Eureka Math Grade 4 Module 4 Lesson 16 Answer Key

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Eureka Math Grade 4 Module 4 Lesson 16 Problem Set Answer Key

Question 1.
On the grid paper, draw at least one quadrilateral to fit the description. Use the given segment as one segment of the quadrilateral. Name the figure you drew using one of the terms below.
Eureka Math Grade 4 Module 4 Lesson 16 Problem Set Answer Key 1
Answer:

Question 2.
On the grid paper, draw at least one quadrilateral to fit the description. Use the given segment as one segment of the quadrilateral. Name the figure you drew using one of the terms below.
Eureka Math Grade 4 Module 4 Lesson 16 Problem Set Answer Key 2
Answer:

Question 3.
Explain the attributes that make a rhombus different from a rectangle.
Answer:

Question 4.
Explain the attribute that makes a square different from a rhombus.
Answer:

Eureka Math Grade 4 Module 4 Lesson 16 Exit Ticket Answer Key

Question 1.
Construct a parallelogram that does not have any right angles on a rectangular grid.
Eureka Math Grade 4 Module 4 Lesson 16 Exit Ticket Answer Key 3
Answer:

Question 2.
Construct a rectangle on a triangular grid.
Eureka Math Grade 4 Module 4 Lesson 16 Exit Ticket Answer Key 4
Answer:

Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key

Use the grid to construct the following. Name the figure you drew using one of the terms in the word box.
Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key 5

Question 1.
Construct a quadrilateral with only one set of parallel sides. Which shape did you create?
Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key 6
Answer:

Question 2.
Construct a quadrilateral with one set of parallel sides and two right angles. Which shape did you create?
Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key 7
Answer:

Question 3.
Construct a quadrilateral with two sets of parallel sides. Which shape did you create?
Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key 8
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Question 4.
Construct a quadrilateral with all sides of equal length. Which shape did you create?
Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key 9
Answer:

Question 5.
Construct a rectangle with all sides of equal length. Which shape did you create?
Eureka Math Grade 4 Module 4 Lesson 16 Homework Answer Key 10
Answer:

Eureka Math Grade 4 Module 4 Lesson 15 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 15 Answer Key

Eureka Math Grade 4 Module 4 Lesson 15 Problem Set Answer Key

Construct the figures with the given attributes. Name the shape you created. Be as specific as possible. Use extra blank paper as needed.

Question 1.
Construct quadrilaterals with at least one set of parallel sides.
Answer:

Question 2.
Construct a quadrilateral with two sets of parallel sides.
Answer:

Question 3.
Construct a parallelogram with four right angles.
Answer:

Question 4.
Construct a rectangle with all sides the same length.
Answer:

Question 5.
Use the word bank to name each shape, being as specific as possible.
Eureka Math Grade 4 Module 4 Lesson 15 Problem Set Answer Key 1

a.
Eureka Math Grade 4 Module 4 Lesson 15 Problem Set Answer Key 2
Answer:

b.
Eureka Math Grade 4 Module 4 Lesson 15 Problem Set Answer Key 3
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c.
Eureka Math Grade 4 Module 4 Lesson 15 Problem Set Answer Key 4
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d.
Eureka Math Grade 4 Module 4 Lesson 15 Problem Set Answer Key 5
Answer:

Question 6.
Explain the attribute that makes a square a special rectangle.
Answer:

Question 7.
Explain the attribute that makes a rectangle a special parallelogram.
Answer:

Question 8.
Explain the attribute that makes a parallelogram a special trapezoid.
Answer:

Eureka Math Grade 4 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.
In the space below, draw a parallelogram.
Answer:

Question 2.
Explain why a rectangle is a special parallelogram.
Answer:

Eureka Math Grade 4 Module 4 Lesson 15 Homework Answer Key

Question 1.
Use the word bank to name each shape, being as specific as possible.
Eureka Math Grade 4 Module 4 Lesson 15 Homework Answer Key 6

a.
Eureka Math Grade 4 Module 4 Lesson 15 Homework Answer Key 7
Answer:

b.
Eureka Math Grade 4 Module 4 Lesson 15 Homework Answer Key 8
Answer:

c.
Eureka Math Grade 4 Module 4 Lesson 15 Homework Answer Key 9
Answer:

d.
Eureka Math Grade 4 Module 4 Lesson 15 Homework Answer Key 10
Answer:

Question 2.
Explain the attribute that makes a square a special rectangle.
Answer:

Question 3.
Explain the attribute that makes a rectangle a special parallelogram.
Answer:

Question 4.
Explain the attribute that makes a parallelogram a special trapezoid.
Answer:

Question 5.
Construct the following figures based on the given attributes. Give a name to each figure you construct. Be as specific as possible.

a. A quadrilateral with four sides the same length and four right angles.
Answer:

b. A quadrilateral with two sets of parallel sides.
Answer:

c. A quadrilateral with only one set of parallel sides.
Answer:

d. A parallelogram with four right angles.
Answer:

Eureka Math Grade 4 Module 4 Lesson 14 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 14 Answer Key

Eureka Math Grade 4 Module 4 Lesson 14 Problem Set Answer Key

1. Draw triangles that fit the following classifications. Use a ruler and protractor. Label the side lengths and angles.
Question a.
Right and isosceles
Answer:

Question b.
Obtuse and scalene
Answer:

Question c.
Acute and scalene
Answer:

Question d.
Acute and isosceles
Answer:

Question 2.
Draw all possible lines of symmetry in the triangles above. Explain why some of the triangles do not have lines of symmetry.
Answer:

Are the following statements true or false? Explain using pictures or words.

Question 3.
If ABC is an equilateral triangle, \(\overline{B C}\) must be 2 cm. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Problem Set Answer Key 1
Answer:

Question 4.
A triangle cannot have one obtuse angle and one right angle. True or False?
Answer:

Question 5.
∆ EFG can be described as a right triangle and an isosceles triangle. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Problem Set Answer Key 2
Answer:

Question 6.
An equilateral triangle is isosceles. True or False?
Answer:

Extension:
In ∆ HIJ, a = b. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Problem Set Answer Key 3
Answer:

Eureka Math Grade 4 Module 4 Lesson 14 Exit Ticket Answer Key

Question 1.
Draw an obtuse isosceles triangle, and then draw any lines of symmetry if they exist.
Answer:

Question 2.
Draw a right scalene triangle, and then draw any lines of symmetry if they exist.
Answer:

Question 3.
Every triangle has at least ____ acute angles.
Answer:

Eureka Math Grade 4 Module 4 Lesson 14 Homework Answer Key

Question 1.
Draw triangles that fit the following classifications. Use a ruler and protractor. Label the side lengths and angles.

a. Right and isosceles
Answer:

b. Right and scalene
Answer:

c. Obtuse and isosceles
Answer:

d. Acute and scalene
Answer:

Question 2.
Draw all possible lines of symmetry in the triangles above. Explain why some of the triangles do not have lines of symmetry.

Are the following statements true or false? Explain.

Question 3.
∆ ABC is an isosceles triangle. \(\overline{A B}\) must be 2 cm. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Homework Answer Key 4
Answer:

Question 4.
A triangle cannot have both an acute angle and a right angle. True or False?
Answer:

Question 5.
∆ XYZ can be described as both equilateral and acute. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Homework Answer Key 5
Answer:

Question 6.
A right triangle is always scalene. True or False?
Answer:

Extension:
In ∆ ABC, x = y. True or False?
Eureka Math Grade 4 Module 4 Lesson 14 Homework Answer Key 6
Answer:

Eureka Math Grade 4 Module 4 Lesson 13 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 13 Answer Key

Eureka Math Grade 4 Module 4 Lesson 13 Practice Sheet Answer Key

Question 1.
Eureka Math Grade 4 Module 4 Lesson 13 Practice Sheet Answer Key 1
Answer:

Eureka Math Grade 4 Module 4 Lesson 13 Problem Set Answer Key

Question 1.
Classify each triangle by its side lengths and angle measurements. Circle the correct names.
Eureka Math Grade 4 Module 4 Lesson 13 Problem Set Answer Key 2
Answer:

Question 2.
∆ ABC has one line of symmetry as shown. What does this tell you about the measures of ∠A and ∠C?
Eureka Math Grade 4 Module 4 Lesson 13 Problem Set Answer Key 3
Answer:

3. ∆ DEF has three lines of symmetry as shown.
Eureka Math Grade 4 Module 4 Lesson 13 Problem Set Answer Key 4
Question a.
How can the lines of symmetry help you to figure out which angles are equal?

Question b.
∆ DEF has a perimeter of 30 cm. Label the side lengths.
Answer:

Question 4.
Use a ruler to connect points to form two other triangles. Use each point only once. None of the triangles may overlap. One or two points will be unused. Name and classify the three triangles below. The first one has been done for you.
Eureka Math Grade 4 Module 4 Lesson 13 Problem Set Answer Key 5

Eureka Math Grade 4 Module 4 Lesson 13 Problem Set Answer Key 6
Answer:

Question 5a.
List three points from the grid above that, when connected by segments, do not result in a triangle.
Answer:

Question 5b.
Why didn’t the three points you listed result in a triangle when connected by segments?
Answer:

Question 6.
Can a triangle have two right angles? Explain.
Answer:

Eureka Math Grade 4 Module 4 Lesson 13 Exit Ticket Answer Key

Use appropriate tools to solve the following problems:

Question 1.
The triangles below have been classified by shared attributes (side length or angle type). Use the words acute, right, obtuse, scalene, isosceles, or equilateral to label the headings to identify the way the triangles have been sorted.
Eureka Math Grade 4 Module 4 Lesson 13 Exit Ticket Answer Key 7
Answer:

Question 2.
Draw lines to identify each triangle according to angle type and side length.
Eureka Math Grade 4 Module 4 Lesson 13 Exit Ticket Answer Key 8
Answer:

Question 3.
Identify and draw any lines of symmetry in the triangles in Problem 2.
Answer:

Eureka Math Grade 4 Module 4 Lesson 13 Homework Answer Key

Question 1.
Classify each triangle by its side lengths and angle measurements. Circle the correct names.
Eureka Math Grade 4 Module 4 Lesson 13 Homework Answer Key 9
Answer:

Question 2a.
∆ ABC has one line of symmetry as shown. Is the measure of ∠A greater tyhan, less than or equal to ∠C?
Eureka Math Grade 4 Module 4 Lesson 13 Homework Answer Key 10
Answer:

Question 2b.
∆ DEF is scalene. What do you observe about its angles? Explain.
Eureka Math Grade 4 Module 4 Lesson 13 Homework Answer Key 11
Answer:

Question 3.
Use a ruler to connect points to form two other triangles. Use each point only once. None of the triangles may overlap. Two points will be unused. Name and classify the three triangles below.
Eureka Math Grade 4 Module 4 Lesson 13 Homework Answer Key 12

Eureka Math Grade 4 Module 4 Lesson 13 Homework Answer Key 13
Answer:

Question 4.
If the perimeter of an equilateral triangle is 15 cm, what is the length of each side?
Answer:

Question 5.
Can a triangle have more than one obtuse angle? Explain.
Answer:

Question 6.
Can a triangle have one obtuse angle and one right angle? Explain.
Answer:

Eureka Math Grade 4 Module 4 Lesson 8 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 8 Answer Key

Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key

Question 1.
Joe, Steve, and Bob stood in the middle of the yard and faced the house. Joe turned 90° to the right. Steve turned 180° to the right. Bob turned 270° to the right. Name the object that each boy is now facing.
Joe ____________________
Steve __________________
Bob ___________________
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 1
Answer:
The object that Joe is now facing is the fence.
The object that Steve is now facing is the tree.
The object that Bob is now facing is the barn.

Explanation:
As Joe, Steve, and Bob stood in the middle of the yard and faced the house then Joe turned 90° to the right and Steve turned 180° to the right and Bob turned 270° to the right. So the object that Joe is now facing is the fence and the object that Steve is now facing is the tree and the object that Bob is now facing is the barn.

Question 2.
Monique looked at the clock at the beginning of class and at the end of class. How many degrees did the minute hand turn from the beginning of class until the end?
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 2
Answer:
The minute hand-turned 270°.

Explanation:
Here, Monique looked at the clock at the beginning of class and at the end of class, and the number of degrees did the minute hand turn from the beginning of class until the end is 270°.

Question 3.
The skater jumped into the air and did a 360. What does that mean?
Answer:
He jumped into the car and turned around the complete circle.

Explanation:
As the skater jumped into the air and did a 360 which means that he jumped into the car and turned around the complete circle and now he is facing the way he started.

Question 4.
Mr. Martin drove away from his house without his wallet. He did a 180. Where is he heading now?
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 3
Answer:
Mr. Martin now is heading to his house.

Explanation:
As Mr. Martin drove away from his house without his wallet and then he did a 180, so Mr. Martin now is heading to his house.

Question 5.
John turned the knob of the shower 270° to the right. Draw a picture showing the position of the knob after he turned it.
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 4
Answer:
As John turned the knob of the shower 270° to the right, so the position of the knob after John turned is shown below.
Eureka-Math-Grade-4-Module-4-Lesson-8-Problem-Set-Answer-Key-4

Question 6.
Barb used her scissors to cut out a coupon from the newspaper. How many quarter-turns does she need to turn the paper in order to stay on the lines?
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 5
Answer:
The number of quarter-turns does Barb needs to turn the paper in order to stay on the lines is four quarter turns.

Explanation:
As Barb used her scissors to cut out a coupon from the newspaper, so the number of quarter-turns does Barb needs to turn the paper in order to stay on the lines is four quarter turns.

Question 7.
How many quarter-turns does the picture need to be rotated in order for it to be upright?
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 6
Answer:
We need to rotate the picture one quarter turn to the left or three quarter turns to the right.

Explanation:
Here, we need to rotate the picture one quarter turn to the left or three quarter turns to the right then the picture will be upright.

Question 8.
Meredith faced north. She turned 90° to the right, and then 180° more. In which direction is she now facing?
Eureka Math Grade 4 Module 4 Lesson 8 Problem Set Answer Key 7
Answer:
the Meredith is now facing West.

Explanation:
As Meredith faced north and then she turned 90° to the right, and then 180° more. So the Meredith is now facing West.

Eureka Math Grade 4 Module 4 Lesson 8 Exit Ticket Answer Key

Question 1.
Marty was doing a handstand. Describe how many degrees his body will turn to be upright again.
Eureka Math Grade 4 Module 4 Lesson 8 Exit Ticket Answer Key 8
Answer:
The number of degrees his body will turn to be upright again is 90°.

Explanation:
As Marty was doing a handstand and the number of degrees his body will turn to be upright again is 90°.

Question 2.
Jeffrey started riding his bike at the Eureka Math Grade 4 Module 4 Lesson 8 Exit Ticket Answer Key 9. He travelled north for 3 blocks, then turned 90° to the right and rode for 2 blocks. In which direction was he headed? Sketch his route on the grid below. Each square unit represents 1 block.
Eureka Math Grade 4 Module 4 Lesson 8 Exit Ticket Answer Key 10
Answer:
Jeffery will head in the East direction.

Explanation:
As Jeffrey started riding his bike at Eureka Math Grade 4 Module 4 Lesson 8 Exit Ticket Answer Key 9and he traveled north for 3 blocks, then turned 90° to the right and rode for 2 blocks. So Jeffery will head in the East direction.
Eureka-Math-Grade-4-Module-4-Lesson-8-Exit-Ticket-Answer-Key-10

Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key

Question 1.
Jill, Shyan, and Barb stood in the middle of the yard and faced the barn. Jill turned 900 to the right. Shyan turned 180° to the left. Barb turned 270° to the left. Name the object that each girl is now facing.
Jill ____________________
Shyan __________________
Barb ___________________
Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key 11
Answer:
The object that Jill is now facing is the house.
The object that Shyan is now facing is the fence.
the object that Barb is now facing is the house.

Explanation:
Jill, Shyan, and Barb stood in the middle of the yard and faced the barn and Jill turned 900 to the right and Shyan turned 180° to the left and Barb turned 270° to the left. So the object that Jill is now facing is the house and the object that Shyan is now facing the fence and the object that Barb is now facing is the house.

Question 2.
Allison looked at the clock at the beginning of class and at the end of class. How many degrees did the minute hand turn from the beginning of class until the end?
Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key 12
Answer:
The minute had turned 360 degrees.

Explanation:
As Allison looked at the clock at the beginning of class and at the end of class, so the number of degrees did the minute hand turn from the beginning of the class until the end is 360 degrees.

Question 3.
The snowboarder went off a jump and did a 180. In which direction was the snowboarder facing when he landed? How do you know?
Answer:
The direction in which the snowboarder facing will be backwards.

Explanation:
As the snowboarder went off a jump and did a 180 degree, so the direction in which the snowboarder facing will be backwards. We came to know that as he did a jump of 180 degree by turning back.

Question 4.
As she drove down the icy road, Mrs. Campbell slammed on her brakes. Her car did a 360. Explain what happened to Mrs. Campbell’s car.
Answer:
Mrs. Campbell’s car did a complete turn and ended up facing the down the icy road

Explanation:
As she drove down the icy road, Mrs. Campbell slammed on her brakes so her car did a 360. Here Mrs. Campbell’s car did a complete turn and ended up facing the down the icy road and the same direction she started in.

Question 5.
Jonah turned the knob of the stove two quarter-turns. Draw a picture showing the position of the knob after he turned it.
Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key 13
Answer:
The image will be as shown below.

Explanation:
As Jonah turned the knob of the stove two quarter-turns, so the image will be as shown below.
Eureka-Math-Grade-4-Module-4-Lesson-8-Homework-Answer-Key-13

Question 6.
Betsy used her scissors to cut out a coupon from the newspaper. How many total quarter-turns will she need to rotate the paper in order to cut out the entire coupon?
Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key 14
Answer:
The number of total quarter-turns will she need to rotate the paper in order to cut out the entire coupon is three quarter turns.

Explanation:
As Betsy used her scissors to cut out a coupon from the newspaper. So the number of total quarter-turns will she need to rotate the paper in order to cut out the entire coupon is three quarter turns.

Question 7.
How many quarter-turns does the picture need to be rotated in order for it to be upright?
Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key 15
Answer:
The picture needs to rotate two quarter turns which will be 180 degrees.

Question 8.
David faced north. He turned 180° to the right, and then 270° to the left. In which direction is he now facing?
Eureka Math Grade 4 Module 4 Lesson 8 Homework Answer Key 16
Answer:
The direction of David now facing will be West.

Explanation:
As David faced north and he turned 180° to the right, and then 270° to the left. So the direction of David now facing will be West.

Eureka Math Grade 4 Module 4 Lesson 12 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 12 Answer Key

Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key

Question 1.
Circle the figures that have a correct line of symmetry drawn.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 1
Answer:

Question 2.
Find and draw all lines of symmetry for the following figures. Write the number of lines of symmetry that you found in the blank underneath the shape.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 2
Answer:

3. Half of each figure below has been drawn. Use the line of symmetry, represented by the dashed line, to complete each figure.

Question a.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 3
Answer:

Question b.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 4
Answer:

Question c.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 5
Answer:

Question d.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 6
Answer:

Question 4.
The figure below is a circle. How many lines of symmetry does the figure have? Explain.
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 7
Answer:

Eureka Math Grade 4 Module 4 Lesson 12 Exit Ticket Answer Key

Question 1.
Is the line drawn a line of symmetry? Circle your choice:
Eureka Math Grade 4 Module 4 Lesson 12 Problem Set Answer Key 8
Answer:

Question 2.
Draw as many lines of symmetry as you can find in the figure below.
Eureka Math Grade 4 Module 4 Lesson 12 Exit Ticket Answer Key 9
Answer:

Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key

Question 1.
Circle the figures that have a correct line of symmetry drawn.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 10
Answer:

Question 2.
Find and draw all lines of symmetry for the following figures. Write the number of lines of symmetry that you found in the blank underneath the shape.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 11
Answer:

3.
Half of each figure below has been drawn. Use the line of symmetry, represented by the dashed line, to complete each figure.

Question a.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 12
Answer:

Question b.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 13
Answer:

Question c.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 14
Answer:

Question d.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 15
Answer:

Question 4.
Is there another shape that has the same number of lines of symmetry as a circle? Explain.
Eureka Math Grade 4 Module 4 Lesson 12 Homework Answer Key 16
Answer:

Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key

Engage NY Eureka Math 4th Grade Module 4 End of Module Assessment Answer Key

Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key

Question 1.
Find and draw all lines of symmetry in the following figures. If there are none, write “none.”
a.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 1
Answer:
Eureka-Math-Grade-4-Module-4-End-of-Module-Assessment-Answer-Key-1
One line of symmetry in the above diagram.

Explanation:

A line of symmetry means it is an imaginary line which can divide an object into two identical halves. By observing the diagram the line can divide into 2 parts. So it has one line of symmetry.

b.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 2
Answer:
There are none.

Explanation:

For the diagram given above, there are no symmetry lines.

c.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 3
Answer:
There are three lines of symmetry lines.
Eureka-Math-Grade-4-Module-4-End-of-Module-Assessment-Answer-Key-3

Explanation:
As we know about the line of symmetry from the above diagram we can identify 3 lines of symmetry.

d.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 4
Answer:
Eureka-Math-Grade-4-Module-4-End-of-Module-Assessment-Answer-Key-4
For square, we have 4 lines of symmetrical lines.

Explanation:
By observing the above diagram we can divide the parts into 4 symmetrical lines. Hence square has 4 lines of symmetry.

e.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 5
Answer:
There are no symmetrical lines.

Explanation:
By observing the above diagram we know that the object cannot divide into two identical halves. Hence it has no symmetrical lines.

f.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 6
Answer:
Eureka-Math-Grade-4-Module-4-End-of-Module-Assessment-Answer-Key-6
As per the rectangle, we have 2 lines of symmetry.

Explanation:
By applying the rules of symmetry the object can be divided into 2 parts. So by observing the diagram we know that the rectangle has 2 lines of symmetrical lines as it can be folded in half horizontally or vertically.

g. For each triangle listed below, state whether it is acute, obtuse, or right and whether it is isosceles, equilateral, or scalene.
Triangle a: _________________________ _________________________
Triangle c: _________________________ _________________________
Triangle e: _________________________ _________________________
Answer:
Triangle a: Obtuse, Isosceles
Traingle c: Acute, Equilateral
Triangle e: Right, Scalene

Explanation:
Triangle a is an obtuse angle that has more than 90° and the triangle is an isosceles triangle, therefore, has both two equal sides and two equal angles.
Triangle c is an acute angle that has less than 90° and here the triangle is an equilateral triangle where all the sides and angles are equal.
Triangle e is right angles triangle i.e is exactly 90° therefore no side is equal to one another.

h. How many lines of symmetry does a circle have? What point do all lines of symmetry for a given circle have in common?
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 1
Answer:
As the circle has an infinite amount of lines of symmetry. All lines of symmetry for a circle pass through the center point.

Explanation:
In a circle, each diameter gives an axis of symmetry to the circle. Since an infinite number of diameters can be drawn in a circle; so a circle has infinite lines of symmetry. But, all the symmetries are look-alike, so a circle has only one symmetry of infinite orientations.

Question 2.
In the following figure, QRST is a rectangle. Without using a protractor, determine the measure of ∠RQS Write an equation that could be used to solve the problem.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 8
Let us assume ∠Q as W°
The equation that could be solved the problem is
24° + W° = 90°
W° = 90°-24° = 66°
∠RQS = 66°

Explanation:
To determine the ∠RQS we need to write an equation to solve. By observing the diagram we could write the equation as 24° + W° = 90°.
Hence we can determine the value of ∠RQS.

For each part below, explain how the measure of the unknown angle can be found without using a protractor.
a. Find the measure of ∠D.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 9
Answer:
To measure the angle we have to subtract 360° from 83° because a circle measures 360°.
so 83°+ ∠D = 360°
∠D=360° – 83°
∠D=277°

Explanation:
A complete trip around the edge of a circle is 360 degrees. So to measure the angle D we need to subtract 360° from 83°. Hence the ∠D=360° – 83°
∠D=277°.

b. In this figure, Q, R, and S lie on a line. Find the measure of ∠QRT
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 10
Answer:
To measure the ∠QRT
58°+ ∠QRT = 180°
∠QRT = 180°-58°
∠QRT = 122°

Explanation:
As we know that the line measures 180°, so we need to add 58° with ∠QRT and hence subtract the resultant with 180°.

c. In this figure, Q, R, and S lie on a line, as do P, R, and T. Find the measure of ∠PRS
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 11
Answer:
First we need to measure ∠TRS
∠VRQ+∠VRT+∠TRS = 180°
48°+74°+∠TRS = 180°
∠TRS = 180°-48°-74°
∠TRS = 180° – 122°
∠TRS = 58°
∠TRS + ∠PRS =180°
58°+ ∠PRS =180°
∠PRS = 180°-58°
∠PRS = 122°

Explanation:
First, to calculate ∠PRS we need to measure ∠TRS. Since Q, R, and S lie on the same line. As we know 48°+74°+∠TRS = 180°. That means we get ∠TRS = 58°.
Also, P, R, and T line on the same line, as we know ∠TRS + ∠PRS = 180°. That means ∠PRS is 122°

Question 3.
Mike drew some two-dimensional figures.
Sketch the figures, and answer each part about the figures that Mike drew.

a. He drew a four-sided figure with four right angles. It is 4 cm long and 3 cm wide.
What type of quadrilateral did Mike draw?
How many lines of symmetry does it have?
Answer:
Mike drew a four-sided figure with four right angles that have 4 cm long and 3 cm wide.
The type of quadrilateral mike drew is a rectangle.
The rectangle has 2 lines of symmetry.

Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-4

Explanation:
As mentioned in the above question mike drew a four-sided figure with four right angles that have 4 cm long and 3 cm wide. A rectangle has two pairs of opposite sides parallel, and four right angles. So the figure mike drew is a rectangle and it has 2 lines of symmetry.

b. He drew a quadrilateral with four equal sides and no right angles.
What type of quadrilateral did Mike draw?
How many lines of symmetry does it have?
Answer:
Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-1
The type of quadrilateral that mike drew is Rhombus.
There are 2 lines of symmetry that the rhombus has.
Explanation:
As the rhombus has four equal-length sides and opposite sides parallel to each other. Hence by observing the diagram we can know that it is a rhombus that has 2 lines of symmetry.

c. He drew a triangle with one right angle and sides that measure 6 cm, 8 cm, and 10 cm.
Classify the type of triangle Mike drew based on side length and angle measure.
How many lines of symmetry does it have?
Answer:

The type of triangle that mike drew is a right-angle triangle. And by observing the dimensions, one can clearly name it as a scalene triangle as all sides have different dimensions.
Explanation:
As the triangle is right angles i.e it has 90°. Hence the triangle mike drew is a right-angled triangle. As the triangle has different dimensions therefore we can name the triangle as a scalene triangle.

d. Using the dimensions given, draw the same shape that Mike drew in Part (c).
Answer:
Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-3

Explanation:
By using the given dimensions mike drew the shape of the right-angled triangle.

e. Mike drew this figure. Without using a protractor, find the sum of ∠FJK, ∠KJH, and ∠HJG.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 12
Answer:
360°- 90° = 270°
∠FJK+∠KJH +∠HJG =270°

Explanation:
Here we need to find the sum of ∠FJK, ∠KJH, and ∠HJG. As we already know that the complete trip around the edge of a circle is 360 degrees. Now we need to subtract 360° from 90° to get a sum of other angles. By doing so we get the sum of angles as 270°.

f. Points F, J, and H lie on a line. What is the measure of ∠KJH if ∠FJK measures 45°? Write an equation that could be used to determine the measure of ∠KJH.
Answer:
The equation to measure ∠KJH is
45°+∠KJH =180°
∠KJH = 180°-45°
∠KJH = 135°

Explanation:
We have ∠FJK and points F, J, and H lie on the same line. So to measure ∠KJH we need to write down an equation that is 45°+∠KJH=180°. Hence ∠KJH is 135°.

g.
Mike used a protractor to measure ∠ABC as shown below and said the result was exactly 130° Do you agree or disagree? Explain your thinking.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 13
Answer:
I disagree. Because mike didn’t use the protractor properly. The 0° should match up with BA but it doesn’t match up.

Explanation:
We can completely disagree with mike’s statement because he doesn’t measure the exact result as he didn’t use the protractor properly. In the diagram, mike lined up the bottom of the protractor with BA instead.

h. Below is half of a line-symmetric figure and its line of symmetry. Use a ruler to complete Mike’s drawing.
Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key 14
Answer:
Eureka-Math-Grade-4-Module-4-End-of-Module-Assessment-Answer-Key-14
Explanation:
As mentioned in the question we have completed mike’s half of a line-symmetric figure and its line of symmetry using the ruler.

Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key

Engage NY Eureka Math 4th Grade Module 4 Mid Module Assessment Answer Key

Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key

Question 1.
Follow the directions below to draw a figure in the box below. Use a straightedge.
a. Draw 2 points, A and B.
b. Draw Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 1.
c. Draw point D that is not on Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 1.
d. Draw \(\overrightarrow{B D}\).
e. Draw \(\overline{A D}\).
f. Name an acute angle.
__________________________
g. Name an obtuse angle. You may have to draw and label another point.
____________________
Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-2
Answer:
The acute angle is <BAD.
Here we labeled C as another point. So the obtuse angle is <DBC

Explanation:
Here, we have to draw two points and then labeled them as A and B.
And use a straight line to draw AB.
Now we have to draw a new point that is not on AB and we label it as D.
Then we will Join BD with a line to get an acute angle BAD.
Use a straight line to join AD.
And for an obtuse angle, we need to label another point as C.
We will use these points and we will label with one angle as <BAD/<DBC

Question 2.
Use your protractor to measure the angle indicated by the arc. Classify each angle as right, acute, or obtuse. Explain how you know each angle’s classification.
a.
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 3
Answer:
The angle indicated by the arc is 30° which is an acute angle.

Explanation:
Here the angle we measured is an acute angle because it indicates 30° which is less than the right angle i.e 90° 

 

b.
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 4
Answer:
The answer is 147° which is an obtuse angle.

Explanation:
Here it is an obtuse angle as it measures 147° which is greater than a right angle and less than 180°

c.
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 5
Answer:
The angle indicated by the arc is 90° which is a right-angled triangle.

Explanation:
This is a right-angled triangle as it exactly measures 90°.

Question 3.
Use the following instructions to draw a figure in the box below.

  • Using a straightedge, draw a line. Label it Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 6.
  • Label a point A on Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 6.
  • Using your protractor and ruler, draw a line perpendicular to Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 6 through point A.
  • Label the perpendicular line Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 13.
  • Label a point B on Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 13, other than point A.
  • Using your protractor and straightedge, draw a line, Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 14, perpendicular to Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 13 through point B.
  • Which lines are parallel in your drawing? Explain why.

Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-7
Answer:
ST||KL
Here ST is parallel to KL because both of them are perpendicular to PQ. It shows us the sides of a rectangle.

Explanation:
By using a straight edge draw a line and label them as KL.
Now label a point on the line of KL and name the point as A.
Next, draw a perpendicular line by using the protractor and ruler through point A to KL.
And label the perpendicular line as PQ.
Name point B on PQ other than point A.
Next is to draw a line ST by using the protractor and straightedge perpendicular to PQ through point B.
Now the answer to the above-mentioned questioned in our drawing is ST is parallel to KL as both of them are perpendicular to PQ.

Question 4.
Use the clock to answer the following:
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 8
a. Use a straightedge to draw the hands as they would appear at 3:00.
Answer:

Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-8

Explanation:
Given in the above diagram draw a line at minutes hand which is 3:00

b. What kind of angle is formed by the clock hands at 3:00?
Answer:
A right angle.

Explanation:
By observing keenly in the above diagram we get a right angle formed by the clock hands at 3:00

c. What time will it be when the minute hand has turned 180°?
Answer:
It will be at 3.30.

Explanation:
By turning the minute hand to 180° the time we get is 3.30

d. How many 90° turns will the minute hand make between 3:00 and 4:00?
Answer:
The minute’s hand will make four 90 turns between 3:00 and 4:00

Explanation:
By observing in the diagram when the clock is rotating in a clockwise direction the minute hand make four 90° turns i.e at 3:00, 3:15, 3:30, 3:45, 4:00

Question 5.
Use the compass rose to answer the following:
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 9
a. Maddy faced East. She turned to her right until she was facing North. How many degrees did she turn?
Answer:
Maddy turned 270°.

Explanation:
Maddy is facing the east side as she turned to her right means she is moving towards the south which is 90. As mentioned in the question Maddy turns right until she reaches north
So to reach north she as to make a move of 90°. Now from East to North she has made 270° i.e ( 90°+90°+90°).

b. Quanisha was facing North. She turned toward her right until she faced East. Alisha was facing South. She turned toward her right until she faced West. What fraction of a full turn did each girl complete? Through how many degrees did each girl turn?
Answer:
Here each girl completed 1/4th of a full turn.
And each girl turned 90°.

Explanation:

As Quanisha was facing north she turned towards the right until she reaches her destination which is East that she needs to face. Another girl Alisha was facing south and turned toward her right until she faces west. So by observing the direction through the compass given above here each girl completed their turn is 1/4th. And also each girl turned 90° while moving to their destination.

Question 6.
The town of Seaford has a large rectangular park with a biking path around its perimeter and two straight-line biking paths that cut across it as shown in the diagram below.
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 10
a. Find the measure of the following angles using a protractor.
∠FGD:
∠GDK:
∠KGN:
Answer:
By using a protractor, here the measuring angle of ∠FGD is 42° ( 180° -138° = 42°)
By doing the same above process the measuring angle of ∠GDK is 138°
The measuring angle of ∠KGN is 42° ( 180° -138° = 42°)

Explanation:
To measure an angle FGD, we place the midpoint of the protractor on the vertex G of the angle. The marking on the inner circle of the protractor is 0° to 180°. Now take the difference of the protractor as 180 and 138° we will get 42°. Repeat the same process to measure the other angles.

b. In the space below, use a protractor to draw an angle with the same measure as ∠DGK.
Answer:

Eureka-Math-Grade-4-Module-4-Mid-Module-Assessment-Answer-Key-2-1

Explanation:
Here the same measure as ∠DGK is 138°. Now, take the protractor to draw an angle 138° and mention the angle as ∠ABC.

c. Below is a sign that bikers may encounter while riding in the park. Using the points in the figure below, identify a line segment, a right angle, an obtuse angle, a set of parallel lines, and a set of perpendicular lines. Write them in the table below.
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 11
Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key 12

Answer:

Line segment: EF
Right Angle: ∠ABD
Obtuse angle: ∠GHJ
Parallel Lines: KL||GH
Perpendicular lines: AC ⊥ BD

Explanation:
Line Segment: A-line that is bounded by two endpoints are in the above diagram are EF.
Right Angle: A right angle is an angle of exactly 90°. So by observing the above diagram we can identify the ∠ABD is a right angle.
Parallel Lines: The two straight lines in a plane that do not meet at any point are called to be parallel lines. By observing the diagram we got to know that KL is parallel to the GH.
Perpendicular Lines: Any two distinct lines meeting each other at 90° or a right angle are perpendicular lines. By keenly observing the diagram AC is perpendicular to BD.

Eureka Math Grade 4 Module 4 Lesson 11 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 11 Answer Key

Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key

Write an equation, and solve for the unknown angle measurements numerically.

Question 1.
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 1
______° + 20° = 30°
d° = ______°

Answer:
The value of d° is 10°.

Explanation:
Given that the value of the angle acute angle is 30° and the value of the other angle is 20° and the value of another angle is d°. So the equation will be
d° + 20°= 30°
so the value of d° is 30° – 20°
= 10°.

Question 2.
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 2
______° + ______°= 360°
c° = ______°

Answer:
The value of c° is 270°.

Explanation:
Here, in the above image, we can see that an arc that represents a complete rotation which means 360°, and the other angle 90°. So the equation will be c°+90°= 360° and the value of c° is
c°= 360°-90°
= 270°.

Question 3.
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 3
______° + ______° + ______° = ______°
e° = ______°

Answer:
The value of e° is 196°.

Explanation:
Here we will measure the angles using a protractor and the values of the angles will be 90° and 196°. So the equation will be 74°+90°+e°= 360° and the value of e is
e° = 360°- 164°
= 196°.
So the value of e° is 196°.

Question 4.
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 4
______° + ______° + ______° = ______°
f° = ______°

Answer:
The value of f° is 110°.

Explanation:
Here we will measure the angles using a protractor and the values of the angles will be 90° and 160°. So the equation will be 90°+160°+f°= 360° and the value of f is
f° = 360°- 250°
= 110°.
So the value of f° is 110°.

Write an equation, and solve for the unknown angles numerically.

Question 5.
O is the intersection of \(\overline{A B}\) and \(\overline{C D}\). ∠DOA is 160°, and ∠AOC is 20°.
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 5
x° = ______°      y° = ______°

Answer:
The value of x° is 160°.
The value of y° is 20°.

Explanation:
In the above image, we can see that the angle COD is 180° and <DOA is 160° and <AOC is 20°. So the equation will be 160° + y°= 180° and the value of y°= 180° – 160°
y°= 20°.
So the value of y° is 20°.
And the other equation will be 20° + x°= 180°
x°= 180° – 20°
= 160°.
So the value of the x°is 160°.

Question 6.
O is the intersection of \(\overline{R S}\) and \(\overline{T V}\). ∠TOS is 125°.
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 6
g° = ______°     h° = ______°       t° = ______°

Answer:
The value of i° is 55°.
The value of h° is 125°.
The value of g° is 55°.

Explanation:
In the above image, we can see that the angle TOV is 180° and angle SOR is 180° and <TOS is 125°. So the equation will be 125° + i°= 180° and the value of i°= 180° – 125°
i°= 55°.
So the value of i° is 55°.
And the other equation will be 55° + h°= 180°
h°= 180° – 55°
= 125°.
So the value of the h°is 125°.
And the other equation will be 125° + g°= 180°
g°= 180° – 125°
= 55°.
So the value of the g°is 55°.

Question 7.
O is the intersection of \(\overline{W X}\), \(\overline{Y Z}\), and \(\overline{U O}\) ∠XOZ is 36°
k° = ______°     m° = ______°     n° = ______°
Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 7
Answer:
The value of m° is 54°.
The value of k° is 36°.
The value of n° is 144°.

Explanation:
In the above image, we can see that the angle WOX is 180° and angle XOZ is 36°, and the value of angle UOX is 90°. So the equation will be 90°+m+ 36°= 180° and the value of m°= 180° – 126°
m°= 54°.
So the value of m° is 54°.
And the other equation will be 54° +90°+k°= 180°
k°= 180° – 144°
= 36°.
So the value of the k°is 36°.
And the other equation will be 36° +n°= 180°
n°= 180° – 36°
= 144°.
So the value of the n°is 144°.

Eureka Math Grade 4 Module 4 Lesson 11 Exit Ticket Answer Key

Write equations using variables to represent the unknown angle measurements. Find the unknown angle measurements numerically.

Eureka Math Grade 4 Module 4 Lesson 11 Problem Set Answer Key 8

Question 1.
x° =
Answer:
The value of x° is 24°.

Explanation:
In the above image, we can see that the angle AEB is 180° and angle AEF is 90°. So the equation will be 90°+x+66= 180° and the value of m°= 180° – 156°
x°= 24°.
So the value of x° is 24°.

Question 2.
y° =
Answer:
The value of y° is 156°.

Explanation:
In the above image, we can see that the angle AEB is 180° and angle AEF is 90°. So the equation will be 24°+y= 180° and the value of y°= 180° – 24°
y°= 156°.
So the value of y° is 156°.

Question 3.
z° =
Answer:
The value of z° is 24°.

Explanation:
In the above image, we can see that the angle AEB is 180° and angle FEB is 90°. So the equation will be 90°+z+66= 180° and the value of m°= 180° – 156°
z°= 24°.
So the value of z° is 24°.

Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key

Write an equation, and solve for the unknown angle measurements numerically.

Question 1.
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 9
__________° + 320° = 360°
a° = ________ °
Answer:
The value of a° is 40°.

Explanation:
Given that the value of the angle is complete rotation which 360° and the value of the other angle is 320° and the value of another angle is a°. So the equation will be
a° + 320°= 360°
so the value of a° is 360° – 320°
= 40°.

Question 2.
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 10
_____________ ° + ____________ ° = 360°
b° = __________ °
Answer:
The value of b° is 315°.

Explanation:
Given that the value of the angle is complete rotation which 360° and the value of the other angle is 45° and the value of another angle is b°. So the equation will be
b° + 45°= 360°
so the value of b° is 360° – 45°
=315°

Question 3.
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 11
_____________ ° + _____________ ° + _____________ ° = _____________ °
c° = _____________ °
Answer:
The value of c° is 145°.

Explanation:
Given that the value of the angle is complete rotation which 360° and the value of the other angle is 115° and the value of another angle is 100° and the value of another angle is c°. So the equation will be c°+115°+100°= 360°.
c° + 215°= 360°
so the value of c° is 360° – 215°
=145°.

Question 4.
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 12
_____________ ° + _____________ ° + _____________ ° = _____________ °
d° = _____________ °
Answer:
The value of d° is 80°.

Explanation:
Given that the value of the angle is complete rotation which 360° and the value of the other angle is 135° and the value of another angle is 145° and the value of another angle is d°. So the equation will be d°+135°+145°= 360°.
d° + 280°= 360°
so the value of c° is 360° – 280°
=80°.

Write an equation, and solve for the unknown angles numerically.

Question 5.
O is the intersection of \(\overline{A B}\) and \(\overline{C D}\). ∠COB is 145°, and ∠AOC is 35°
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 13
e° = _____________ °        f° = _____________ °
Answer:
The value of e° is 145°.
The value of y° is 35°.

Explanation:
In the above image, we can see that the angle COD is 180° and <COB is 145° and <AOC is 35°. So the equation will be 35° + e°= 180° and the value of e°= 180° – 35°
e°= 145°.
So the value of e° is 145°.
And the other equation will be 145° + f°= 180°
x°= 180° – 145°
= 35°.

Question 6.
O is the intersection of \(\overline{Q R}\) and \(\overline{S T}\). ∠QOS is 55°
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 14
g° = _____________ °         h° = _____________ °        i° = _____________ °
Answer:
The value of g° is 125°.
The value of h° is 125°.
The value of i° is 55°.

Explanation:
In the above image, we can see that the angle SOT is 180° and angle QOR is 180° and <QOS is 55°. So the equation will be 55° + g°= 180° and the value of g°= 180° – 55°
g°= 125°.
So the value of g° is 125°.
And the other equation will be 55° + h°= 180°
h°= 180° – 55°
= 125°.
So the value of the h°is 125°.
And the other equation will be 125° +g°= 180°
g°= 180° – 125°
= 55°.
So the value of the g°is 55°.

Question 7.
O is the intersection of \(\overline{U V}\), \(\overline{W X}\), and \(\overline{Y O}\). ∠VOX is 46°
Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key 15
J° = _____________ °         K° = _____________ °         m° = _____________ °
Answer:
The value of j° is 44°.
The value of k° is 46°.
The value of m° is 134°.

Explanation:
In the above image, we can see that the angle WOX is 180° and angle UOV is 180° and <VOX is 46°. So the equation will be 46° +90°+ j°= 180° and the value of j°= 180° – 136°
j°= 44°.
So the value of j° is 44°.
And the other equation will be 44° +90°+ k°= 180°
k°= 180° – 134°
= 46°.
So the value of the k°is 46°.
And the other equation will be 46° +m°= 180°
m°= 180° – 46°
= 134°.
So the value of the m°is 134°.

Eureka Math Grade 4 Module 4 Lesson 10 Answer Key

Engage NY Eureka Math 4th Grade Module 4 Lesson 10 Answer Key

Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key

Write an equation, and solve for the measure of ∠x. Verify the measurement using a protractor.

Question 1.
∠CBA is a right angle.
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 1
45° + ________ = 90°
x° = __________
Answer:
The value of x is 45°.

Explanation:
Given that <CBA is a right angle and angle B is 45°, so the angle CBA= 90° which is 45°+x= 90°, so the value of x is
x°= 90° – 45°
= 45°
So the value of x° is 45°.

Question 2.
∠GFE is a right angle.
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 2
_______ + ________ = _________
x° = __________
Answer:
The value of x is 70°.

Explanation:
Given that <GFE is a right angle which is 90° and another angle is 20°, so the angle GFE= 90° which is 20°+x= 90°, so the value of x is x°= 90° – 20°
= 70°.
So the value of x° is 70°.

Question 3.
∠IJK is a straight angle
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 3
___________ + 70° = 180°
x° = ____________
Answer:
The value of x is 110°

Explanation:
Given that <IJK is a straight angle which is 180° and another angle is 70°, so the angle IJK= 180° which is 70°+x= 180°, so the value of x is
x°= 180° – 70°
= 110°.
So the value of x° is 110°.

Question 4.
∠MNO is a straight angle
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 4
_________ + _________ = __________
x° = ___________
Answer:
The value of x is 97°

Explanation:
Given that <MNO is a straight angle which is 180° and another angle is 83°, so the angle MNO= 180° which is 83°+x= 180°, so the value of x is
x°= 180° – 83°
= 97°.
So the value of x° is 97°.

Solve for the unknown angle measurements. Write an equation to solve.

Question 5.
Solve for the measurement of ∠TRU. ∠QRS is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 5
Answer:
The value of <TRU is 144°.

Explanation:
Given that <QRS is a straight angle which is 180° and another angle is 36°, so the angle MNO= 180° which is 36°+x= 180°, so the value of x is
x°= 180° – 36°
= 144°.
So the value of <TRU is 144°.

Question 6.
Solve for the measurement of ∠ZYV. ∠XYZ is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 6
Answer:
The value of <ZYV is 12°.

Explanation:
Given that <XYZ is a straight angle which is 180° and another two angle is 108°+60° which is 168°, so the angle XYZ= 180° which is 168°+x= 180°, so the value of x is
x°= 180° – 168°
= 12°.
So the angle ZYV is 12°.

Question 7.
In the following figure, ACDE ¡s a rectangle. Without using a protractor, determine the measurement of ∠DEB. Write an equation that could be used to solve the problem.
Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 7
Answer:
The value of x° is 63°.

Explanation:
Here, in the above image we can see that a rectangle where every angle is a right angle, so let’s take one angle as 90° and then we should find out the other angle and the other angle be x. So the equation will be 90°= 27°+x° and the value of x is
x= 90° – 27°
= 63°.
So the value of x° is 63°.

Question 8.
Complete the following directions in the space to the right.
a. Draw 2 points: M and N. Using a straightedge, draw Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 8.
b. Plot a point O somewhere between points M and N.
c. Plot a point P, which is not on Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 8.
d. Draw \(\overline{O P}\).
e. Find the measure of ∠MOP and ∠NOP.
f. Write an equation to show that the angles add to the measure of a straight angle.
Answer:
The equation of the angles add to the measure of a straight angle is <MOP+<PON= 180°.

Explanation:
Here, we have plotted 2 points which are M and N using a straightedge, and have constructed Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 8.
Then we plotted a point O somewhere between points M and N.
Then we need to plot a point P, which is not on Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key 8.
And now we will construct \(\overline{O P}\).
So, now we need to find the measure of ∠MOP and ∠NOP.
As we have plotted a straight line which is <MON is 180°, which is <MOP+<PON= 180°.
So the equation of the angles add to the measure of a straight angle is <MOP+<PON= 180°.
Eureka-Math-Grade-4-Module-4-Lesson-9-Homework-Answer-Key-8-5

Eureka Math Grade 4 Module 4 Lesson 10 Exit Ticket Answer Key

Write an equation, and solve for x. ∠TUV is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Exit Ticket Answer Key 9
Equation = ______________
x° = ______________
Answer:
The value of x is 60°.

Explanation:
Given that <TUV is a straight angle which is 180° and another two angle is 53°+67° which is 120°, as the angle TUV= 180° which is 168°+x= 180°, so the value of x is
x°= 180° – 120°
= 60°.
So the angle x is 60°.

Write an equation, and solve for the measurement of ∠x. Verify the measurement using a protractor
Question 1.
∠DCB is a right angle
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 10
___________ + 35° = 90°
x° = ___________
Answer:
The value of x° is 55°.

Explanation:
Given that <DCB is a right angle and other angle is 35°, so the angle DCB is 90° which is 35°+x= 90°, so the value of x is
x°= 90° – 35°
= 55°
So the value of x° is 55°.

Question 2.
∠HGF is a right angle
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 11
___________ + ___________ = ___________
x° = ___________

Answer:
The value of x° is 28°.

Explanation:
Given that <HGF is a right angle and other angle is 62°, so the angle HGF is 90° which is 62°+x= 90°, so the value of x is
x°= 90° – 62°
= 28°
So the value of x° is 28°.

Question 3.
∠JKL is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 12
145° + ___________ = 180°
x° = ___________
Answer:
The value of x is 35°.

Explanation:
Given that <JKL is a straight angle which is 180° and another angle is 145° and the angle JKL is 180° which is 145°+ x= 180°, so the value of x is
x°= 180° – 145°
= 35°.
So the angle x is 35°.

Question 4.
∠PQR is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 13
___________ + ___________ = ___________
x° = ___________
Answer:
The value of x is 164°.

Explanation:
Given that <PQR is a straight angle which is 180° and another angle is 16° as the angle PQR is 180° which is 16°+x= 180°, so the value of x is
x°= 180° – 16°
= 164°.
So the angle x is 164°.

Write an equation, and solve for the unknown angle measurements.

Question 5.
Solve for the measurement of ∠USW. ∠RST is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 14
Answer:
The value of x is 75°.

Explanation:
Given that <RST is a straight angle which is 180° and another two angle is 70°+35° which is 105°, as the angle RST= 180° and the equation is 105°+x= 180°, so the value of x is
x°= 180° – 105°
= 75°.
So the angle x is 75°.

Question 6.
Solve for the measurement of ∠OML. ∠LMN is a straight angle.
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 15
Answer:
The value of <OML is 35°.

Explanation:
Given that <LMN is a straight angle which is 180° and another two angle is 72°+73° which is 145°, as the angle LMN= 180° and the equation is 145°+x= 180°, so the value of x is
x°= 180° – 145°
= 35°.
So the angle <OML is 35°.

Question 7.
In the following figure, DEFH is a rectangle. Without using a protractor, determine the measurement of ∠GEF Write an equation that could be used to solve the problem.
Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 16
Answer:
The value of x° is 16° and the equation is 90°= 74°+x°.

Explanation:
Here, in the above image we can see that a rectangle where every angle is a right angle, so let’s take one angle as 90° and then we should find out the other angle and the other angle be x. So the equation will be 90°= 74°+x° and the value of x is
x= 90° – 74°
= 16°.
So the value of x° is 16° and the equation is 90°= 74°+x°.

Question 8.
Complete the following directions in the space to the right.
a. Draw 2 points: Q and R. Using a straightedge, draw Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 17.
b. Plot a point S somewhere between points Q and R.
c. Plot a point T, which is not on Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 17
d. Draw \(\overline{T S}\).
e. Find the measure of ∠QST and ∠RST.
f. Write an equation to show that the angles add to the measure of a straight angle.
Answer:
The equation of the angles add to the measure of a straight angle is <QST+<STR= 180°.

Explanation:
Here, we need to draw 2 points which is Q and R using a straightedge and then we need to construct Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 17.
Then we need to plot a point S somewhere between points Q and R.
And then we need to plot a point T, which is not on Eureka Math Grade 4 Module 4 Lesson 10 Homework Answer Key 17
Then we will construct \(\overline{T S}\).
Then we need to find the measure of ∠QST and ∠RST.
As we have plotted a straight line which is <QSR is 180°, which is <QST+<STR= 180°.
So the equation of the angles add to the measure of a straight angle is <QST+<STR= 180°.
Eureka-Math-Grade-4-Module-4-Lesson-9-Homework-Answer-Key-8-6