of each side. 5. m ∠ R = 90° 5. _____________________________ 6. m ∠ C < m ∠ R 6. Converse of the Hinge Theorem (Theorem 6.13) 7. m ∠ C < 90° 7. _____________________________ 8. ∠C is an acute angle. 8. _____________________________ 9. ∆ABC is an acute triangle. 9. _____________________________

Question 43.
PROVING A THEOREM
Prove the Pythagorean Inequalities Theorem (Theorem 9.3) when c2 > a2 + b2. (Hint: Look back at Exercise 42.)

Maintaining Mathematical Proficiency

Simplify the expression by rationalizing the denominator.

Question 44.
$$\frac{7}{\sqrt{2}}$$

7/√2 = 7√2 /2.

Explanation:
In the above-given question,
given that,
$$\frac{7}{\sqrt{2}}$$ = 7/√2.
7/√2 = 7/√2  x √2 /√2 .
7 √2 /√4.
7√2 /2.

Question 45.
$$\frac{14}{\sqrt{3}}$$

Question 46.
$$\frac{8}{\sqrt{2}}$$

8/√2 = 8√2 /2.

Explanation:
In the above-given question,
given that,
$$\frac{8}{\sqrt{2}}$$ = 8/√2.
8/√2 = 8/√2  x √2 /√2 .
8 √2 /√4.
8√2 /2.

Question 47.
$$\frac{12}{\sqrt{3}}$$

### 9.2 Special Right Triangles

Exploration 1

Side Ratios of an Isosceles Right Triangle

Work with a partner:

a. Use dynamic geometry software to construct an isosceles right triangle with a leg length of 4 units.

b. Find the acute angle measures. Explain why this triangle is called a 45° – 45° – 90° triangle.

c. Find the exact ratios of the side lenghts (using square roots).
$$\frac{A B}{A C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.

d. Repeat parts (a) and (c) for several other isosceles right triangles. Use your results to write a conjecture about the ratios of the side lengths of an isosceles right triangle.

Exploration 2

Work with a partner.

a. Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° (a 30° – 60° – 90° triangle), where the shorter leg length is 3 units.

b. Find the exact ratios of the side lengths (using square roots).
$$\frac{A B}{A C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________

C. Repeat parts (a) and (b) for several other 30° – 60° – 90° triangles. Use your results to write a conjecture about the ratios of the side lengths of a 30° – 60° – 90° triangle.

Question 3.
What is the relationship among the side lengths of 45°- 45° – 90° triangles? 30° – 60° – 90° triangles?

### Lesson 9.2 Special Right Triangles

Monitoring Progress

Find the value of the variable. Write your answer in simplest form.

Question 1.

x = 4

Explanation:
(2√2)² = x² + x²
8 = 2x²
x² = 4
x = 4

Question 2.

y = 2

Explanation:
y² = 2 + 2
y² = 4
y = 2

Question 3.

x = 3, y = 2√3

Explanation:
longer leg = shorter leg • √3
x = √3 • √3
x = 3
hypotenuse = shorter leg • 2
= √3 • 2 = 2√3

Question 4

h = 2√3

Explanation:
longer leg = shorter leg • √3
h = 2√3

Question 5.
The logo on a recycling bin resembles an equilateral triangle with side lengths of 6 centimeters. Approximate the area of the logo.

Area is $$\frac { 1 }{ 3√3 }$$

Explanation:
Area = $$\frac { √3 }{ 4 }$$ a²
= $$\frac { √3 }{ 4 }$$(6)²
= $$\frac { 1 }{ 3√3 }$$

Question 6.
The body of a dump truck is raised to empty a load of sand. How high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle?

28/3 ft high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle.

Explanation:
Height of body at 90 degrees = 14 ft
Height of body at 1 degree = 14/90
Height of body at 60 degrees = 14 x 60/90
= 14 x 2/3
= 28/3 ft

### Exercise 9.2 Special Right Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Name two special right triangles by their angle measures.

Question 2.
WRITING
Explain why the acute angles in an isosceles right triangle always measure 45°.

Because the acute angles of a right isosceles triangle must be congruent by the base angles theorem and complementary, their measures must be 90°/2 = 45°.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x. Write your answer in simplest form.

Question 3.

Question 4.

x = 10

Explanation:
hypotenuse = leg • √2
x = 5√2 • √2
x = 10

Question 5.

Question 6.

x = $$\frac { 9 }{ √2 }$$

Explanation:
hypotenuse = leg • √2
9 = x • √2
x = $$\frac { 9 }{ √2 }$$

In Exercises 7 – 10, find the values of x and y. Write your answers in simplest form.

Question 7.

Question 8.

x = 3, y = 6

Explanation:
hypotenuse = 2 • shorter leg
y = 2 • 3
y = 6
longer leg = √3 • shorter leg
3√3 = √3x
x = 3

Question 9.

Question 10.

x = 18, y = 6√3

Explanation:
hypotenuse = 2 • shorter leg
12√3 = 2y
y = 6√3
longer leg = √3 • shorter leg
x = √3 . 6√3
x = 18

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in finding the length of the hypotenuse.

Question 11.

Question 12.

hypotenuse = leg • √2 = √5 . √2 = √10

In Exercises 13 and 14. sketch the figure that is described. Find the indicated length. Round decimal answers to the nearest tenth.

Question 13.
The side length of an equilateral triangle is 5 centimeters. Find the length of an altitude.

Question 14.
The perimeter of a square is 36 inches. Find the length of a diagonal.

The length of a diagonal is 9√2

Explanation:
Side of the square = 36/4 = 9
square diagonal = √2a = √2(9) = 9√2

In Exercises 15 and 16, find the area of the figure. Round decimal answers to the nearest tenth.

Question 15.

Question 16.

Area is 40√(1/3) sq m

Explanation:
longer leg = √3 • shorter leg
4 = √3 • shorter leg
shorter leg = 4/√3
h² = 16/3 + 16
h² = 16(4/3)
h = 8√(1/3)
Area of the parallelogram = 5(8√(1/3)) = 40√(1/3) sq m

Question 17.
PROBLEM SOLVING
Each half of the drawbridge is about 284 feet long. How high does the drawbridge rise when x is 30°? 45°? 60°?

Question 18.
MODELING WITH MATHEMATICS
A nut is shaped like a regular hexagon with side lengths of 1 centimeter. Find the value of x. (Hint: A regular hexagon can be divided into six congruent triangles.)

Side length = 1cm
A regular hexagon has six equal the side length. A line drawn from the centre to any vertex will have the same length as any side.
This implies the radius is equal to the side length.
As a result, when lines are drawn from the centre to each of the vertexes, a
regular hexagon is said to be made of six equilateral triangles.
From the diagram, x = 2× apothem
Apothem is the distance from the centre of a regular polygon to the midpoint of side.
Using Pythagoras theorem, we would get the apothem
1² = apothem² + (½)²
Apothem = √(1² -(½)²)
= √(1-¼) = √¾
Apothem = ½√3
x = 2× Apothem = 2 × ½√3
x = √3

Question 19.
PROVING A THEOREM
Write a paragraph proof of the 45°- 45°- 90° Triangle Theorem (Theorem 9.4).
Given ∆DEF is a 45° – 45° – 90° triangle.
Prove The hypotenuse is √2 times as long as each leg.

Question 20.
HOW DO YOU SEE IT?
The diagram shows part of the wheel of Theodorus.

a. Which triangles, if any, are 45° – 45° – 90° triangles?

b. Which triangles, if any, are 30° – 60° – 90° triangles?

Question 21.
PROVING A THEOREM
Write a paragraph proof of the 30° – 60° – 90° Triangle Theorem (Theorem 9.5).
(Hint: Construct ∆JML congruent to ∆JKL.)
Given ∆JKL is a 30° 60° 9o° triangle.
Prove The hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.

Question 22.
THOUGHT PROVOKING
A special right triangle is a right triangle that has rational angle measures and each side length contains at most one square root. There are only three special right triangles. The diagram below is called the Ailles rectangle. Label the sides and angles in the diagram. Describe all three special right triangles.

Question 23.
WRITING
Describe two ways to show that all isosceles right triangles are similar to each other.

Question 24.
MAKING AN ARGUMENT
Each triangle in the diagram is a 45° – 45° – 90° triangle. At Stage 0, the legs of the triangle are each 1 unit long. Your brother claims the lengths of the legs of the triangles added are halved at each stage. So, the length of a leg of a triangle added in Stage 8 will be $$\frac{1}{256}$$ unit. Is your brother correct? Explain your reasoning.

Question 25.
USING STRUCTURE
ΔTUV is a 30° – 60° – 90° triangle. where two vertices are U(3, – 1) and V( – 3, – 1), $$\overline{U V}$$ is the hypotenuse. and point T is in Quadrant I. Find the coordinates of T.

Maintaining Mathematical Proficiency

Find the Value of x.

Question 26.
ΔDEF ~ ΔLMN

x = 18

Explanation:
$$\frac { DE }{ LM }$$ = $$\frac { DF }{ LN }$$
$$\frac { 12 }{ x }$$ = $$\frac { 20 }{ 30 }$$
$$\frac { 12 }{ x }$$ = $$\frac { 2 }{ 3 }$$
x = 12($$\frac { 3 }{ 2 }$$) = 18

Question 27.
ΔABC ~ ΔQRS

### 9.3 Similar Right Triangles

Exploration 1

Writing a Conjecture

a. Use dynamic geometry software to construct right ∆ABC, as shown. Draw $$\overline{C D}$$ so that it is an altitude from the right angle to the hypotenuse of ∆ABC.

b. The geometric mean of two positive numbers a and b is the positive number x that satisfies
$$\frac{a}{x}=\frac{x}{b}$$
x is the geometric mean of a and b.
Write a proportion involving the side lengths of ∆CBD and ∆ACD so that CD is the geometric mean of two of the other side lengths. Use similar triangles to justify your steps.

c. Use the proportion you wrote in part (b) to find CD.

d. Generalize the proportion you wrote in part (b). Then write a conjecture about how the geometric mean is related to the altitude from the right angle to the hypotenuse of a right triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.

Exploration 2

Comparing Geometric and Arithmetic Means

Work with a partner:
Use a spreadsheet to find the arithmetic mean and the geometric mean of several pairs of positive numbers. Compare the two means. What do you notice?

Question 3.
How are altitudes and geometric means of right triangles related?

### Lesson 9.3 Similar Right Triangles

Monitoring progress

Identify the similar triangles.

Question 1.

△QRS ~ △ QST

Question 2.

△EFG ~ △ EHG

Question 3.

△EGH ~ △EFG
$$\frac { EF }{ EG }$$ = $$\frac { GF }{ GH }$$
$$\frac { 5 }{ 3 }$$ = $$\frac { 4 }{ x }$$
x = 2.4

Question 4.

△JLM ~ △LMK
$$\frac { JL }{ LM }$$ = $$\frac { JM }{ KM }$$
$$\frac { 13 }{ 5 }$$ = $$\frac { 12 }{ x }$$
x = 4.615

Find the geometric mean of the two numbers.

Question 5.
12 and 27

x = √(12 x 27)
x = √324 = 18

Question 6.
18 and 54

x = √(18 x 54) = √(972)
x = 31.17

Question 7.
16 and 18

x = √(16 x 18) = √(288)
x = 16.970

Question 8.
Find the value of x in the triangle at the left.

x = √(9 x 4)
x = 6

Question 9.
WHAT IF?
In Example 5, the vertical distance from the ground to your eye is 5.5 feet and the distance from you to the gym wall is 9 feet. Approximate the height of the gym wall.

9² = 5.5 x w
81 = 5.5 x w
w = 14.72
The height of the wall = 14.72 + 5.5 = 20.22

### Exercise 9.3 Similar Right Triangles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and ____________ .

Question 2.
WRITING
In your own words, explain geometric mean.

The geometric mean is the average value or mean that signifies the central tendency of set of numbers by finding the product of their values.

Monitoring progress and Modeling with Mathematics

In Exercises 3 and 4, identify the similar triangles.

Question 3.

Question 4.

△LKM ~ △LMN ~ △MKN

In Exercises 5 – 10, find the value of x.

Question 5.

Question 6.

x = 9.6

Explanation:
$$\frac { QR }{ SR }$$ = $$\frac { SR }{ TS }$$
$$\frac { 20 }{ 16 }$$ = $$\frac { 12 }{ x }$$
1.25 = $$\frac { 12 }{ x }$$
x = 9.6

Question 7.

Question 8.

x = 14.11

Explanation:
$$\frac { AB }{ AC }$$ = $$\frac { BD }{ BC }$$
$$\frac { 16 }{ 34 }$$ = $$\frac { x }{ 30 }$$
x = 14.11

Question 9.

Question 10.

x = 2.77

Explanation:
$$\frac { 5.8 }{ 3.5 }$$ = $$\frac { 4.6 }{ x }$$
x = 2.77

In Exercises 11 – 18, find the geometric mean of the two numbers.

Question 11.
8 and 32

Question 12.
9 and 16

x = √(9 x 16)
x = 12

Question 13.
14 and 20

Question 14.
25 and 35

x = √(25 x 35)
x = 29.5

Question 15.
16 and 25

Question 16.
8 and 28

x = √(8 x 28)
x = 14.96

Question 17.
17 and 36

Question 18.
24 and 45

x = √(24 x 45)
x = 32.86

In Exercises 19 – 26. find the value of the variable.

Question 19.

Question 20.

y = √(5 x 8)
y = √40
y = 2√10

Question 21.

Question 22.

10 • 10 = 25 • x
100 = 25x
x = 4

Question 23.

Question 24.

b² = 16(16 + 6)
b² = 16(22) = 352
b = 18.76

Question 25.

Question 26.

x² = 8(8 + 2)
x² = 8(10) = 80
x = 8.9

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in writing an equation for the given diagram.

Question 27.

Question 28.

d² = g • e

MODELING WITH MATHEMATICS
In Exercises 29 and 30, use the diagram.

Question 29.
You want to determine the height of a monument at a local park. You use a cardboard square to line up the top and bottom of the monument, as shown at the above left. Your friend measures the vertical distance from the ground to your eye and the horizontal distance from you to the monument. Approximate the height of the monument.

Question 30.
Your classmate is standing on the other side of the monument. She has a piece of rope staked at the base of the monument. She extends the rope to the cardboard square she is holding lined up to the top and bottom of the monument. Use the information in the diagram above to approximate the height of the monument. Do you get the same answer as in Exercise 29? Explain your reasoning.

MATHEMATICAL CONNECTIONS
In Exercises 31 – 34. find the value(s) of the variable(s).

Question 31.

Question 32.

$$\frac { 6 }{ b + 3 }$$ = $$\frac { 8 }{ 6 }$$
36 = 8(b + 3)
36 = 8b + 24
8b = 12
b = $$\frac { 3 }{ 2 }$$

Question 33.

Question 34.

x = 42.66, y = 40, z = 53

Explanation:
$$\frac { 24 }{ 32 }$$ = $$\frac { 32 }{ x }$$
0.75 = $$\frac { 32 }{ x }$$
x = 42.66
y = √24² + 32²
y = √576 + 1024 = 40
z = √42.66² + 32² = √1819.87 + 1024 = 53

Question 35.
REASONING
Use the diagram. Decide which proportions are true. Select all that apply.

(A) $$\frac{D B}{D C}=\frac{D A}{D B}$$
(B) $$\frac{B A}{C B}=\frac{C B}{B D}$$
(C) $$\frac{C A}{B \Lambda}=\frac{B A}{C A}$$
(D) $$\frac{D B}{B C}=\frac{D A}{B A}$$

Question 36.
ANALYZING RELATIONSHIPS
You are designing a diamond-shaped kite. You know that AD = 44.8 centimeters, DC = 72 centimeters, and AC = 84.8 centimeters. You Want to use a straight crossbar $$\overline{B D}$$. About how long should it be? Explain your reasoning.

BD = 76.12

Explanation:
AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm
Two disjoint pairs of consecutive sides are congruent.
So, AD = AB = 44.8 cm
DC = BC = 72 cm
The diagonals are perpendicular.
So, AC ⊥ BD
AC = AO + OC
AX = x + y = 84.8 — (i)
Perpendicular bisects the diagonal BD into equal parts let it be z.
BD = BO + OD
BD = z + z
Using pythagorean theorem
44.8² = x² + z² —- (ii)
72² = y² + z² —– (iii)
Subtract (ii) and (iii)
72² – 44.8² = y²+ z² – x² – z²
5184 – 2007.04 = (x + y) (x – y)
3176.96 = (84.8)(x – y)
37.464 = x – y —- (iv)
x + y + x – y = 84.8 + 37.464
2x = 122.264
x = 61.132
x + y = 84.8
61.132 + y = 84.8
y = 23.668
44.8² = x² + z²
z = 38.06
BD = z + z
BD = 76.12

Question 37.
ANALYZING RELATIONSHIPS
Use the Geometric Mean Theorems (Theorems 9.7 and 9.8) to find AC and BD.

Question 38.
HOW DO YOU SEE IT?
In which of the following triangles does the Geometric Mean (Altitude) Theorem (Theorem 9.7) apply?
(A)

(B)

(C)

(D)

Question 39.
PROVING A THEOREM
Use the diagram of ∆ABC. Copy and complete the proof of the Pythagorean Theorem (Theorem 9. 1).

Given In ∆ABC, ∆BCA is a right angle.
Prove c2 = a2 + b2

 Statements Reasons 1. In ∆ABC, ∠BCA is a right angle. 1. ________________________________ 2. Draw a perpendicular segment (altitude) from C to $$\overline{A B}$$. 2. Perpendicular Postulate (Postulate 3.2) 3. ce = a2 and cf = b2 3. ________________________________ 4. ce + b2 = ___  + b2 4. Addition Property of Equality 5. ce + cf = a2 + b2 5. ________________________________ 6. c(e + f) a2 + b2 6. ________________________________ 7. e + f = ________ 7. Segment Addition Postulate (Postulate 1.2) 8. c  • c = a2 + b2 8. ________________________________ 9. c2 = a2 + b2 9. Simplify.

Question 40.
MAKING AN ARGUMENT
Your friend claims the geometric mean of 4 and 9 is 6. and then labels the triangle, as shown. Is your friend correct? Explain your reasoning.

G.M = √(4 x 9) = √36 = 6
My friend is correct.

In Exercises 41 and 42, use the given statements to prove the theorem.

Gien ∆ABC is a right triangle.
Altitude $$\overline{C D}$$ is dravn to hypotenuse $$\overline{A B}$$.

Question 41.
PROVING A THEOREM
Prove the Geometric Mean (Altitude) Theorem (Theorem 9.7) b showing that CD2 = AD • BD.

Question 42.
PROVING A THEOREM
Prove the Geometric Mean ( Leg) Theorem (Theorem 9.8) b showing that CB2 = DB • AB and AC2 = AD • AB.

Question 43.
CRITICAL THINKING
Draw a right isosceles triangle and label the two leg lengths x. Then draw the altitude to the hypotenuse and label its length y. Now, use the Right Triangle Similarity Theorem (Theorem 9.6) to draw the three similar triangles from the image and label an side length that is equal to either x or y. What can you conclude about the relationship between the two smaller triangles? Explain your reasoning.

Question 44.
THOUGHT PROVOKING
The arithmetic mean and geometric mean of two nonnegative numbers x and y are shown.
arithmetic mean = $$\frac{x+y}{2}$$
geometric mean = $$\sqrt{x y}$$

Question 45.
PROVING A THEOREM
Prove the Right Triangle Similarity Theorem (Theorem 9.6) by proving three similarity statements.
Given ∆ABC is a right triangle.
Altitude $$\overline{C D}$$ is drawn to hvpotenuse $$\overline{A B}$$.
Prove ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC,
∆CBD ~ ∆ACD

Maintaining Mathematical proficiency

Solve the equation for x.

Question 46.
13 = $$\frac{x}{5}$$

13 = $$\frac{x}{5}$$
x = 65

Question 47.
29 = $$\frac{x}{4}$$

Question 48.
9 = $$\frac{78}{x}$$

9 = $$\frac{78}{x}$$
9x = 78
x = 8.6

Question 49.
30 = $$\frac{115}{x}$$

### 9.1 to 9.3 Quiz

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 1.

x = 15

Explanation:
x² = 9² + 12²
x² = 81 + 144
x² = 225
x = 15

Question 2.

x = 10.63

Explanation:
x² = 7² + 8² = 49 + 64
x = √113
x = 10.63

Question 3.

x = 4√3

Explanation:
8² = x² + 4²
64 = x² + 16
x² = 48
x = 4√3

Verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
(Section 9.1)
Question 4.
24, 32, and 40

Triangle is a right angle trinagle.

Explanation:
40² = 1600
24² + 32² = 576 + 1024 = 1600
40² = 24² + 32²
So, the triangle is a right angle trinagle.

Question 5.
7, 9, and 13

Triangle is an obtuse trinagle.

Explanation:
13² = 169
7² + 9² = 49 + 81 = 130
13² > 7² + 9²
So, the triangle is an obtuse trinagle.

Question 6.
12, 15, and 10√3

Triangle is an acute trinagle.

Explanation:
15² = 225
12² + (10√3)² = 144 + 300 = 444
15² < 12² + (10√3)²
So, the triangle is an acute trinagle.

Find the values of x and y. Write your answers in the simplest form.

Question 7.

x = 6, y = 6√2

Explanation:
x = 6
hypotenuse = leg • √2
y = 6√2

Question 8.

y = 8√3, x = 16

Explanation:
longer leg = shorter leg • √3
y = 8√3
x² = 8² + (8√3)² = 64 + 192
x = 16

Question 9.

x = 5√2, y = 5√6

Explanation:
longer leg = shorter leg • √3
y = x√3
y = 5√6
hypotenuse = shorter leg • 2
10√2 = 2x
x = 5√2

Find the geometric mean of the two numbers.
Question 10.
6 and 12

G.M = √(6 • 12) = 6√2

Question 11.
15 and 20

G.M = √(15 • 20) = 10√3

Question 12.
18 and 26

G.M = √(18 • 26) = 6√13

Identify the similar right triangles. Then find the value of the variable.

Question 13.

x = √(8 • 4)
x = 4√2

Question 14.

y = √(9 • 6) = 3√6

Question 15.

Question 16.
Television sizes are measured by the length of their diagonal. You want to purchase a television that is at least 40 inches. Should you purchase the television shown? Explain your reasoning.

x² = 20.25² + 36²
x² = 410.0625 + 1296 = 1706.0625
x = 41.30
Yes, i will purchase the television.

Question 17.
Each triangle shown below is a right triangle.

a. Are any of the triangles special right triangles? Explain your reasoning.
A is a similar triangle.

b. List all similar triangles. if any.
B, C and D, E are similar triangles.

c. Find the lengths of the altitudes of triangles B and C.
B altitude = √(9 + 27) = 6
C altitude = √(36 + 72) = 6√3

### 9.4 The Tangent Ratio

Exploration 1

Calculating a Tangent Ratio

Work with a partner

a. Construct ∆ABC, as shown. Construct segments perpendicular to $$\overline{A C}$$ to form right triangles that share vertex A and arc similar to ∆ABC with vertices, as shown.

b. Calculate each given ratio to complete the table for the decimal value of tan A for each right triangle. What can you Conclude?

Exploration 2

Using a calculator

Work with a partner: Use a calculator that has a tangent key to calculate the tangent of 36.87°. Do you get the same result as in Exploration 1? Explain.
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.

Question 3.
Repeat Exploration 1 for ∆ABC with vertices A(0, 0), B(8, 5), and C(8, 0). Construct the seven perpendicular segments so that not all of them intersect $$\overline{A C}$$ at integer values of x. Discuss your results.

Question 4.
How is a right triangle used to find the tangent of an acute angle? Is there a unique right triangle that must be used?

### Lesson 9.4 The Tangent Ratio

Monitoring progress

Find tan J and tan K. Write each answer as a fraction and as a decimal rounded to four places.

Question 1.

tan K = $$\frac { opposite side }{ adjacent side }$$ = $$\frac { JL }{ KL }$$
= $$\frac { 32 }{ 24 }$$ = $$\frac { 4 }{ 3 }$$ = 1.33
tan J = $$\frac { KL }{ JL }$$ = $$\frac { 24 }{ 32 }$$ = $$\frac { 3 }{ 4 }$$ = 0.75

Question 2.

tan K = $$\frac { LJ }{ LK }$$ = $$\frac { 15 }{ 8 }$$
tan J = $$\frac { LK }{ LJ }$$ = $$\frac { 8 }{ 15 }$$

Find the value of x. Round your answer to the nearest tenth.

Question 3.

Tan 61 = $$\frac { 22 }{ x }$$
1.804 = $$\frac { 22 }{ x }$$
x = 12.1951

Question 4.

tan 56 = $$\frac { x }{ 13 }$$
1.482 = $$\frac { x }{ 13 }$$
x = 19.266

Question 5.
WHAT IF?
In Example 3, the length of the shorter leg is 5 instead of 1. Show that the tangent of 60° is still equal to √3.

longer leg = shorter leg • √3
= 5√3
tan 60 = $$\frac { 5√3 }{ 5 }$$
= √3

Question 6.
You are measuring the height of a lamppost. You stand 40 inches from the base of the lamppost. You measure the angle ot elevation from the ground to the top of the lamppost to be 70°. Find the height h of the lamppost to the nearest inch.

tan 70 = $$\frac { h }{ 40 }$$
2.7474 = $$\frac { h }{ 40 }$$
h = 109.896 in

### Exercise 9.4 The Tangent Ratio

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The tangent ratio compares the length of _________ to the length of ___________ .

Question 2.
WRITING
Explain how you know the tangent ratio is constant for a given angle measure.

When two triangles are similar, the corresponding sides are proportional which makes the ratio constant for a given acute angle measurement.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 3.

Question 4.

tan F = $$\frac { DE }{ EF }$$ = $$\frac { 24 }{ 7 }$$
tan D = $$\frac { EF }{ DE }$$ = $$\frac { 7 }{ 24 }$$

Question 5.

Question 6.

tan K = $$\frac { JL }{ LK }$$ = $$\frac { 3 }{ 5 }$$
tan J = $$\frac { LK }{ JL }$$ = $$\frac { 5 }{ 3 }$$

In Exercises 7 – 10, find the value of x. Round your answer to the nearest tenth.

Question 7.

Question 8.

tan 27 = $$\frac { x }{ 15 }$$
0.509 = $$\frac { x }{ 15 }$$
x = 7.635

Question 9.

Question 10.

tan 37 = $$\frac { 6 }{ x }$$
0.753 = $$\frac { 6 }{ x }$$
x = 7.968

ERROR ANALYSIS
In Exercises 11 and 12, describe the error in the statement of the tangent ratio. Correct the error if possible. Otherwise, write not possible.

Question 11.

Question 12.

In Exercises 13 and 14, use a special right triangle to find the tangent of the given angle measure.

Question 13.
45°

Question 14.
30°

tan 30° = $$\frac { 1 }{ √3 }$$

Question 15.
MODELING WITH MATHEMATICS
A surveyor is standing 118 Feet from the base of the Washington Monument. The surveyor measures the angle of elevation from the ground to the top of the monument to be 78°. Find the height h of the Washington Monument to the nearest foot.

Question 16.
MODELING WITH MATHEMATICS
Scientists can measure the depths of craters on the moon h looking at photos of shadows. The length of the shadow cast by the edge of a crater is 500 meters. The angle of elevation of the rays of the Sun is 55°. Estimate the depth d of the crater.

tan 55 = $$\frac { d }{ 500 }$$
1.428 = $$\frac { d }{ 500 }$$
d = 714 m
The depth of the crater is 714 m

Question 17.
USING STRUCTURE
Find the tangent of the smaller acute angle in a right triangle with side lengths 5, 12, and 13.

Question 18.
USING STRUCTURE
Find the tangent 0f the larger acute angle in a right triangle with side lengths 3, 4, and 5.

tan x = $$\frac { 4 }{ 3 }$$

Question 19.
REASONING
How does the tangent of an acute angle in a right triangle change as the angle measure increases? Justify your answer.

Question 20.
CRITICAL THINKING
For what angle measure(s) is the tangent of an acute angle in a right triangle equal to 1? greater than 1? less than 1? Justify your answer.

In order for the tangent of an angle to equal 1, the opposite and adjacent sides of a right triangle must be the same. This means the right triangle is an isosceles right triangle so the angles are 45 – 45 – 90. The acute angle must be 1. In order for the tangent to be greater than 1, the opposite side must be greater than the adjacent side. This means the angle must be between 45 and 90 degrees. If the tangent is less than 1, this means the opposite side must be smaller than the adjacent side. The acute angle must be between 0 and 45.

Question 21.
MAKING AN ARGUMENT
Your family room has a sliding-glass door. You want to buy an awning for the door that will be just long enough to keep the Sun out when it is at its highest point in the sky. The angle of elevation of the rays of the Sun at this points is 70°, and the height of the door is 8 feet. Your sister claims you can determine how far the overhang should extend by multiplying 8 by tan 70°. Is your sister correct? Explain.

Question 22.
HOW DO YOU SEE IT?
Write expressions for the tangent of each acute angle in the right triangle. Explain how the tangent of one acute angle is related to the tangent of the other acute angle. What kind of angle pair is ∠A and ∠B?

tan A = $$\frac { BC }{ AC }$$ = $$\frac { a }{ b }$$
tan B = $$\frac { AC }{ BC }$$ = $$\frac { b }{ a }$$

Question 23.
REASONING
Explain why it is not possible to find the tangent of a right angle or an obtuse angle.

Question 24.
THOUGHT PROVOKING
To create the diagram below. you begin with an isosceles right triangle with legs 1 unit long. Then the hypotenuse of the first triangle becomes the leg of a second triangle, whose remaining leg is 1 unit long. Continue the diagram Until you have constructed an angle whose tangent is $$\frac{1}{\sqrt{6}}$$. Approximate the measure of this angle.

Question 25.
PROBLEM SOLVING
Your class is having a class picture taken on the lawn. The photographer is positioned 14 feet away from the center of the class. The photographer turns 50° to look at either end of the class.

a. What is the distance between the ends of the class?
b. The photographer turns another 10° either way to see the end of the camera range. If each student needs 2 feet of space. about how many more students can fit at the end of each row? Explain.

Question 26.
PROBLEM SOLVING
Find the perimeter of the figure. where AC = 26, AD = BF, and D is the midpoint of $$\overline{A C}$$.

Maintaining Mathematical proficiency

Find the value of x.

Question 27.

Question 28.

longer side = shorter side • √3
7 = x√3
x = $$\frac { 7 }{ √3 }$$
x = 4.04

Question 29.

### 9.5 The Sine and Cosine Ratios

Exploration 1

Work with a partner: Use dynamic geometry software.

a. Construct ∆ABC, as shown. Construct segments perpendicular to $$\overline{A C}$$ to form right triangles that share vertex A arid are similar to ∆ABC with vertices, as shown.

b. Calculate each given ratio to complete the table for the decimal values of sin A and cos A for each right triangle. What can you conclude?

Question 2.
How is a right triangle used to find the sine and cosine of an acute angle? Is there a unique right triangle that must be used?

Question 3.
In Exploration 1, what is the relationship between ∠A and ∠B in terms of their measures’? Find sin B and cos B. How are these two values related to sin A and cos A? Explain why these relationships exist.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.

### Lesson 9.5 The Sine and Cosine Ratios

Monitoring Progress

Question 1.
Find sin D, sin F, cos D, and cos F. Write each answer as a fraction and as a decimal rounded to four places.

sin D = $$\frac { 7 }{ 25 }$$
sin F = $$\frac { 24 }{ 25 }$$
cos D = $$\frac { 24 }{ 25 }$$
cos F = $$\frac { 7 }{ 25 }$$

Explanation:
sin D = $$\frac { EF }{ DF }$$ = $$\frac { 7 }{ 25 }$$
sin F = $$\frac { DE }{ DF }$$ = $$\frac { 24 }{ 25 }$$
cos D = $$\frac { DE }{ DF }$$ = $$\frac { 24 }{ 25 }$$
cos F = $$\frac { EF }{ DF }$$ = $$\frac { 7 }{ 25 }$$

Question 2.
Write cos 23° in terms of sine.

cos X = sin(90 – X)
cos 23° = sin (90 – 23)
= sin(67)
So, cos 23° = sin 67°

Question 3.
Find the values of u and t using sine and cosine. Round your answers to the nearest tenth.

t = 7.2, u = 3.3

Explanation:
sin 65 = $$\frac { t }{ 8 }$$
0.906 = $$\frac { t }{ 8 }$$
t = 7.2
cos 65 = $$\frac { u }{ 8 }$$
0.422 x 8 = u
u = 3.3

Question 4.
Find the sine and cosine of a 60° angle.

sin 60° = $$\frac { √3 }{ 2 }$$
cos 60° = $$\frac { 1 }{ 2 }$$

Question 5.
WHAT IF?
In Example 6, the angle of depression is 28°. Find the distance x you ski down the mountain to the nearest foot.

sin 28° = $$\frac { 1200 }{ x }$$
x = $$\frac { 1200 }{ 0.469 }$$
x = 2558.6

### Exercise 9.5 The Sine and Cosine Ratios

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The sine raio compares the length of ______________ to the length of _____________

Question 2.
WHICH ONE DOESN’T BELONG?
Which ratio does not belong with the other three? Explain your reasoning.

sin B

sin B = $$\frac { AC }{ BC }$$

cos C

cos C = $$\frac { AC }{ BC }$$

tan B

tan B = $$\frac { AC }{ AB }$$

$$\frac{A C}{B C}$$

$$\frac{A C}{B C}$$ = sin B

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find sin D, sin E, cos D, and cos E. Write each answer as a Fraction and as a decimal rounded to four places.

Question 3.

Question 4.

sin D = $$\frac { 35 }{ 37 }$$
sin E = $$\frac { 12 }{ 37 }$$
cos D = $$\frac { 12 }{ 37 }$$
cos E = $$\frac { 35 }{ 37 }$$

Question 5.

Question 6.

sin D = $$\frac { 36 }{ 45 }$$
sin E = $$\frac { 27 }{ 45 }$$
cos D = $$\frac { 27 }{ 45 }$$
cos E = $$\frac { 36 }{ 45 }$$

Question 7.

Question 8.

sin D = $$\frac { 8 }{ 17 }$$
sin E = $$\frac { 15 }{ 17 }$$
cos D = $$\frac { 15 }{ 17 }$$
cos E = $$\frac { 8 }{ 17 }$$

In Exercises 9 – 12. write the expression in terms of cosine.

Question 9.
sin 37°

Question 10.
sin 81°

sin 81° = cos(90° – 81°) = cos9°

Question 11.
sin 29°

Question 12.
sin 64°

sin 64° = cos(90° – 64°) = cos 26°

In Exercise 13 – 16, write the expression in terms of sine.

Question 13.
cos 59°

Question 14.
cos 42°

cos 42° = sin(90° – 42°) = sin 48°

Question 15.
cos 73°

Question 16.
cos 18°

cos 18° = sin(90° – 18°) = sin 72°

In Exercises 17 – 22, find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Question 17.

Question 18.

p = 30.5, q = 14.8

Explanation:
sin 64° = $$\frac { p }{34 }$$
p = 0.898 x 34
p = 30.5
cos 64° = $$\frac { q }{ 34 }$$
q = 0.4383 x 34
q = 14.8

Question 19.

Question 20.

s = 17.7, r = 19

Explanation:
sin 43° = $$\frac { s }{26 }$$
s = 0.681 x 26
s = 17.7
cos 43° = $$\frac { r }{ 26 }$$
r = 0.731 x 26
r = 19

Question 21.

Question 22.

m = 6.7, n = 10.44

Explanation:
sin 50° = $$\frac { 8 }{n }$$
0.766 = $$\frac { 8 }{n }$$
n = 10.44
cos 50° = $$\frac { m }{ n }$$
0.642 = $$\frac { m }{ 10.44 }$$
m = 6.7

Question 23.
REASONING
Which ratios are equal? Select all that apply.

sin X

cos X

sin Z

cos Z

Question 24.
REASONING
Which ratios arc equal to $$\frac{1}{2}$$ Select all

sin L

sin L = $$\frac { 2 }{ 4 }$$ = $$\frac { 1 }{ 2 }$$

cos L

cos L = $$\frac { 2√3 }{ 4 }$$ = $$\frac { √3 }{ 2 }$$

sin J

sin J = $$\frac { 2√3 }{ 4 }$$ = $$\frac { √3 }{ 2 }$$

cos J

cos J = $$\frac { 2 }{ 4 }$$ = $$\frac { 1 }{ 2 }$$

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding sin A.

Question 26.
WRITING
Explain how to tell which side of a right triangle is adjacent to an angle and which side is the hypotenuse.

Question 27.
MODELING WITH MATHEMATICS
The top of the slide is 12 feet from the ground and has an angle of depression of 53°. What is the length of the slide?

Question 28.
MODELING WITH MATHEMATICS
Find the horizontal distance x the escalator covers.

cos 41 = $$\frac { x }{ 26 }$$
0.754 = $$\frac { x }{ 26 }$$
x = 19.6 ft

Question 29.
PROBLEM SOLVING
You are flying a kite with 20 feet of string extended. The angle of elevation from the spool of string to the kite is 67°.
a. Draw and label a diagram that represents the situation.
b. How far off the ground is the kite if you hold the spool 5 feet off the ground? Describe how the height where you hold the spool affects the height of the kite.

Question 30.
MODELING WITH MATHEMATICS
Planes that fly at high speeds and low elevations have radar s sterns that can determine the range of an obstacle and the angle of elevation to the top of the obstacle. The radar of a plane flying at an altitude of 20,000 feet detects a tower that is 25,000 feet away. with an angle of elevation of 1°

a. How many feet must the plane rise to pass over the tower?

sin 1 = $$\frac { h }{ 25000 }$$
0.017 = $$\frac { h }{ 25000 }$$
h = 425 ft
425 ft the plane rise to pass over the tower

b. PIanes Caillot come closer than 1000 feet vertically to any object. At what altitude must the plane fly in order to pass over the tower?

Question 31.
MAKING AN ARGUMENT
Your friend uses che equation sin 49° = $$\frac{x}{16}$$ to find BC. Your cousin uses the equation cos 41° = $$\frac{x}{16}$$ to find BC. Who is correct? Explain your reasoning.

Question 32.
WRITING
Describe what you must know about a triangle in order to use the sine ratio and what you must know about a triangle in order to use the cosine ratio

sin = $$\frac { opposite side }{ hypotenuse }$$
cos = $$\frac { adjacent side }{ hypotenuse }$$

Question 33.
MATHEMATICAL CONNECTIONS
If ∆EQU is equilateral and ∆RGT is a right triangle with RG = 2, RT = 1. and m ∠ T = 90°, show that sin E = cos G.

Question 34.
MODELING WITH MATHEMATICS
Submarines use sonar systems, which are similar to radar systems, to detect obstacles, Sonar systems use sound to detect objects under water.

a. You are traveling underwater in a submarine. The sonar system detects an iceberg 4000 meters a head, with an angle of depression of 34° to the bottom of the iceberg. How many meters must the submarine lower to pass under the iceberg?

tan 34 = $$\frac { x }{ 4000 }$$
.674 = $$\frac { x }{ 4000 }$$
x = 2696

b. The sonar system then detects a sunken ship 1500 meters ahead. with an angle of elevation of 19° to the highest part of the sunken ship. How many meters must the submarine rise to pass over the sunken ship?

tan 19 = $$\frac { x }{ 1500 }$$
0.344 = $$\frac { x }{ 1500 }$$
x = 516 m

Question 35.
ABSTRACT REASONING
Make a conjecture about how you could use trigonometric ratios to find angle measures in a triangle.

Question 36.
HOW DO YOU SEE IT?
Using only the given information, would you use a sine ratio or a cosine ratio to find the length of the hypotenuse? Explain your reasoning.

sin 29 = $$\frac { 9 }{ x }$$
0.48 = $$\frac { 9 }{ x }$$
x = 18.75
The length of hypotenuse is 18.75

Question 37.
MULTIPLE REPRESENTATIONS
You are standing on a cliff above an ocean. You see a sailboat from your vantage point 30 feet above the ocean.

a. Draw and label a diagram of the situation.
b. Make a table showing the angle of depression and the length of your line of sight. Use the angles 40°, 50°, 60°, 70°, and 80°.
c. Graph the values you found in part (b), with the angle measures on the x-axis.
d. Predict the length of the line of sight when the angle of depression is 30°.

Question 38.
THOUGHT PROVOKING
One of the following infinite series represents sin x and the other one represents cos x (where x is measured in radians). Which is which? Justify your answer. Then use each series to approximate the sine and cosine of $$\frac{\pi}{6}$$.
(Hints: π = 180°; 5! = 5 • 4 • 3 • 2 • 1; Find the values that the sine and cosine ratios approach as the angle measure approaches zero).
a.

For x = 0
0 – $$\frac { 0³ }{ 3! }$$ + $$\frac { 0⁵ }{ 5! }$$ – $$\frac { 0⁷ }{ 7! }$$ + . . . = 0
sin x = x – $$\frac { x³ }{ 3! }$$ + $$\frac { x⁵ }{ 5! }$$ – $$\frac { x⁷ }{ 7! }$$ + . . .
sin $$\frac { π }{ 6 }$$ = 0.5

b.

1 – $$\frac { 1² }{ 2! }$$ + $$\frac { 1⁴ }{ 4! }$$ – $$\frac { 1⁶ }{ 6! }$$ + . .  = 1
cos x =x1 – $$\frac { x² }{ 2! }$$ + $$\frac { x⁴ }{ 4! }$$ – $$\frac { x⁶ }{ 6! }$$ + . .
cos $$\frac { π }{ 6 }$$ = 0.86

Question 39.
CRITICAL THINKING
Let A be any acute angle of a right triangle. Show that
(a) tan A = $$\frac{\sin A}{\cos A}$$ and
(b) (sin A)2 + (cos A)2 = 1.

Question 40.
CRITICAL THINKING
Explain why the area ∆ ABC in the diagram can be found using the formula Area = $$\frac{1}{2}$$ ab sin C. Then calculate the area when a = 4, b = 7, and m∠C = 40°:

Area = $$\frac{1}{2}$$ ab sin C
= $$\frac{1}{2}$$ (4 x 7) sin 40°
= 14 x 0.642
= 8.988

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 41.

Question 42.

x = 12√2

Explanation:
c² = a² + b²
x² = 12² + 12²
x² = 144 + 144
x² = 288
x = 12√2

Question 43.

Question 44.

x = 6√2

Explanation:
c² = a² + b²
9² = x² + 3²
81 = x² + 9
x² = 81 – 9
x = 6√2

### 9.6 Solving Right Triangles

Exploration 1

Solving Special Right Triangles

Work with a partner. Use the figures to find the values of the sine and cosine of ∠A and ∠B. Use these values to find the measures of ∠A and ∠B. Use dynamic geometry software to verify your answers.
a.

b.

Exploration 2

Solving Right Triangles

Work with a partner: You can use a calculator to find the measure of an angle when you know the value of the sine, cosine, or tangent of the rule. Use the inverse sine, inverse cosine, 0r inverse tangent feature of your calculator to approximate the measures of ∠A and ∠B to the nearest tenth of a degree. Then use dynamic geometry software to verify your answers.
ATTENDING TO PRECISION
To be proficient in math, you need to calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context.
a.

b.

Question 3.
When you know the lengths of the sides of a right triangle, how can you find the measures of the two acute angles?

Question 4.
A ladder leaning against a building forms a right triangle with the building and the ground. The legs of the right triangle (in meters) form a 5-12-13 Pythagorean triple. Find the measures of the two acute angles to the nearest tenth of a degree.

### Lesson 9.6 Solving Right Triangles

Determine which of the two acute angles has the given trigonometric ratio.

Question 1.
The sine of the angle is $$\frac{12}{13}$$.

Sin E = $$\frac{12}{13}$$
m∠E = sin-1($$\frac{12}{13}$$) = 67.3°

Question 2.
The tangent of the angle is $$\frac{5}{12}$$

tan F = $$\frac{5}{12}$$
m∠F = tan-1($$\frac{5}{12}$$) = 22.6°

Let ∠G, ∠H, and ∠K be acute angles. Use a calculator to approximate the measures of ∠G, ∠H, and ∠K to the nearest tenth of a degree.

Question 3.
tan G = 0.43

∠G = inverse tan of 0.43 = 23.3°

Question 4.
sin H = 0.68

∠H = inverse sin of 0.68 = 42.8°

Question 5.
cos K = 0.94

∠K = inverse cos of 0.94 = 19.9°

Solve the right triangle. Round decimal answers to the nearest tenth.

Question 6.

DE = 29, ∠D = 46.05°, ∠E = 42.84°

Explanation:
c² = a² + b²
x² = 20² + 21²
x² = 400 + 441
x² = 841
x = 29
sin D = $$\frac { 21 }{ 29 }$$
∠D = 46.05
sin E = $$\frac { 20 }{ 29 }$$
∠E = 42.84

Question 7.

GJ = 60, ∠G = 56.09°, ∠H = 33.3°

Explanation:
c² = a² + b²
109² = 91² + x²
x² = 11881 – 8281
x² = 3600
x = 60
sin G = $$\frac { 91 }{ 109 }$$
∠G = 56.09
sin H = $$\frac { 60 }{ 109 }$$
∠H = 33.3

Question 8.
Solve the right triangle. Round decimal answers to the nearest tenth.

XY = 13.82, YZ = 6.69, ∠Y = 37.5

Explanation:
cos 52 = $$\frac { 8.5 }{ XY }$$
0.615 = $$\frac { 8.5 }{ XY }$$
XY = 13.82
sin 52 = $$\frac { YZ }{ XY }$$
0.788 = $$\frac { YZ }{ 8.5 }$$
YZ = 6.69
sin Y = $$\frac { 8.5 }{ 13.82 }$$
∠Y = 37.5

Question 9.
WHAT IF?
In Example 5, suppose another raked stage is 20 feet long from front to back with a total rise of 2 feet. Is the raked stage within your desired range?

x = inverse sine of $$\frac { 2 }{ 20 }$$
x = 5.7°

### Exercise 9.6 Solving Right Triangles

Question 1.
COMPLETE THE SENTENCE
To solve a right triangle means to find the measures of all its ________ and _______ .

Question 2.
WRITING
Explain when you can use a trigonometric ratio to find a side length of a right triangle and when you can use the Pythagorean Theorem (Theorem 9.1 ).

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. determine which of the two acute angles has the given trigonometric ratio.

Question 3.
The cosine of the angle is $$\frac{4}{5}$$

Question 4.
The sine of the angle is $$\frac{5}{11}$$

Sin(angle) = $$\frac { opposite }{ hypo }$$
sin A = $$\frac{5}{11}$$
The acute angle that has a sine of the angle is $$\frac{5}{11}$$ is ∠A.

Question 5.
The sine of the angle is 0.95.

Question 6.
The tangent of the angle is 1.5.

tan(angle) = $$\frac { opposite }{ adjacent }$$
1.5 = $$\frac { 18 }{ 12 }$$
tan C = 1.5
The acute angle that has a tangent of the angle is 1.5 ∠C.

In Exercises 7 – 12, let ∠D be an acute angle. Use a calculator to approximate the measure of ∠D to the nearest tenth of a degree.

Question 7.
sin D = 0.75

Question 8.
sin D = 0.19

sin D = 0.19
∠D = inverse sine of 0.19
∠D = 10.9°

Question 9.
cos D = 0.33

Question 10.
cos D = 0.64

cos D = 0.64
∠D = inverse cos of 0.64
∠D = 50.2°

Question 11.
tan D = 0.28

Question 12.
tan D = 0.72

tan D = 0.72
∠D = inverse tan of 0.72
∠D = 35.8°

In Exercises 13 – 18. solve the right triangle. Round decimal answers to the nearest tenth.

Question 13.

Question 14.

ED = 2√65, ∠E = 59.3, ∠D = 29.7

Explanation:
c² = 8² + 14²
x² = 64 + 196
x² = 260
x = 2√65
sin E = $$\frac { 14 }{ 2√65 }$$
∠E = 59.3
sin D = $$\frac { 8 }{ 2√65 }$$
∠D = 29.7

Question 15.

Question 16.

HJ = 2√15, ∠G = 28.9, ∠J = 61

Explanation:
c² = a² + b²
16² = 14² + x²
x² = 256 – 196
x² = 60
x = 2√15
sin G = $$\frac { 2√15 }{ 16 }$$
∠G = 28.9
sin J = $$\frac { 14 }{ 16 }$$
∠J = 61

Question 17.

Question 18.

RT = 17.8, RS = 9.68, ∠T = 32.8

Explanation:
sin 57 = $$\frac { 15 }{ x }$$
0.838 = $$\frac { 15 }{ x }$$
x = 17.899
RT = 17.8
cos 57 = $$\frac { x }{ 17.8 }$$
0.544 = $$\frac { x }{ 17.8 }$$
x = 9.68
RS = 9.68
sin T = $$\frac { 9.68 }{ 17.8 }$$
∠T = 32.8

Question 19.
ERROR ANALYSIS
Describe and correct the error in using an inverse trigonometric ratio.

Question 20.
PROBLEM SOLVING
In order to unload clay easily. the body of a dump truck must be elevated to at least 45° The body of a dump truck that is 14 feet long has been raised 8 feet. Will the clay pour out easily? Explain your reasoning.

Angle of elevation: sin x = $$\frac { 8 }{ 14 }$$
x = inverse sine of $$\frac { 8 }{ 14 }$$ = 34.9
The clay will not pour out easily.

Question 21.
PROBLEM SOLVING
You are standing on a footbridge that is 12 feet above a lake. You look down and see a duck in the water. The duck is 7 feet away from the footbridge. What is the angle of elevation from the duck to you

Question 22.
HOW DO YOU SEE IT?
Write three expressions that can be used to approximate the measure of ∠A. Which expression would you choose? Explain your choice.

Three expressions are ∠A = inverse tan of ($$\frac { 15 }{ 22 }$$) = 34.2°
∠A = inverse sine of ($$\frac { 15 }{ BA }$$)
∠A = inverse cos of ($$\frac { 22 }{ BA }$$)

Question 23.
MODELING WITH MATHEMATICS
The Uniform Federal Accessibility Standards specify that awheel chair ramp may not have an incline greater than 4.76. You want to build a ramp with a vertical rise of 8 inches. you want to minimize the horizontal distance taken up by the ramp. Draw a diagram showing the approximate dimensions of your ramp.

Question 24.
MODELING WITH MATHEMATICS
The horizontal part of a step is called the tread. The vertical part is called the riser. The recommended riser – to – tread ratio is 7 inches : 11 inches.

a. Find the value of x for stairs built using the recommended riser-to-tread ratio.

b. you want to build stairs that are less steep than the stairs in part (a). Give an example of a riser – to – tread ratio that you could use. Find the value of x for your stairs.

Question 25.
USING TOOLS
Find the measure of ∠R without using a protractor. Justify your technique.

Question 26.
MAKING AN ARGUMENT
Your friend claims that tan-1x = $$\frac{1}{\tan x}$$. Is your friend correct? Explain your reasoning.

No
For example
tan-1(√3) = 60
$$\frac{1}{\tan √3}$$ = 33.1

USING STRUCTURE
In Exercises 27 and 28, solve each triangle.

Question 27.
∆JKM and ∆LKM

Question 28.
∆TUS and ∆VTW

TS = 8.2, UT = 7.3, ∠T = 28.6
TV = 13.2, TW = 9.6, ∠V = 46, ∠T = 42.84

Explanation:
tan 64 = $$\frac { TS }{ 4 }$$
TS = 2.05 x 4
TS = 8.2
sin 64 = $$\frac { UT }{ 8.2 }$$
0.898= $$\frac { UT }{ 8.2 }$$
UT = 7.3
sin T = $$\frac { 4 }{ 8.2 }$$
∠T = 28.6
TV = TS + SV
TV = 8.2 + 5 = 13.2
13.2² = TW² + 9²
TW² = 174.24 – 81
TW = 9.6
sin V = $$\frac { 9.6 }{ 13.2 }$$
∠V = 46
sin T = $$\frac { 9 }{ 13.2 }$$
∠T = 42.84

Question 29.
MATHEMATICAL CONNECTIONS
Write an expression that can be used to find the measure of the acute angle formed by each line and the x-axis. Then approximate the angle measure to the nearest tenth of a degree.
a. y = 3x
b. y = $$\frac{4}{3}$$x + 4

Question 30.
THOUGHT PROVOKING
a. sin-1 (sin x)

sin-1 (sin x) = x

b. tan(tan-1 y)

tan(tan-1 y) = y

C. cos(cos-1 z)

cos(cos-1 z) = z

Question 31.
REASONING
Explain why the expression sin-1 (1.2) does not make sense.

Question 32.
USING STRUCTURE
The perimeter of the rectangle ABCD is 16 centimeters. and the ratio of its width to its length is 1 : 3. Segment BD divides the rectangle into two congruent triangles. Find the side lengths and angle measures of these two triangles.

The perimeter of the rectangle ABCD is 16 centimeters
2(l + b) = 16
l + b = 8
b : l = 1 : 3
4x = 8
x = 2
l = 6, b = 2
BD = √(6² + 2²) = √40 = 2√10
sin B = $$\frac { AD }{ BD }$$ = $$\frac { 2 }{ 2√10 }$$
∠ABD = 18.4
∠CBD = 71.6
sin D = $$\frac { AB }{ BD }$$ = $$\frac { 6 }{ 2√10 }$$
∠CDB = 19

Maintaining Mathematical Proficiency

Solve the equation

Question 33.
$$\frac{12}{x}=\frac{3}{2}$$

Question 34.
$$\frac{13}{9}=\frac{x}{18}$$

$$\frac{13}{9}=\frac{x}{18}$$
x = 1.44 x 18
x = 26

Question 35.
$$\frac{x}{2.1}=\frac{4.1}{3.5}$$

Question 36.
$$\frac{5.6}{12.7}=\frac{4.9}{x}$$

$$\frac{5.6}{12.7}=\frac{4.9}{x}$$
0.44 = 4.9/x
x = 11.13

### 9.7 Law of Sines and Law of Cosines

Exploration 1

Discovering the Law of Sines

Work with a partner.

a. Copy and complete the table for the triangle shown. What can you conclude?

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about the relationship between the sines of the angles and the lengths of the sides of a triangle.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to compare predictions with data.

Exploration 2

Discovering the Law of Cosines

Work with a partner:

a. Copy and complete the table for the triangle in Exploration 1 (a). What can you conclude?

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about what you observe in the completed tables.

Question 3.
What are the Law of Sines and the Law of Cosines?

Question 4.
When would you use the Law of Sines to solve a triangle? When would you use the Law of Cosines to solve a triangle?

### Lesson 9.7 Law of Sines and Law of Cosines

Monitoring Progress

Use a calculator to find the trigonometric ratio. Round your answer to four decimal places.

Question 1.
tan 110°

tan 110° = -2.7474

Question 2.
sin 97°

sin 97° = 0.9925

Question 3.
cos 165°

cos 165° = -0.9659

Find the area of ∆ABC with the given side lengths and included angle. Round your answer to the nearest tenth.

Question 4.
m ∠ B = 60°, a = 19, c = 14

Area = 155.18

Explanation:
Area = $$\frac { 1 }{ 2 }$$ ac sin B
= $$\frac { 1 }{ 2 }$$ (19 x 14) sin 60°
= 133 x 0.866
= 155.18

Question 5.
m ∠ C = 29°, a = 38, b = 31

Area = 282.72

Explanation:
Area = $$\frac { 1 }{ 2 }$$ ab sin C
= $$\frac { 1 }{ 2 }$$ (38 x 31) sin 29
= 598 x 0.48
= 282.72

Solve the triangle. Round decimal answers to the nearest tenth.

Question 6.

∠C = 46.6, ∠B = 82.4, AC = 23.57

Explanation:
Using law of sines
$$\frac { c }{ sin C }$$ = $$\frac { a }{ sin A }$$
$$\frac { 17 }{ sin C }$$ = $$\frac { 18 }{ sin 51 }$$
$$\frac { 17 }{ sin C }$$ = $$\frac { 18 }{ 0.77 }$$
sin C = 0.7274
∠C = 46.6
∠A + ∠B + ∠C  = 180
51 + 46.6 + ∠B = 180
∠B = 82.4
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { 0.99 }{ b }$$ = $$\frac { 0.77 }{ 18 }$$
b = 23.57

Question 7.

∠B = 31.3, ∠C  = 108.7, c = 23.6

Explanation:
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin B }{ 13 }$$ = $$\frac { sin 40 }{ 16 }$$
sin B = $$\frac { .64 }{ 16 }$$ x 13
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C  = 180
40 + 31.3 + ∠C  = 180
∠C  = 108.7
$$\frac { c }{ sin C }$$ = $$\frac { a }{ sin A }$$
$$\frac { c }{ sin 108.7 }$$ = $$\frac {16 }{ sin 40 }$$
c = $$\frac {16 }{ .64 }$$ x 0.947
c = 23.6

Solve the triangle. Round decimal answers to the nearest tenth.

Question 8.

∠C = 66, a = 4.36, c = 8.27

Explanation:
∠A + ∠B + ∠C = 180
29 + 85 + ∠C = 180
∠C = 66
$$\frac { a }{ sin A }$$ = $$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { a }{ sin 29 }$$ = $$\frac { 9 }{ sin 85 }$$
$$\frac { a }{ 0.48 }$$ = $$\frac { 9 }{ 0.99 }$$
a = 4.36
$$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { 9 }{ sin 85 }$$ = $$\frac { c }{ sin 66 }$$
$$\frac { 9 }{ 0.99 }$$ = $$\frac { c }{ 0.91 }$$
c = 8.27

Question 9.

∠A = 29, b = 19.37, c = 20.41

Explanation:
∠A + ∠B + ∠C  = 180
∠A + 70 + 81 = 180
∠A = 29
$$\frac { a }{ sin A }$$ = $$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { 10 }{ sin 29 }$$ = $$\frac { b }{ sin 70 }$$
$$\frac { 10 }{ 0.48 }$$ = $$\frac { b }{ 0.93 }$$
b = 19.37
$$\frac { a }{ sin A }$$ = $$\frac { c }{ sin C }$$
$$\frac { 10 }{ sin 29 }$$ = $$\frac { c }{ sin 81 }$$
$$\frac { 10 }{ 0.48 }$$ = $$\frac { c }{ 0.98 }$$
c = 20.41

Question 10.
WHAT IF?
In Example 5, what would be the length of a bridge from the South Picnic Area to the East Picnic Area?

The length of a bridge from the South Picnic Area to the East Picnic Area is 188 m.

Explanation:
$$\frac { a }{ sin A }$$ = $$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { a }{ sin 71 }$$ = $$\frac { 150 }{ sin 49 }$$
a = 188

Solve the triangle. Round decimal answers to the nearest tenth.

Question 11.

b = 61.3, ∠A = 46, ∠C  = 46

Explanation:
b² = a² + c² − 2ac cos B
b² = 45² + 43² – 2(45)(43) cos 88
b² = 2025 + 1849 – 3870 x 0.03 = 3757.9
b = 61.3
$$\frac { sin A }{ a }$$ = $$\frac { sin B }{ b }$$
$$\frac { sin A }{ 45 }$$ = $$\frac { sin 88 }{ 61.3 }$$
sin A = 0.72
∠A = 46
∠A + ∠B + ∠C  = 180
46 + 88 + ∠C  = 180
∠C  = 46

Question 12.

a = 41.1, ∠C  = 35.6, ∠B = 30.4

Explanation:
a² = b² + c² − 2bc cos A
a² = 23² + 26² – 2(23)(26) cos 114
a² = 529 + 676 – 1196 x -0.406
a² = 1690.5
a = 41.1
$$\frac { sin 114 }{ 41.1 }$$ = $$\frac { sin B }{ 23 }$$
0.02 = $$\frac { sin B }{ 23 }$$
sin B = 0.507
∠B = 30.4
∠A + ∠B + ∠C  = 180
114 + 30.4 + ∠C  = 180
∠C  = 35.6

Question 13.

∠A = 41.4, ∠B = 81.8, ∠C  = 56.8

Explanation:
a² = b² + c² − 2bc cos A
4² = 6² + 5² – 2(6)(5) cos A
16 = 36 + 25 – 60 cos A
-45 = – 60 cos A
cos A = 0.75
∠A = 41.4
$$\frac { sin 41.4 }{ 4 }$$ = $$\frac { sin B }{ 6 }$$
0.165 = $$\frac { sin B }{ 6 }$$
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C  = 180
41.4 + 81.8 + ∠C  = 180
∠C  = 56.8

Question 14.

∠B = 81.8, ∠A = 58.6, ∠C  = 39.6

Explanation:
a² = b² + c² − 2bc cos A
23² = 27² + 16² – 2(27)(16) cos A
529 = 729 + 256 – 864 cos A
456 = 864 cos A
cos A = 0.52
∠A = 58.6
$$\frac { sin 58.6 }{ 23 }$$ = $$\frac { sin B }{ 27 }$$
0.03 = $$\frac { sin B }{ 27 }$$
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C  = 180
58.6 + 81.8 + ∠C  = 180
∠C  = 39.6

### Exercise 9.7 Law of Sines and Law of Cosines

Vocabulary and Core Concept Check

Question 1.
WRITING
What type of triangle would you use the Law of Sines or the Law of Cosines to solve?

Question 2.
VOCABULARY
What information do you need to use the Law of Sines?

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 8, use a calculator to find the trigonometric ratio, Round your answer to four decimal places.

Question 3.
sin 127°

Question 4.
sin 98°

sin 98° = 0.9902

Question 5.
cos 139°

Question 6.
cos 108°

cos 108° = -0.309

Question 7.
tan 165°

Question 8.
tan 116°

tan 116° = -2.0503

In Exercises 9 – 12, find the area of the triangle. Round your answer to the nearest tenth.

Question 9.

Question 10.

Area = $$\frac{1}{2}$$bc sin A
Area = $$\frac{1}{2}$$(28)(24) sin83
Area = 332.64

Question 11.

Question 12.

Area = $$\frac{1}{2}$$ab sin C
Area = $$\frac{1}{2}$$(15)(7) sin 96
Area = 51.9

In Exercises 13 – 18. solve the triangle. Round decimal answers to the nearest tenth.

Question 13.

Question 14.

∠B = 38.3, ∠A = 37.7, a = 15.7

Explanation:
$$\frac { sin B }{16 }$$ = $$\frac { sin 104 }{ 25 }$$
sin B = 0.62
∠B = 38.3
∠A + ∠B + ∠C = 180
∠A + 38.3 + 104 = 180
∠A = 37.7
$$\frac { sin 37.7 }{ a }$$ = $$\frac { sin 104 }{ 25 }$$
$$\frac { 0.61 }{ a }$$ = 0.0388
a = 15.7

Question 15.

Question 16.

∠B = 65, b = 33.55, a = 24.4

Explanation:
∠A + ∠B + ∠C = 180
42 + 73 + ∠B = 180
∠B = 65
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 65 }{ b }$$ = $$\frac { sin 73 }{ 34 }$$
b = 33.55
$$\frac { sin A }{ a }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 42 }{ a }$$ = $$\frac { sin 73 }{ 34 }$$
a = 24.4

Question 17.

Question 18.

∠C = 90, b = 39.56, a = 17.6

Explanation:
∠A + ∠B + ∠C = 180
24 + 66 + ∠C = 180
∠C = 90
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 66 }{ b }$$ = $$\frac { sin 90 }{ 43 }$$
$$\frac { .91 }{ b }$$ = 0.023
b = 39.56
$$\frac { sin A }{ a }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 24 }{a }$$ = $$\frac { sin 90 }{ 43 }$$
$$\frac { 0.406 }{a }$$ = 0.023
a = 17.6

In Exercises 19 – 24, solve the triangle. Round decimal answers to the nearest tenth.

Question 19.

Question 20.

b = 29.9, ∠A = 26.1, ∠C = 15.07

Explanation:
b² = a² + c² – 2ac cos B
b² = 20² + 12² – 2(12)(20) cos 138
b² = 400 + 144 – 480 (-0.74)
b² = 899.2
b = 29.9
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin 138 }{ 29.9 }$$ = $$\frac { sin A }{ 20 }$$
$$\frac { 0.66 }{ 29.9 }$$ = $$\frac { sin A }{ 20 }$$
sin A = 0.44
∠A = 26.1
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 138 }{ 29.9 }$$ = $$\frac { sin C }{ 12 }$$
sin C = 0.26
∠C = 15.07

Question 21.

Question 22.

∠A = 107.3, ∠B = 51.6, ∠C = 21.1

Explanation:
b² = a² + c² – 2ac cos B
28² = 18² + 13² – 2(18)(13) cos B
784 = 324 + 169 – 468 cos B
291 = 468 cos B
cos B = 0.62
∠B = 51.6
$$\frac { sin C }{ c }$$ = $$\frac { sin B }{ b }$$
$$\frac { sin C }{ 13 }$$ = $$\frac { sin 51.6 }{ 28 }$$
sin C = 0.36
∠C = 21.1
∠A + ∠B + ∠C = 180
51.6 + 21.1 + ∠A = 180
∠A = 107.3

Question 23.

Question 24.

∠A = 23, ∠B = 132.1, ∠C = 24.9

Explanation:
b² = a² + c² – 2ac cos B
5² = 12² + 13² – 2(12)(13) cos B
25 = 144 + 169 – 312 cos B
288 = 312 cos B
cos B = 0.92
∠B = 23
$$\frac { sin C }{ c }$$ = $$\frac { sin B }{ b }$$
$$\frac { sin C }{ 13 }$$ = $$\frac { sin 23 }{ 5 }$$
sin C = 1.014
∠C = 24.9
∠A + ∠B + ∠C = 180
23 + 24.9 + ∠B = 180
∠B = 132.1

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ C.

Question 26.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ A in ∆ABC when a = 19, b = 21, and c = 11.

a² = b² + c² – 2bc cos A
19² = 21² + 11² – 2(21)(11) cos A
361 = 441 + 121 – 462 cosA
201 = 462 cosA
cos A = 0.43
∠A = 64.5

COMPARING METHODS

In Exercise 27 – 32. tell whether you would use the Law of Sines, the Law of Cosines. or the Pythagorean Theorem (Theorem 9.1) and trigonometric ratios to solve the triangle with the given information. Explain your reasoning. Then solve the triangle.

Question 27.
m ∠ A = 72°, m ∠ B = 44°, b = 14

Question 28.
m ∠ B = 98°, m ∠ C = 37°, a = 18

∠A = 45, b = 25.38, c = 15.38

Explanation:
∠A + ∠B + ∠C = 180
∠A + 98 + 37 = 180
∠A = 45
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin 98 }{ b }$$ = $$\frac { sin 45 }{ 18 }$$
$$\frac { 0.99 }{ b }$$ = 0.039
b = 25.38
$$\frac { sin A }{ a }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 45 }{ 18 }$$ = $$\frac { sin 37 }{ c }$$
0.039 = $$\frac { sin 37 }{ c }$$
c = 15.38

Question 29.
m ∠ C = 65°, a = 12, b = 21

Question 30.
m ∠ B = 90°, a = 15, c = 6

b = 3√29, ∠A = 66.9, ∠C = 23.1

Explanation:
b² = a² + c²- 2ac cos B
b² = 15² + 6² – 2(15)(6) cos 90
= 225 + 36 – 180(0)
b² = 261
b = 3√29
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin 90 }{ 3√29 }$$ = $$\frac { sin A }{ 15 }$$
sin A = 0.92
∠A = 66.9
∠A + ∠B + ∠C = 180
66.9 + 90 + ∠C = 180
∠C = 23.1

Question 31.
m ∠ C = 40°, b = 27, c = 36

Question 32.
a = 34, b = 19, c = 27

∠B = 33.9, ∠A = 78.5, ∠C = 67.6

Explanation:
b² = a² + c²- 2ac cos B
19² = 34² + 27²- 2(34)(27) cos B
361 = 1156 + 729 – 1836 cos B
cos B = 0.83
∠B = 33.9
$$\frac { sin 33.9 }{ 19 }$$ = $$\frac { sin A }{ 34 }$$
sin A = 0.98
∠A = 78.5
∠A + ∠B + ∠C = 180
78.5 + 33.9 + ∠C = 180
∠C = 67.6

Question 33.
MODELING WITH MATHEMATICS
You and your friend are standing on the baseline of a basketball court. You bounce a basketball to your friend, as shown in the diagram. What is the distance between you and your friend?

Question 34.
MODELING WITH MATHEMATICS
A zip line is constructed across a valley, as shown in the diagram. What is the width w of the valley?

w = 92.5 ft

Explanation:
w² = 25² + 84² – 2(25)(84) cos 102
w² = 7681 – 4200 cos 102
w = 92.5 ft

Question 35.
MODELING WITH MATHEMATICS
You are on the observation deck of the Empire State Building looking at the Chrysler Building. When you turn 145° clockwise, you see the Statue of Liberty. You know that the Chrysler Building and the Empire Slate Building arc about 0.6 mile apart and that the Chrysler Building and the Statue of Liberty are about 5.6 miles apart. Estimate the distance between the Empire State Building and the Statue of Liberty.

Question 36.
MODELING WITH MATHEMATICS
The Leaning Tower of Pisa in Italy has a height of 183 feet and is 4° off vertical. Find the horizontal distance d that the top of the tower is off vertical.

Question 37.
MAKING AN ARGUMENT
Your friend says that the Law of Sines can be used to find JK. Your cousin says that the Law of Cosines can be used to find JK. Who is correct’? Explain your reasoning.

Question 38.
REASONING
Use ∆XYZ

a. Can you use the Law of Sines to solve ∆XYZ ? Explain your reasoning.

b. Can you use another method to solve ∆XYZ ? Explain your reasoning.

Question 39.
MAKING AN ARGUMENT
Your friend calculates the area of the triangle using the formula A = $$\frac{1}{2}$$qr sin S and says that the area is approximately 208.6 square units. Is your friend correct? Explain your reasoning.

Question 40.
MODELING WITH MATHEMATICS
You are fertilizing a triangular garden. One side of the garden is 62 feet long, and another side is 54 feet long. The angle opposite the 62-foot side is 58°.
a. Draw a diagram to represent this situation.
b. Use the Law of Sines to solve the triangle from part (a).
c. One bag of fertilizer covers an area of 200 square feet. How many bags of fertilizer will you need to cover the entire garden?

C = 47.6, A = 74.4, a = 70.4
9 bags of fertilizer.

Question 41.
MODELING WITH MATHEMATICS
A golfer hits a drive 260 yards on a hole that is 400 yards long. The shot is 15° off target.

a. What is the distance x from the golfer’s ball to the hole?
b. Assume the golfer is able to hit the ball precisely the distance found in part (a). What is the maximum angle θ (theta) by which the ball can be off target in order to land no more than 10 yards fr0m the hole?

Question 42.
COMPARING METHODS
A building is constructed on top of a cliff that is 300 meters high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is 72° and the angle of elevation to the top of the cliff is 63°.
a. How far away is the person from the base of the cliff?

b. Describe two different methods you can use to find the height of the building. Use one of these methods to find the building’s height.

Consider △SYZ and evaluate d using tangent function
tan SYZ = $$\frac { 300 }{ d }$$
d = $$\frac { 300 }{ tan 63 }$$
d = 152.86
The person is standing 152.86 m away from the base of the cliff.
Consider △XYS and evaluate h + 300
tan XYZ = $$\frac { h + 300 }{ d }$$
h = 152.86 x tan 72 – 300
h = 170.45
The building is 170.45 m high.

Question 43.
MATHEMATICAL CONNECTIONS
Find the values of x and y.

Question 44.
HOW DO YOU SEE IT?
Would you use the Law of Sines or the Law of Cosines to solve the triangle?

Question 45.
REWRITING A FORMULA
A Simplify the Law of Cosines for when the given angle is a right angle.

Question 46.
THOUGHT PROVOKING
Consider any triangle with side lengths of a, b, and c. Calculate the value of s, which is half the perimeter of the triangle. What measurement of the triangle is represented by $$\sqrt{s(s-a)(s-b)(s-c)} ?$$

Question 47.
ANALYZING RELATIONSHIPS
The ambiguous case of the Law of Sines occurs when you are given the measure of one acute angle. the length of one adjacent side, and the length of the side opposite that angle, which is less than the length of the adjacent side. This results in two possible triangles. Using the given information, find two possible solutions for ∆ABC
Draw a diagram for each triangle.
(Hint: The inverse sine function gives only acute angle measures. so consider the acute angle and its supplement for ∠B.)

a. m ∠ A = 40°, a = 13, b = 16
b. m ∠ A = 21°, a = 17, b = 32

Question 48.
ABSTRACT REASONING
Use the Law of Cosines to show that the measure of each angle of an equilateral triangle is 60°. Explain your reasoning.

a² = b² + c²- 2bc cos A
a² = a² + a² – 2 aa cos A
a² = 2a² coas A
cos A = 1/2
∠A = 60

Question 49.
CRITICAL THINKING
An airplane flies 55° east of north from City A to City B. a distance of 470 miles. Another airplane flies 7° north of east from City A to City C. a distance of 890 miles. What is the distance between Cities B and C?

Question 50.
REWRITING A FORMULA
Follow the steps to derive the formula for the area of a triangle.
Area = $$\frac{1}{2}$$ab sin C.

a. Draw the altitude from vertex B to $$\overline{A C}$$. Label the altitude as h. Write a formula for the area of the triangle using h.

b. Write an equation for sin C

c. Use the results of parts (a) and (b) to write a formula for the area of a triangle that does not include h.

Question 51.
PROVING A THEOREM
Follow the steps to use the formula for the area of a triangle to prove the Law of Sines (Theorem 9.9).

a. Use the derivation in Exercise 50 to explain how to derive the three related formulas for the area of a triangle.
Area = $$\frac{1}{2}$$bc sin A,
Area = $$\frac{1}{2}$$ac sin B,
Area = $$\frac{1}{2}$$ab sin C
b. why can you use the formulas in part (a) to write the following statement?
$$\frac{1}{2}$$bc sin A = $$\frac{1}{2}$$ac sin B = $$\frac{1}{2}$$ab sin C
c. Show how to rewrite the statement in part (b) to prove the Law of Sines. Justify each step.

Question 52.
PROVING A THEOREM
Use the given information to complete the two – column proof of the Law of Cosines (Theorem 9.10).

Given $$\overline{B D}$$ is an altitude of ∆ABC.
Prove a2 = b2 + c2 – 2bc cos A

 Statements Reasons 1. $$\overline{B D}$$ is an altitude of ∆ABC. 1. Given 2. ∆ADB and ∆CDB are right triangles. 2. _______________________ 3. a2 = (b – x)2 + h2 3. _______________________ 4. _______________________ 4. Expand binomial. 5. x2 + h2 = c2 5. _______________________ 6. _______________________ 6. Substitution Property of Equality 7. cos A = $$\frac{x}{c}$$ 7. _______________________ 8. x = c cos A 8. _______________________ 9. a2 = b2 + c2 – 2bc Cos A 9. _______________________

Maintaining Mathematical Proficiency

Find the radius and diameter of the circle.

Question 53.

Question 54.

The radius is 10 in and the diameter is 20 in.

Question 55.

Question 56.

The radius is 50 in and the diameter is 100 in.

### 9.1 The Pythagorean Theorem

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 1.

x = 2√34
The sides will not form a Pythagorean triple.

Explanation:
x² = 6² + 10²
x²= 36 + 100
x = 2√34

Question 2.

x = 12
The sides form a Pythagorean triple.

Explanation:
20² = 16²+ x²
400 = 256 + x²
x² = 144
x = 12

Question 3.

x = 2√30
The sides will not form a Pythagorean triple.

Explanation:
13² = 7² + x²
169 = 49 + x²
x = 2√30

Verify that the segments lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 4.
6, 8, and 9

9² = 81
6² + 8² = 36 + 64 = 100
9² < 6² + 8²
So, the triangle is acute

Question 5.
10, 2√2, and 6√3

10² = 100
(2√2)² + (6√3)² = 8 + 108 = 116
So, the triangle is acute.

Question 6.
13, 18 and 3√55

18² = 324
13² + (3√55)² = 169 + 495 = 664
So, the triangle is acute.

### 9.2 Special Right Triangles

Question 7.

hypotenuse = leg • √2
x = 6√2

Question 8.

longer leg = shorter leg • √3
14 = x • √3
x = 8.08

Question 9.

longer leg = shorter leg • √3
x = 8√3 • √3
x = 24

### 9.3 Similar Right Triangles

Identify the similar triangles. Then find the value of x.

Question 10.

9 = √(6x)
81 = 6x
x = $$\frac { 27 }{ 2 }$$

Question 11.

x = √(6 x 4)
x = 2√6

Question 12.

$$\frac { RP }{ RQ }$$ = $$\frac { SP }{ RS }$$
$$\frac { 9 }{ x }$$ = $$\frac { 6 }{ 3 }$$
x = 3.5

Question 13.

$$\frac { SU }{ ST }$$ = $$\frac { SV }{ VU }$$
$$\frac { 16 }{ 20 }$$ = $$\frac { x }{ (x – 16) }$$
4(x – 16) = 5x
x = 16

Find the geometric mean of the two numbers.

Question 14.
9 and 25

mean = √(9 x 25)
= 15

Question 15.
36 and 48

mean = √(36 x 48)
= 24√3

Question 16.
12 and 42

mean = √(12 x 42)
= 6√14

#### 9.4 The Tangent Ratio

Find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 17.

tan J = $$\frac { Opposite side }{ Adjacent side }$$
tan J = $$\frac { LK }{ JK }$$ = $$\frac { 11 }{ 60 }$$
tan L = $$\frac { JK }{ LK }$$ = $$\frac { 60 }{ 11 }$$

Question 18.

tan P = $$\frac { MN }{ MP }$$ = $$\frac { 35 }{ 12 }$$
tan N = $$\frac { MP }{ MN }$$ = $$\frac { 12 }{ 35 }$$

Question 19.

tan A = $$\frac { BC }{ AC }$$ = $$\frac { 7 }{ 4√2 }$$
tan B = $$\frac { AC }{ BC }$$ = $$\frac { 4√2 }{ 7 }$$

Find the value of x. Round your answer to the nearest tenth.

Question 20.

tan 54 = $$\frac { x }{ 32 }$$
1.37 = $$\frac { x }{ 32 }$$
x = 43.8

Question 21.

tan 25 = $$\frac { x }{ 20 }$$
0.46 x 20 = x
x = 9.2

Question 22.

tan 38 = $$\frac { 10 }{ x }$$
x = 12.82

Question 23.
The angle between the bottom of a fence and the top of a tree is 75°. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot.

tan 75 = $$\frac { x }{ 4 }$$
x = 14.92

### 9.5 The Sine and Cosine Ratios

Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 24.

sin X = $$\frac { 3 }{ 5 }$$
sin Z = $$\frac { 4 }{ 5 }$$
cos X = $$\frac { 4 }{ 5 }$$
cos Z = $$\frac { 3 }{ 5 }$$

Question 25.

sin X = $$\frac { 7 }{ √149 }$$
sin Z = $$\frac { 10 }{ √149 }$$
cos X = $$\frac { 10 }{ √149 }$$
cos Z = $$\frac { 7 }{ √149 }$$

Question 26.

sin X = $$\frac { 55 }{ 73 }$$
sin Z = $$\frac { 48 }{ 73 }$$
cos X = $$\frac { 48 }{ 73 }$$
cos Z = $$\frac { 55 }{ 73 }$$

Find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Question 27.

sin 23 = $$\frac { t }{ 34 }$$
t = 13.26
cos 23 = $$\frac { s }{ 34 }$$
s = 31.28

Question 28.

sin 36 = $$\frac { s }{ 5 }$$
s = 2.9
cos 36 = $$\frac { r }{ 5 }$$
r = 4

Question 29.

sin 70 = $$\frac { v }{ 10 }$$
v = 9.39
cos 70 = $$\frac { w }{ 10 }$$
w = 3.42

Question 30.
Write sin 72° in terms of cosine.

sin 72 = cos(90 – 72)
= cos 18 = 0.95

Question 31.
Write cos 29° in terms of sine.

sin 29 = cos(90 – 29)
= cos 61 = 0.48

### 9.6 Solving Right Triangles

Let ∠Q be an acute angle. Use a calculator to approximate the measure of ∠Q to the nearest tenth of a degree.

Question 32.
cos Q = 0.32

cos Q = 0.32
∠Q = inverse cos of .32
∠Q = 71.3

Question 33.
sin Q = 0.91

sin Q = 0.91
∠Q = inverse sin of 0.91
∠Q = 65.5

Question 34.
tan Q = 0.04

tan Q = 0.04
∠Q = inverse tan of 0.04
∠Q = 2.29

Solve the right triangle. Round decimal answers to the nearest tenth.

Question 35.

a = 5√5, ∠A = 47.7, ∠B = 42.3

Explanation:
c² = a² + b²
15² = a² + 10²
a² = 125
a = 5√5
sin A = $$\frac { 5√5 }{ 15 }$$ = 0.74
∠A = 47.7
∠A + ∠B + ∠C = 180
47.7 + ∠B + 90 = 180
∠B = 42.3

Question 36.

NL = 7.59, ∠L = 53, ML = 4.55

Explanation:
cos 37 = $$\frac { 6 }{ NL }$$
NL = 7.59
∠N + ∠M + ∠L = 180
37 + 90 + ∠L = 180
∠L = 53
sin 37 = $$\frac { ML }{ 7.59 }$$
ML = 4.55

Question 37.

XY = 17.34, ∠X = 46, ∠Z = 44

Explanation:
c² = a² + b²
25² = 18²+ b²
b² = 301
b = 17.34
sin X = $$\frac { 18 }{ 25 }$$
∠X = 46
sum of angles = 180
46 + 90 + ∠Z = 180
∠Z = 44

### 9.7 Law of Sines and Law of Cosines

Find the area of ∆ABC with the given side lengths and included angle.

Question 38.
m ∠ B = 124°, a = 9, c = 11

Area = $$\frac { 1 }{ 2 }$$ ac sin B
= $$\frac { 1 }{ 2 }$$ (9 x 11) sin 124
= 40.59

Question 39.
m ∠ A = 68°, b = 13, c = 7

Area = $$\frac { 1 }{ 2 }$$ bc sin A
= $$\frac { 1 }{ 2 }$$ (13 x 7) sin 68
= 41.86

Question 40.
m ∠ C = 79°, a = 25 b = 17

Area = $$\frac { 1 }{ 2 }$$ ab sin C
= $$\frac { 1 }{ 2 }$$ (25 x 17) sin 79
= 208.25

Solve ∆ABC. Round decimal answers to the nearest tenth.

Question 41.
m ∠ A = 112°, a = 9, b = 4

∠B = 24, ∠C = 44, c = 6.76

Explanation:
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin B }{ 4 }$$ = $$\frac { sin 112 }{ 9 }$$
sin B = 0.408
∠B = 24
∠A + ∠B + ∠C = 180
112 + 24 + ∠C = 180
∠C = 44
$$\frac { sin 112 }{ 9 }$$ = $$\frac { sin 44 }{ c }$$
c = 6.76

Question 42.
m ∠ 4 = 28°, m ∠ B = 64°, c = 55

∠C = 88, b = 49.4, a = 25.5

Explanation:
∠A + ∠B + ∠C = 180
28 + 64 + ∠C = 180
∠C = 88
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 64 }{ b }$$ = $$\frac { sin 88 }{ 55 }$$
$$\frac { sin 64 }{ b }$$ = 0.018
b = 49.4
$$\frac { sin 28 }{ a }$$ = $$\frac { sin 88 }{ 55 }$$
a = 25.5

Question 43.
m ∠ C = 48°, b = 20, c = 28

∠B = 31.3, ∠A = 100.7, a = 37.6

Explanation:
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin B }{ 20 }$$ = $$\frac { sin 48 }{ 28 }$$
$$\frac { sin B }{ 20 }$$ = 0.026
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C = 180
∠A + 31.3 + 48 = 180
∠A = 100.7
$$\frac { sin 100.7 }{ a }$$ = $$\frac { sin 48 }{ 28 }$$
a = 37.6

Question 44.
m ∠ B = 25°, a = 8, c = 3

b = 5.45, ∠A = 37.5, ∠C = 117.5

Explanation:
b² = a² + c²- 2ac cos B
b² = 8² + 3² – 2(8 x 3) cos 25 = 73 – 43.2 = 29.8
b = 5.45
$$\frac { sin A }{ 8 }$$ = $$\frac { sin 25 }{ 5.45 }$$
sin A = 0.61
∠A = 37.5
∠A + ∠B + ∠C = 180
37.5 + 25 + ∠C = 180
∠C = 117.5

Question 45.
m ∠ B = 102°, m ∠ C = 43°, b = 21

∠A = 35, c = 14.72, a = 12.3

Explanation:
∠A + ∠B + ∠C = 180
∠A + 102 + 43 = 180
∠A = 35
$$\frac { sin 102 }{ 21 }$$ = $$\frac { sin 43 }{ c }$$
0.046 = $$\frac { sin 43 }{ c }$$
c = 14.72
$$\frac { sin 102 }{ 21 }$$ = $$\frac { sin 35 }{ a }$$
0.046 = $$\frac { sin 35 }{ a }$$
a = 12.3

Question 46.
a = 10, b = 3, c = 12

∠B = 11.7, ∠C = 125.19, ∠A = 43.11

Explanation:
b² = a² + c²- 2ac cos B
3² = 10² + 12²- 2(10 x 12) cos B
9 = 100 + 144 – 240 cos B
cos B = 0.979
∠B = 11.7
a² = b² + c²- 2bc cos A
100 = 9 + 144 – 72 cos A
cos A = 0.73
∠A = 43.11
∠A + ∠B + ∠C = 180
43.11 + 11.7 + ∠C = 180
∠C = 125.19

### Right Triangles and Trigonometry Test

Find the value of each variable. Round your answers to the nearest tenth.

Question 1.

sin 25 = $$\frac { t }{ 18 }$$
t = 7.5
cos 25 = $$\frac { s }{ 18 }$$
s = 16.2

Question 2.

sin 22 = $$\frac { 6 }{ x }$$
x = 16.21
cos 22 = $$\frac { y }{ 16.21 }$$
y = 14.91

Question 3.

tan 40 = $$\frac { k }{ 10 }$$
k = 8.3
cos 40 = $$\frac { 10 }{ j }$$
j = 13.15

Verity that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 4.
16, 30, and 34

34²= 16² + 30²
So, the triangle is a right-angled triangle.

Question 5.
4, √67, and 9

9² = 81
4² + (√67)² = 83
So the triangle is acute

Question 6.
√5. 5. and 5.5

5.5² = 30.25
√5² + 5² = 30
So the triangle is obtuse

Solve ∆ABC. Round decimal answers to the nearest tenth.

Question 7.

c = 12.08, ∠A = 24.22, ∠C = 65.78

Explanation:
tan A = $$\frac { 5 }{ 11 }$$
∠A = 24.22
c² = 11²+ 5²
c = 12.08
24.22 + 90 + ∠C = 180
∠C = 65.78

Question 8.

∠B = 35.4, ∠C = 71.6, c = 17.9

Explanation:
$$\frac { sin 73 }{ 18 }$$ = $$\frac { sin B }{ 11 }$$
sin B = 0.58
∠B = 35.4
73 + 35.4 + ∠C = 180
∠C = 71.6
$$\frac { sin 73 }{ 18 }$$ = $$\frac { sin 71.6 }{ c }$$
c = 17.9

Question 9.

BC = 4.54, ∠B = 59.3, ∠A = 30.7

Explanation:
9.2² = 8² + x²
x = 4.54
$$\frac { sin 90 }{ 9.2 }$$ = $$\frac { sin B }{ 8 }$$
sin B = 0.86
∠B = 59.3
∠A + 59.3 + 90 = 180
∠A = 30.7

Question 10.
m ∠ A = 103°, b = 12, c = 24

a = 29, ∠B = 53.5

Explanation:
a² = b² + c²- 2bc cos A
a² = 144 + 24² – 2(12 x 24) cos 103
a = 29
$$\frac { sin B }{ 12 }$$ = $$\frac { sin 103 }{ 29 }$$
∠B = 23.5
∠C = 180 – (103 + 23.5) = 53.5

Question 11.
m ∠ A = 26°, m ∠ C = 35°, b = 13

∠B = 119, a = 6.42, c = 8.5

Explanation:
∠B + 26 + 35 = 180
∠B = 119
$$\frac { sin 119 }{ 13 }$$ = $$\frac { sin 26 }{ a }$$
a = 6.42
$$\frac { sin 119 }{ 13 }$$ = $$\frac { sin 35 }{ c }$$
c = 8.5

Question 12.
a = 38, b = 31, c = 35

∠B=50.2, ∠C = 59.8, ∠A = 70

Explanation:
b² = a² + c²- 2ac cos B
31² = 38²+ 35²- 2(35 x 38) cos B
cos B = 0.64
∠B=50.2
a² = b² + c²- 2bc cos A
38² = 31²+ 35²- 2(31 x 35) cos A
cos A = 0.341
∠A = 70
∠C = 59.8

Question 13.
Write cos 53° in terms of sine.

cos 53° = sin (90 – 53) = sin 37

Find the value of each variable. Write your answers in simplest form.

Question 14.

sin 45 = $$\frac { 16 }{ q }$$
q = 22.6
cos 45 = $$\frac { r }{ q }$$
r = 16

Question 15.

Question 16.

sin 30 = $$\frac { f }{ 9.2 }$$
f = 4.6
cos 30 = $$\frac { 8 }{ h }$$
h = 9.2

Question 17.
In ∆QRS, m ∠ R = 57°, q = 9, and s = 5. Find the area of ∆QRS.

Area = $$\frac { 1 }{ 2 }$$ qs sin R
= $$\frac { 1 }{ 2 }$$ (9 x 5) sin 57 = 18.675

Question 18.
You are given the measures of both acute angles of a right triangle. Can you determine the side lengths? Explain.

No.

Question 19.
You are at a parade looking up at a large balloon floating directly above the street. You are 60 feet from a point on the street directly beneath the balloon. To see the top of the balloon, you look up at an angle of 53°. To see the bottom of the balloon, you look up at an angle of 29°. Estimate the height h of the balloon.

Question 20.
You warn to take a picture of a statue on Easter Island, called a moai. The moai is about 13 feet tall. Your camera is on a tripod that is 5 feet tall. The vertical viewing angle of your camera is set at 90°. How far from the moai should you stand so that the entire height of the moai is perfectly framed in the photo?

### Right Triangles and Trigonometry Cummulative Assessment

Question 1.
The size of a laptop screen is measured by the length of its diagonal. you Want to purchase a laptop with the largest screen possible. Which laptop should you buy?
(A)

(B)

(C)

(D)

(B)

Explanation:
(a) d = √9² + 12² = 15
(b) d = √11.25² + 20² = 22.94
(c) d = √12² + 6.75² = 13.76
(d) d = √8² + 6² = 10

Question 2.
In ∆PQR and ∆SQT, S is between P and Q, T is between R and Q, and  What must be true about $$\overline{S T}$$ and $$\overline{P R}$$? Select all that apply.
$$\overline{S T}$$ ⊥ $$\overline{P R}$$      $$\overline{S T}$$ || $$\overline{P R}$$    ST = PR        ST = $$\frac{1}{2}$$PR

Question 3.
In the diagram, ∆JKL ~ ∆QRS. Choose the symbol that makes each statement true.

<      =       >
sin J ___________ sin Q                 sin L ___________ cos J                   cos L ___________ tan Q
cos S ___________ cos                  J cos J ___________ sin S                 tan J ___________ tan Q
tan L ___________ tan Q               tan S ___________ cos Q                sin Q ___________ cos L
sin J = sin Q                 sin L = cos J                   cos L = tan Q
cos S > cos J                  cos J > sin S                 tan J = tan Q
tan L < tan Q               tan S > cos Q                sin Q = cos L

Question 4.
A surveyor makes the measurements shown. What is the width of the river.

tan 34 = $$\frac { AB }{ 84 }$$
AB = 56.28

Question 5.
Create as many true equations as possible.

____________ = ______________

sin X            cos X           tan x           $$\frac{X Y}{X Z}$$           $$\frac{Y Z}{X Z}$$

Sin Z            cos Z           tan Z           $$\frac{X Y}{Y Z}$$           $$\frac{Y Z}{X Y}$$

sin X = $$\frac{Y Z}{X Z}$$ = cos Z
cos X = $$\frac{X Y}{X Z}$$ = sin Z
tan x = $$\frac{Y Z}{X Y}$$
tan Z = $$\frac{X Y}{Y Z}$$

Question 6.
Prove that quadrilateral DEFG is a kite.
Given $$\overline{H E} \cong \overline{H G}$$, $$\overline{E G}$$ ⊥ $$\overline{D F}$$
Prove $$\overline{F E} \cong \overline{F G}$$, $$\overline{D E} \cong \overline{D G}$$

Question 7.
What are the coordinates of the vertices of the image of ∆QRS after the composition of transformations show?

(A) Q’ (1, 2), R'(5, 4), S'(4, -1)
(B) Q'(- 1, – 2), R’ (- 5, – 4), S’ (- 4, 1)
(C) Q'(3, – 2), R’ (- 1, – 4), S’ (0, 1)
(D) Q’ (-2, 1), R'(- 4, 5), S'(1, 4)

Question 8.
The Red Pyramid in Egypt has a square base. Each side of the base measures 722 feet. The height of the pyramid is 343 fee.

a. Use the side length of the base, the height of the pyramid, and the Pythagorean Theorem to find the slant height, AB, of the pyramid.

343² = h² + 722²
h = 635.3

b. Find AC.
AC = 343

c. Name three possible ways of finding m ∠ 1. Then, find m ∠ 1.
Three possible ways are sin 1, cos 1 and tan 1
tan 1 = $$\frac { 722 }{ 635.3 }$$
∠1 = 48

## Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume

### Circumference, Area, and Volume Maintaining Mathematical Proficiency

Find the surface area of the prism.

Question 1.

The surface area of the prism = 158.

Explanation:
In the above-given question,
given that,
l = 5 ft, w = 8 ft, and h = 3 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(5×8 + 5×3 + 8×3).
surface area = 2(40 + 15 + 24).
surface area = 2(79).
surface area = 158.

Question 2.

The surface area of the triangular prism = 68m.

Explanation:
In the above-given question,
given that,
l = 10 m, p = 4 m, and h = 10.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 8) + 4(10).
surface area = 2(14) + 40.
surface area = 28 + 40.
surface area = 68m.

Question 3.

The surface area of the triangular prism = 42m.

Explanation:
In the above-given question,
given that,
w = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 5) + 4(5).
surface area = 2(11) + 20.
surface area = 22 + 20.
surface area = 42cm.

Find the missing dimension.

Question 4.
A rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?

The length of the rectangle = 9 in.

Explanation:
In the above-given question,
given that,
A rectangle has a perimeter of 28 inches and a width of 5 inches.
length of the rectangle = p/2 – w.
length = 28/2 – 5.
where perimeter = 28 in, and w = 5 in.
length = 14 – 5.
length = 9.
so the length of the rectangle = 9 in.

Question 5.
A triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?

The base of the triangle = 2 cm.

Explanation:
In the above-given question,
given that,
A triangle has an area of 12 sq cm and a height of 12 cm.
The base of the triangle = 2(A)/h.
base = 2(12)/12.
base = 24/12.
base = 2cm.
so the base of the triangle = 2 cm.

Question 6.
A rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?

The length of the rectangle = 12 ft.

Explanation:
In the above-given question,
given that,
A rectangle has an area of 84 sq ft and a width of 7 feet.
area of the rectangle = l x b.
84 = l x 7.
l = 84/7.
l = 12.
so the length of the rectangle = 12 ft.

Question 7.
ABSTRACT REASONING
Write an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?

The surface area of a prism = 2(lw + wh + lh).

Explanation:
In the above-given question,
given that,
length = l, width = w, and height = x inches.
the surface area of the prism = 2(lw + wh + lh).
the solid figure does the prism represent the rectangular prism.

### Circumference, Area, and Volume Monitoring Progress

Draw a net of the three-dimensional figure. Label the dimensions.

Question 1.

The surface area of the prism = 64 cm.

Explanation:
In the above-given question,
given that,
l = 2 ft, w = 4 ft, and h = 4 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(2×4 + 4×4 + 4×2).
surface area = 2(8 + 16 + 8).
surface area = 2(32).
surface area = 64.

Question 2.

The surface area of the prism = 392 m.

Explanation:
In the above-given question,
given that,
l =8 m, w = 12 m, and h = 5 m.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(8×12 + 12×5 + 5×8).
surface area = 2(96 + 60 + 40).
surface area = 2(196).
surface area = 392.

Question 3.

The surface area of the triangular prism = 170 in.

Explanation:
In the above-given question,
given that,
B = 10 in, p = 15 in, and h = 15 in, l = 10 in.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(10) + 15(10).
surface area = 2(10) + 150.
surface area = 20 + 150.
surface area = 170 in.

11.1 Circumference and Arc Length

Exploration 1

Finding the Length of a Circular Arc

Work with a partner: Find the length of each red circular arc.

a. entire circle

b. one-fourth of a circle

c. one-third of a circle

d. five-eights of a circle

Exploration 2

Using Arc Length

Work with a partner: The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. The diameter of the tire is 25 inches. Is the front tire still in contact with the painted box? Explain.

Question 3.
How can you find the length of a circular arc?
The length of a circular arc = 2
LOOKING FOR REGULARITY IN REPEATED REASONING
To be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.

Question 4.
A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution?

### Lesson 11.1 Circumference and Arc Length

Monitoring Progress

Question 1.
Find the circumference of a circle with a diameter of 5 inches.

Circumference C = πd
C = 3.14x 5 = 15.7 in

Question 2.
Find the diameter of a circle with a circumference of 17 feet.

Diameter d = C/π
d = 17/π = 5.41 ft

Find the indicated measure.

Question 3.
arc length of $$\widehat{P Q}$$

arc length of $$\widehat{P Q}$$ is 5.887

Explanation:
$$\widehat{P Q}$$ = $$\frac { 75 }{ 360 }$$ . π(9)
= 5.887

Question 4.
circumference of ⊙N

arc length of LM/C = LM/360
61.26/C = 270/360
C = 81.68

Question 5.

arc length of EF = $$\frac { 60 }{ 360 }$$ • 2πr
10.5 = $$\frac { 1 }{ 6 }$$ • 2πr
r = 10.02

Question 6.
A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?

The car tire have to make 69 revolutions to travel 500 ft.

Explanation:
Circumference C = 2πr = πd
C = 28π
Distance travelled = number of revolutions x C
500 x 12 = number of revolutions x 28π
number of revolutions = 68.2

Question 7.
In Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.

Question 8.

15° = 15 . $$\frac { π radians }{ 180° }$$ = $$\frac { π }{ 12 }$$ radians

Question 9.
Convert $$\frac{4 \pi}{3}$$ radians to degrees.

$$\frac{4 \pi}{3}$$ radians = $$\frac{4 \pi}{3}$$ radians . $$\frac { 180° }{ π radians }$$ = 240 degrees

### Exercise 11.1 Circumference and Arc Length

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The circumference of a circle with diameter d is C = _______ .

Question 2.
WRITING
Describe the difference between an arc measure and an arc length.

An arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 10, find the indicated measure.

Question 3.
circumference of a circle with a radius of 6 inches

Question 4.
diameter of a circle with a circumference of 63 feet

C = 63 ft
πd = 63
d = 20.05

Question 5.
radius of a circle with a circumference of 28π

Question 6.
exact circumference of a circle with a diameter of 5 inches

C = πd
C = 5π = 15.707

Question 7.
arc length of $$\widehat{A B}$$

Question 8.
m$$\widehat{D E}$$

$$\frac { arc length of DE }{ 2πr }$$ = $$\frac { DE }{ 360 }$$
$$\frac { 8.73 }{ 2π(10) }$$ = $$\frac { DE }{ 360 }$$
DE = 50.01°

Question 9.
circumference of ⊙C

Question 10.

$$\frac { arc length of LM }{ 2πr }$$ = $$\frac { LM }{ 360 }$$
$$\frac { 38.95 }{ 2πr }$$ = $$\frac { 260 }{ 360 }$$
r = 8.583

Question 11.
ERROR ANALYSIS
Describe and correct the error in finding the circumference of ⊙C.

Question 12.
ERROR ANALYSIS
Describe and correct the error in finding the length of $$\widehat{G H}$$.

$$\frac { arc length of GH }{ 2πr }$$ = $$\frac { m GH }{ 360 }$$
$$\widehat{G H}$$. = $$\frac { 5 }{ 24 }$$ . 2π(10)
= 13.08

Question 13.
PROBLEM SOLVING
A measuring wheel is used to calculate the length of a path. The diameter of the wheel is 8 inches. The wheel makes 87 complete revolutions along the length of the path. To the nearest foot, how long is the path?

Question 14.
PROBLEM SOLVING
You ride your bicycle 40 meters. How many complete revolutions does the front wheel make?

Circumference of the front wheel = 2π(32.5)
= 65π cm
Distance covered = 40 m = 40 x 100 = 4000 cm
Number of revolutions = $$\frac { 4000 }{ 65π }$$ = 19.58

In Exercises 15-18 find the perimeter of the shaded region.

Question 15.

Question 16.

Two horizontal edges are 2 . 3 = 6
Circumference of circle = 2π(3) = 6π
The perimeter of the shaded region = 6 + 6π

Question 17.

Question 18.

In Exercises 19 – 22, convert the angle measure.

Question 19.

Question 20.

300 • ($$\frac { π }{ 180 }$$) = $$\frac { 5π }{ 3 }$$ radian

Question 21.
Convert $$\frac{11 \pi}{12}$$ radians to degrees.

Question 22.
Convert $$\frac{\pi}{8}$$ radian to degrees.

$$\frac { π }{ 8 }$$ • $$\frac { 180 }{ π }$$) = 22.5°

Question 23.
PROBLEM SOLVING
The London Eye is a Ferris wheel in London, England, that travels at a speed of 0.26 meter per second. How many minutes does it take the London Eye to complete one full revolution?

Question 24.
PROBLEM SOLVING
You are planning to plant a circular garden adjacent to one of the corners of a building, as shown. You can use up to 38 feet of fence to make a border around the garden. What radius (in feet) can the garden have? Choose all that apply. Explain your reasoning.

(A) 7
(B) 8
(C) 9
(D) 10

C = 38 ft
2πr = 38
r = 6.04

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of π

Question 25.
x2 + y2 = 16

Question 26.
(x + 2)2 + (y – 3)2 = 9

The radius of circle (x + 2)² + (y – 3)² = 9 is 3
C = 2πr = 2π(3) = 6π
The circumference of the circle is 6π units.

Question 27.
USING STRUCTURE
A semicircle has endpoints (- 2, 5) and (2, 8). Find the arc length of the semicircle.

Question 28.
REASONING
$$\widehat{E F}$$ is an arc on a circle with radius r. Let x° be the measure of $$\widehat{E F}$$. Describe the effect on the length of $$\widehat{E F}$$ if you (a) double the radius of the circle, and (b) double the measure of $$\widehat{E F}$$.

Question 29.
MAKING AN ARGUMENT
Your friend claims that it is possible for two arcs with the same measure to have different arc lengths. Is your friend correct? Explain your reasoning.

Question 30.
PROBLEM SOLVING
Over 2000 years ago, the Greek scholar Eratosthenes estimated Earth’s circumference by assuming that the Sun’s rays were Parallel. He chose a day when the Sun shone straight down into a well in the city of Syene. At noon, he measured the angle the Sun’s rays made with a vertical stick in the city of Alexandria. Eratosthenes assumed that the distance from Syene to Alexandria was equal to about 575 miles. Explain how Eratosthenes was able to use this information to estimate Earth’s circumference. Then estimate Earth’s circumference.

Question 31.
ANALYZING RELATIONSHIPS
In ⊙C the ratio of the length of $$\widehat{P Q}$$ to the length of $$\widehat{R S}$$ is 2 to 1. What is the ratio of m∠PCQ to m∠RCS?
(A) 4 to 1
(B) 2 to 1
(C) 1 to 4
(D) 1 to 2

Question 32.
ANALYZING RELATIONSHIPS
A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. What is the ratio of the radius r1 of ⊙C to the radius r2 of ⊙P? Explain your reasoning.

Question 33.
PROBLEM SOLVING
How many revolutions does the smaller gear complete during a single revolution of the larger gear?

Question 34.
USING STRUCTURE
Find the circumference of each circle.
a. a circle circumscribed about a right triangle whose legs are 12 inches and 16 inches long

c² = a² + b²
c² = 12²+ 16²= 400
c = 20 in
circumference c = dπ
= 20π = 62.83 in

b. a circle circumscribed about a square with a side length of 6 centimeters

d² = 6²+ 6² = 72
d = 8.49 cm
C = dπ
C = 8.49π
C = 26.67 cm

c. a circle inscribed in an equilateral triangle with a side length of 9 inches

r = $$\frac { a√3 }{ 3 }$$
r = $$\frac { 9√3 }{ 3 }$$
r = 3√3 = 5.2
C = 2πr
C = 2π (5.2) = 32.67 in

Question 35.
REWRITING A FORMULA
Write a formula in terms of the measure θ (theta) of the central angle in radians) that can he used to find the length of an arc of a circle. Then use this formula to find the length of an arc of a circle with a radius of 4 inches and a central angle of $$\frac{3 \pi}{4}$$ radians.

Question 36.
HOW DO YOU SEE IT?
Compare the circumference of ⊙P to the length of $$\widehat{D E}$$. Explain your reasoning.

Question 37.
MAKING AN ARGUMENT
In the diagram. the measure of the red shaded angle is 30°. The arc length a is 2. Your classmate claims that it is possible to find the circumference of the blue circle without finding the radius of either circle. Is your classmate correct? Explain your reasoning.

Question 38.
MODELING WITH MATHEMATICS
What is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning.
a. 1 : 30 P.M.

3π/4

b. 3:15 P.M.

π/24

Question 39.
MATHEMATICAL CONNECTIONS
The sum of the circumferences of circles A, B, and C is 63π. Find AC.

Question 40.
THOUGHT PROVOKING
Is π a rational number? Compare the rational number $$\frac{355}{113}$$ to π. Find a different rational number that is even closer π.

π is not a rational number as it can not be represented as an equivalent fraction. π = 3.14 and 355/113 = 3.14. This fraction resembles that value of π. Therefore a more accurate fraction will be starting by the value of 7 decimla places of π, therefore 3.1415926 x x = a.

Question 41.
PROOF
The circles in the diagram are concentric and $$\overrightarrow{F G}$$ ≅ $$\overrightarrow{G H}$$ Prove that $$\widehat{J K}$$ and $$\widehat{N G}$$ have the same length.

Question 42.
REPEATED REASONING
$$\overline{A B}$$ is divided into four congruent segments, and semicircles with radius r are drawn.

a. What is the sum of the four arc lengths?

b. What would the sum of the arc lengths be if $$\overline{A B}$$ was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning.

Maintaining Mathematical Proficiency

Find the area of the polygon with the given vertices.

Question 43.
X(2, 4), Y(8, – 1), Z(2, – 1)

Question 44.
L(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)

LP = √(-3 + 3)² + (-5 – 1)² = 6
PN = √(4 + 3)² + (-5 + 5)² = 7
MN = √(4 – 4)² + (-5 – 1)² = 6
LM = √(4 + 3)² + (1 – 1)²= 7
Area = 6 x 7 = 42 units

### 11.2 Areas of Circles and Sectors

Exploration 1

Finding the Area of a Sector of a Circle

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.

a. entire circle

b. one – fourth of a circle

c. seven – eighths of a circle.

d. two – thirds of a circle

Exploration 2

Finding the Area of a Circular Sector

Work with a partner: A center pivot irrigation system consists of 400 meters of sprinkler equipment that rotates around a central pivot point at a rate of once every 3 days to irrigate a circular region with a diameter of 800 meters. Find the area of the sector that is irrigated by this system in one day.
REASONING ABSTRACTLY
To be proficient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution.

Question 3.
How can you find the area of a sector of a circle?

Question 4.
In Exploration 2, find the area of the sector that is irrigated in 2 hours.

### Lesson 11.2 Areas of Circles and Sectors

Monitoring progress

Question 1.
Find the area of a circle with a radius of 4.5 meters.

Circle area = πr²
A = π(4.5)² = 20.25π

Question 2.
Find the radius of a circle with an area of 176.7 square feet.

Circle area = πr²
176.7 = πr²
r² = 56.24
r = 7.499

Question 3.
About 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.

The population density is about 4615.49 people per square mile.

Explanation:
A = πr² = π • 2² = 4π
Population density = $$\frac { number of people }{ area of land }$$
= $$\frac { 58000 }{ 4π }$$ = 4615.49

Question 4.
A region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.

The number of people who live in the region are 28274.

Explanation:
A = πr² = π • 3² = 9π
Population density = $$\frac { number of people }{ area of land }$$
Number of people = 1000 x 9π = 28274

Find the indicated measure

Question 5.
area of red sector

The area of red sector = 205.25

Explanation:
m∠FDE = 120°, FE = 120° and FGE = 360° – 120° = 240°
Area of red sector = $$\frac { FE }{ 360° }$$ • πr²
= $$\frac { 120 }{ 360° }$$ • π(14²)
= 205.25

Question 6.
area of blue sector

Area of blue sector = 410.5

Explanation:
Area of blue sector = $$\frac { FGE }{ 360° }$$ • πr²
= $$\frac { 240 }{ 360° }$$ • π(14²) = 410.5

Question 7.
Find the area of ⊙H.

Area of ⊙H = 907.92 sq cm

Explanation:
Area of sector FHG =$$\frac { FG }{ 360° }$$ • Area of ⊙H
214.37 = $$\frac { 85 }{ 360° }$$ • Area of ⊙H
Area of ⊙H = 907.92 sq cm

Question 8.
Find the area of the figure.

Area of triangle = $$\frac { 1 }{ 2 }$$ • 7 • 7
= 24.5 sq m
Area of semi circle = πr²/2
= π(3.5)²/2
= 19.242255
Area of the figure = 24.5 + 19.24 = 43.74 sq m

Question 9.
If you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.

### Exercise 11.2 Areas of Circles and Sectors

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ____________ of a circle is the region bounded by two radii of the circle and their intercepted arc.

Question 2.
WRITING
The arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.

Yes

Explanation:
Area of sector with arc measure x and radius r is s = π/180(xr)
If x becomes doube, then s1 = π/180(2xr) = 2s
This means that if the arc measure doubles, area of the sector also doubles.

Monitoring Progress and Modeling with Mathematics

In Exercise 3 – 10, find the indicated measure,

Question 3.
area of ⊙C

Question 4.
area of ⊙C

Area A = πr²
A = π(10)² = 100π sq in

Question 5.
area of a circle with a radius of 5 inches

Question 6.
area of a circle with a diameter of 16 feet

d = 2r
Circle area = πr² = (π/4)d²
= (π/4)16² = 64π

Question 7.
radius of a circle with an area of 89 square feet

Question 8.
radius of a circle with an area of 380 square inches

A = πr²
380 = πr²
r = 10.99

Question 9.
diameter of a circle with an area of 12.6 square inches

Question 10.
diameter of a circle with an area of 676π square centimeters

Area A = 676π square centimeters
(π/4)d² = 676π
d² = 2704
d = 52

In Exercises 11 – 14, find the indicated measure.

Question 11.
About 210,000 people live in a region with a 12-mile radius. Find the population density in people per square mile.

Question 12.
About 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.

The population density is about 5747 people per square mile.

Explanation:
Area of region = π(6)² = 36π
Population density = $$\frac { Number of people }{ area of land }$$
= $$\frac { 650,000 }{ 36π }$$ = 5747.2

Question 13.
A region with a 4-mile radius has a population density of about 6366 people per square mile. Find the number of people who live in the region.

Question 14.
About 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.

The radius of the region is 7

Explanation:
Population density = $$\frac { Number of people }{ area of land }$$
513 = $$\frac { 79,000 }{ πr² }$$
πr² = 153.99
r = 7

In Exercises 15-18 find the areas of the sectors formed by∠DFE.

Question 15.

Question 16.

Area of sector = $$\frac { 104° }{ 360° }$$ • π(14)²
= 177.88
Area of red region is 177.88 sq cm
Area of blue region = $$\frac { 256° }{ 360° }$$ • π(14)²
= 437.86 sq cm

Question 17.

Question 18.

Area of red region is 10.471 sq ft
Area of the blue region is 39.79 sq ft

Explanation:
Area of sector = $$\frac { 75° }{ 360° }$$ • π(4)²
= 10.471
Area of red region is 10.471 sq ft
Area of blue region = $$\frac { 285° }{ 360° }$$ • π(4)²
= 39.79 sq ft

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the circle.

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding the area of sector XZY when the area of ⊙Z is 255 square feet.

Area of ⊙Z is 255 square feet
πr² = 255
r = 9
Area of sector XZY = $$\frac { 115 }{ 360 }$$ • 255
n = 81.458 sq ft

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.

Question 21.
area of ⊙M

Question 22.

Explanation:
Area of region = $$\frac { 89 }{ 360 }$$ . Area of ⊙M
12.36 = $$\frac { 89 }{ 360 }$$ . Area of ⊙M
Area of ⊙M = 49.99
πr² = 49.99
r = 3.98

In Exercises 23 – 28, find the area of the shaded region.

Question 23.

Question 24.

The area of the shaded region is 85.840 sq in.

Explanation:
Area of square = 20² = 400
Diameter of one circle = 10
radius of one circle = 5 in
Area of one circle = π(5)² = 78.53
Areas of four circle = 314.159
Area of shaded region = 400 – 314.159 = 85.840

Question 25.

Question 26.

Area of smaller semicircle = $$\frac { 1 }{ 2 }$$(π(8)²) = 100.53
Area of lager semicircle = $$\frac { 1 }{ 2 }$$(π(16)²) = 402.123