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Worksheet on Comparison on Ratios | Comparing Ratios Worksheet with Answers PDF

December 27, 2021

In this worksheet on comparison on ratios, students and teachers can find various ways of comparing ratios problems between two or more ratios. The term ratio defines the quantitative relationship of two numbers or amounts when there are three or more quantities that come together, then a comparison of ratios is required.

To compare two ratios, we need to adapt them into equivalent such as fractions. Every student should aware of the procedure of comparing ratios, which makes it easier to compare the ratios. Refer to this free worksheet for ratio comparison and solve the basic to complex problems with ease. The Printable Worksheet on Comparing Ratios is fun to practice and gives you deeper insight into the concept.

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How to Compare Ratios?

Firstly, check the second terms of the given ratios.
If they are not the same take the L.C.M of both the second terms and divide the L.C.M with the consequent of both the ratios.
Now, multiply the quotient with both the terms of ratio.
After that, we decide which ratio of the first term or antecedent is greater than that of the new ratios obtained.

Practice the questions of comparison on ratios in detail from the below comparing ratios worksheet with answers.

Comparing Ratios Worksheet PDF

Example 1.
Compare the ratio 4: 7 and 2: 14.

Solution:

The given ratio to compare is 4: 7 and 2: 14.
Now, take the L.C.M of both the second terms 7 and 14 is 14.
Divide the L.C.M with the second term. i.e., 14$\div$7=2 and 14$\div$14=1.
Next, multiply the quotient with the antecedent and the consequent of the ratios.
Therefore, $\frac{4}{7}$ = $\frac{4×2}{7×2}$ = $\frac{8}{14}$.
$\frac{2}{14}$ = $\frac{2×1}{14×1}$ = $\frac{2}{14}$.
As we can see, 8 > 2, $\frac{4}{7}$ > $\frac{2}{14}$ i.e., 4: 7 > 2: 14.
Hence, 4: 7 is greater than the 2: 14 by the comparison rules of ratio.

Example 2.
Which ratio is greater 2$\frac{1}{3}$: 1$\frac{2}{5}$ and 5: 2.

Solution:

Given ratios is 2$\frac{1}{3}$: 1$\frac{2}{5}$ and 5: 2.
Convert the mixed fractions into proper fractions. i.e., 2$\frac{1}{3}$: 1$\frac{2}{5}$=$\frac{7}{3}$: $\frac{7}{5}$.
$\frac{7}{3}$: $\frac{7}{5}$ and $\frac{50}{10}$: $\frac{20}{10}$
= $\frac{7}{3}$×15: $\frac{7}{5}$×15 and $\frac{50}{10}$×10: $\frac{20}{10}$×10
= 35: 21 and 50: 20
= $\frac{35}{21}$ and $\frac{50}{20}$
= $\frac{5×7}{3×7}$ and $\frac{5×10}{2×10}$
= $\frac{5}{3}$ and $\frac{5}{2}$
= 5: 3 and 5: 2
Now, compare the ratios 5:3 and 5: 2.
The L.C.M of the second terms 3 and 2 is 6.
Now, divide with the second terms of both the ratios. ie., 6$\div$3=2 and 6$\div$2=3.
Therefore, $\frac{5}{3}$ = $\frac{5×2}{3×2}$ = $\frac{10}{6}$.
$\frac{5}{2}$ = $\frac{5×3}{2×3}$ = $\frac{15}{6}$.
So, 15>10, $\frac{5}{2}$ > $\frac{5}{3}$, 5: 3 < 5: 2.
Therefore, the ratios 5: 2 > 2$\frac{1}{3}$: 1$\frac{2}{5}$.

Example 3.
Arrange the following ratios in descending order.
(i) 1: 2, 4: 2, and 3:5
(ii) 4:1, 3:6, 7:5, and 2:4

Solution:

(i) Given ratios to arrange in descending order are 1: 2, 4: 2, and 3:5.
The ratios can also be written as $\frac{1}{2}$, $\frac{4}{2}$, and $\frac{3}{5}$.
L.C.M of all the denominators 2, 2, and 5 is 10.
Now, divide the L.C.M
10$\div$2=5, 10$\div$2=5, and 10$\div$5=2.
Multiply the quotients with the ratios.
$\frac{1}{2}$ = $\frac{1×5}{2×5}$ = $\frac{5}{10}$,
$\frac{4}{2}$ = $\frac{4×5}{2×5}$ = $\frac{20}{10}$,
and $\frac{3}{5}$ = $\frac{3×2}{5×2}$ = $\frac{6}{10}$.
Therefore, $\frac{20}{10}$ > $\frac{6}{10}$ > $\frac{5}{10}$.
⇒ $\frac{4}{2}$ > $\frac{3}{5}$ > $\frac{1}{2}$.
The ratios in descending order is 4: 2, 3: 5, and 1: 2.

(ii) Given ratios to arrange in descending order is 4:1, 3:5, 7:5, and 2:4.
The L.C.M of the second terms of the ratios 1, 5, 5, and 4 = 20.
Divide the L.C.M with all the consequent of the ratios. i.e., 20$\div$1=20, 20$\div$6=, 20$\div$5=4, and 20$\div$4=5.
Now,
$\frac{4}{1}$ = $\frac{4×20}{1×20}$ = $\frac{80}{20}$,
$\frac{3}{5}$ = $\frac{3×6}{6×6}$ = $\frac{18}{36}$,
$\frac{7}{5}$ = $\frac{7×4}{5×4}$ = $\frac{28}{20}$,
and $\frac{2}{4}$ = $\frac{2×5}{4×5}$ =$\frac{10}{20}$.
Here, compare the numerators for which ratio is greater.
Since, $\frac{80}{20}$ > $\frac{28}{20}$ > $\frac{18}{36}$ > $\frac{10}{20}$
⇒ $\frac{4}{1}$ > $\frac{7}{5}$ > $\frac{3}{5}$ > $\frac{2}{4}$.
Therefore, the ratios in descending order are 4: 1, 7: 5, 3: 5, and 2: 4.

Example 4.
Write and arrange the following ratios in ascending order.
(i) 11: 4, 5: 1, and 9: 3
(ii) 4: 5, 5: 7, and 7: 9

Solution:

(i) Given ratios are 11: 4, 5: 1, and 9: 3.
L.C.M of the second terms 4, 1, and 3 is 12.
Now,
$\frac{11}{4}$ = $\frac{11×3}{4×3}$ = $\frac{33}{12}$,
$\frac{5}{1}$ = $\frac{5×12}{1×12}$ = $\frac{60}{12}$,
and $\frac{9}{3}$ = $\frac{9×4}{3×4}$ = $\frac{36}{12}$.
Since, $\frac{33}{12}$ < $\frac{36}{12}$ < $\frac{60}{12}$
⇒ $\frac{11}{4}$ < $\frac{9}{3}$ < $\frac{5}{1}$.
Thus, the ascending order of the ratio is 11: 4, 9: 3, and 5: 1.

(ii) Given ratios are 4: 5, 5: 7, and 7: 9
As per the comparison rules of ratio, we take the L. C. M of the second term. i.e., 5, 7, and 9 is 315.
Next, divide the L.C.M with the second term of all the ratios.
315$\div$5=63, 315$\div$7=45, and 315$\div$9=35.
Now, multiply the quotient with both the terms of the ratios. We get,
$\frac{4}{5}$ = $\frac{4×63}{5×63}$ = $\frac{252}{315}$,
$\frac{5}{7}$ = $\frac{5×45}{7×45}$ = $\frac{225}{315}$,
and $\frac{7}{9}$ = $\frac{7×35}{9×35}$ = $\frac{245}{315}$.
Since, $\frac{225}{315}$ < $\frac{245}{315}$ < $\frac{252}{315}$
⇒ $\frac{5}{7}$ < $\frac{7}{9}$ < $\frac{4}{5}$
⇒ 5: 7, 7: 9, and 4: 5
Therefore, the ratios in ascending order are 5: 7, 7: 9, and 4: 5.

Worksheet on Types of Ratios | Types of Ratios Worksheet with Answers PDF

December 27, 2021

Kids who are looking for various types of ratio numbers to practice can take the help from this Worksheet on Types of Ratios. Here, we will discuss different types of ratios that make students calculate the ratios in any manner. Without knowing the types of ratios, it becomes difficult to solve their relative sizes so go for these types of ratios practice worksheets and make your calculations so perfect.

The ratio is the mathematical expression of two or more quantities of the same units by dividing the one ratio by the other part of the ratio and must have the same unit measurements. The following different types of ratios worksheet with answers helps the students & teachers to perceive the various expression easily. Practice and express the below ratios in the respected kind of ratio.

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Different Types of Ratios Worksheet with Answers

Example 1.
Find the compound ratio of the following ratios.
(i) 4: 5 and 3: 2
(ii) b: d and g: f
(iii) 12: 8 and 7: 15 and 5: 9

Solution:

(i) Given ratio is 4: 5 and 3: 2
To find the compound ratio, if we multiply two or more ratios termwise, then the ratio is obtained as a compound ratio.
Now, 4: 5 and 3: 2 = 4×3 : 5×2 = 12: 10.
So, the compound ratio is 12: 10.

(ii) Given ratio is b: d and g: f.
The compound ratio for b: d and g: f = b×g : d×f = bg : df.
Thus, the ratio is bg: df.

(iii) Given ratio is 12: 8 and 7: 15 and 5: 9.
To find the compound ratio for three ratios, we use the ratio formula
if m: n, p: q, and r: s, the compound ratio is (m× p× r) : (n× q× s).
Now, the ratio of 12: 8 and 7: 15 and 5: 9 = (12× 7× 5) : (8× 15× 9) = 420: 1080 = 7: 18.
Therefore, the compound ratio is 7: 18.

Example 2.
Get the duplicate ratio of the following expressions.
(i) √3: √4
(ii) a²: b²
(iii) 5√p: √20q

Solution:

(i) Given ratio is √3: √4
The duplicate ratio is the ratio of two equal ratios. For example, m: n = m²: n².
Now, the ratio for √3: √4 = (√3)²: (√4)² = 3: 4.
Thus, the duplicate ratio is 3: 4.

(ii) Given ratio is a²: b²Now, duplicate ratio a²: b² = (a²)² : (b²)² = a⁴: b⁴.
a⁴: b⁴ is the duplicate ratio of a²: b².

(iii) Given ratio is 5√p: √20q.
The duplicate ratio of 5√p: √20q = (5√p)²: (√20q)² = 5²×√p² : √20q×√20q = 25p: 20q = 5p: 4q.
So, the duplicate ratio is 5p: 4q.

Example 3.
Find the triplicate ratio of the following expressions.
(i) ³√9m: 4
(ii) a/3: ³√a
(iii) p/2: q/4

Solution:

(i) Given ratio is ³√9m: 4.
The triplicate ratio is the compound ratio of three equal ratios. For example, m: n = m³: n³.
Now, ³√9m: 4 = (³√9m)³ : 4³ = 9m: 64.
The triplicate ratio is 9m: 64.

(ii) Given ratio is a/3: ³√a.
To find the triplicate ratio we apply the formula a: b = a³: b³.
a/3: ³√a = (a/3)³: (³√a)³ = a³/3³: (³√a)³ =a³/9: a = a²: 9.
So, the ratio is a²: 9.

(iii) Given ratio is p/2: q/4.
Now, the triplicate ratio of p/2: q/4 = (p/2)³: (q/4)³ = p³/8: q³/64 = 64p³: 8q³ = 8p³: q³.
Therefore, the triplicate ratio is 8p³: q³.

Example 4.
Find the subduplicate ratio of (a+b)²: (a-b)⁴.

Solution:

Given ratio (a+b)²: (a-b)⁴.
The subduplicate ratio of p: q is √p: √q.
Now, we find the subduplicate ratio of (a+b)²: (a-b)⁴.
(a+b)²: (a-b)⁴= √(a+b)²: √(a-b)⁴= (a+b): (a-b)².
Thus, the subduplicate ratio of (a+b)²: (a-b)⁴ is (a+b): (a-b)².

Example 5.
Find the subtriplicate ratio of 343x³: 8y³.

Solution:

The given ratio is 343x³: 8y³.
The subtriplicate ratio of a: b is ³√a: ³√b.
Now, the subtriplicate ratio of 343x³: 8y³
343x³: 8y³ = ³√(343x³): ³√(8y³) = 7x: 2y (since, 7³=343, 2³=8).
Hence, the subtriplicate ratio of 343x³: 8y³ is 7x: 2y.

Example 6.
Find the reciprocal ratio of a/9: b/6.

Solution:

The given ratio is a/9: b/6.
The reciprocal ratio of p: q (p≠0, q≠0) is the ratio 1/p: 1/q = q: p.
Now, to find the reciprocal ratio of a/9: b/6
a/9: b/6 = 1/(a/9): 1/(b/6) = 9/a: 6/b = 9b: 6a = 3b: 2a.
Simply, we can say the reciprocal ratio is the inverse ratio. For, example 4: 5 is the inverse ratio of 5: 4.
Thus, the reciprocal ratio of a/9: b/6 is 3b: 2a.

Example 7.
Find x if (2x + 1) : (3x + 6) is the duplicate ratio of 4 : 6.

Solution:

(2x + 1) : (3x + 6) is the duplicate ratio of 4: 6.
Also, the duplicate ratio of 4: 6 is 4²: 6² = 16: 36.
Now, (2x+1)/(3x+6) = 16/36
⇒ 36(2x+1) = 16(3x+6)
⇒ 72x+36 = 48x+96
⇒ 72x- 48x = 96-36
⇒ 24x = 60
⇒ x = 60/24
⇒ x = 2.5
Therefore, the value of x is 2.5.

Worksheet on Ratio in Lowest Term | Simplifying Ratios Worksheet with Answers PDF

December 18, 2021

In this Worksheet on the Ratio in Lowest Term or the simplest form, all students can learn many practice questions to make the topic easy. This reducing ratios to lowest form worksheet provide various types of lowest term ratios questions with solutions. In short, we can say that ratio shows their relative sizes.

The ratio of two or more quantities of the same units of measurement is a comparison got by dividing one quantity by the other and must have the same units of measurement. The following Worksheet on Ratio in Simplest Form helps the students and teachers to perceive the ratio in the lowest term. Practice the questions on express the following ratios in the simplest form worksheet with answers pdf.

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Reduce Ratios to Lowest Term Worksheet with Answers

1. Find the ratio of 30 min and 1½ hr in the simplest form.

Solution:

The given ratio is 30: 1½.
First, we have to convert the time 1½ hr to minutes.
1½ hr = 60min + 30min = 90min
Therefore, the ratio = 30min : 90min.
Dividing by 30, we get the ratio of 1:3.
Hence, the simplest form of the given ratio is 1:3.

2. Find the ratios of the following in the simplest form.
(i) 2.4 kg to 400g
(ii) 425 and 220
(iii) 4 dozens to 2 scores
(iv) 1kg 100g and 2kg 300g

Solution:

(i) Given 2.4 kg to 400g
Convert the kilograms to grams
We know 1kg = 1000g.
Now, 2.4kg = 2.4 × 1000 = 2400gms.
The ratio of 2400g: 400g is reduced to the lowest term by diving both the numbers by 400.
When we divide the ratio by 400, we get
2400g : 400g = 6:1.
So, the simplest form is 6:1.

(ii) Given 425: 220
Divide the ratio by 5, we get
425/5: 220/5 = 85: 44.
The simplest form of ratio 425: 220 is 85: 44.

(iii) Given 4dozens and 2scores
We know, 4dozens = 4×12 = 48.
2 scores = 2×20 = 40 (since, 1 score = 20 of something).
Now, the ratio of 4dozens and 2scores is 48: 40
Divide by 8, we get
48/8 : 40/8 = 6:5.
The simplest form of ratio is 6:5.

(iv) Given 1kg 100g and 2kg 300g.
First, have to convert them into grams.
1kg 100g = 1000g × 100g = 1100g.
2kg 300g = 2000g × 300g = 2300g.
Now, the ratio is 1100g: 2300g.
The G.C.F of 1100 and 2300 is 100.
When we divide by 100, we get
1100g/100 : 2300g/100 = 11:23.
The simplest form of given ratio 1kg 100g and 2kg 300g is 11:23.

3. Find the ratio of $ 2.15: $ 6.25 in the simplest form?

Solution:

The given ratio is $ 2.15: $ 6.25.
Where, $ 2.15= 215 cents and $ 6.25= 625 cents
Therefore, the required ratio is 215 cents: 625 cents.
Divide the ratio by 5, we get
215cents/5 : 625cents/5 = 43:125.
Thus, the simplest ratio is 43:125.

4. Simplify the following ratios
(i) 4/5: 7/5
(ii) 2 1/3: 4
(iii) 0.04: 0.26
(iv) 4 1/5: 2 3/10: 6/10
(v) 1.8: 2.2

Solution:

(i) Given ratio 4/5: 7/5.
By multiplying both the parts of the ratio by 5, we get
4/5: 7/5 = (4/5)×5: (7/5)×5 = 4: 7.
So, the simplest form of the ratio is 4: 7.

(ii) Given ratio 2 1/3: 4.
First, convert 2 1/3 as an improper fraction, we get 7/3.
Now, multiply both the parts by 3,
7/3: 4 = (7/3)×3: 4×3 = 7: 12.
Thus, the simplest ratio is 7:12.

(iii) Given ratio 0.04: 0.26
As we know, after the decimal part we have two units. So, multiply both the parts by 100.
Then, 4: 26
Now, divide the ratio by 2
4: 26 = 4/2: 26/2 2: 13.
Hence, the simplest ratio is 2: 13.

(iv) Given ratio 4 1/5: 2 3/10: 6/10
Convert, the mixed numbers into improper fractions,
4 1/5 = 21/5 and 2 3/10 = 33/10.
Now, the proper ratio is 21/5: 33/10: 6/10.
The L.C.D is 10.
We now have,
21/5: 33/10: 6/10 = 42/10: 33/10: 4/10 = 42: 33: 6.
Try to reduce the ratio in the simplest form.
The G.C.F for 42: 33: 6 is 3.
We get, 42/3: 33/3: 6/3 = 14: 11: 2.
Therefore, the simplest form of the given ratio is 14: 11: 2.

(v) Given ratio 1.8: 2.2.
After the decimal part we have one unit, multiply both the parts by 10.
1.8 = 18 and 2.2 = 22.
Divide by 2, we get
18: 22 = 18/2: 22/2 = 9: 11.
The simplest ratio is 9:11.

5. Reduce the following ratios to the lowest terms
(i) 3hours: 1hour 40min
(ii) 2years 2months: 4years 4months
(iii) 5weeks: 25days

Solution:

(i) Given 3hours: 1hour 40min
Convert hours into minutes.
3hrs = 3×60 = 180min.
1hr 40min = 60min + 40min = 120min.
The required ratio is 180min: 120min = 180min/60: 120min/60 = 3:2.
So, the ratio in the lowest term is 3:2.

(ii) Given 2years 2months: 4years 4months.
Convert years into months,
2years 2months = 2×12 + 2 = 24+2 = 26months.
4years 4months = 4×12 + 4 = 48+4 = 52months.
The ratio = 26months: 52months = 26/26: 52/26 = 1: 2.
The simplest ratio is 1:2.

(iii) Given the ratio 5weeks: 25days
Convert weeks into days, as we know one week = 7days.
5weeks = 5 × 7 = 35days.
Now, the ratio is 35days: 25days = 35/5: 25/5 = 7: 5.
So, the lowest term of the given ratio is 7:5.

Collinear Points Proved by Midpoint Theorem – Statement & Proof | How do you prove that Points are Collinear?

December 18, 2021

In this article, students can acquire more about the collinear points and the proof of the Collinear Points by Midpoint Theorem. Collinear points are the group of three or more points that lie on the same straight line. These points may exist on different planes but not on different lines. The property of the points being collinear is known as collinearity.

Let us learn the concept of collinear points and see how we construct the theorem to prove the given statement. We hope the information provided on this page will help you all to understand how the points are collinear and proved by the mid-point theorem.

Collinear Points – Definition

The word collinear is a mix of two Latin words ‘col’ and ‘linear’. The word ‘col’ means together and the word ‘linear’ means line. Thus, collinear points mean points together in the same or a single line. In geometry, the set of three or more points that lies on the same straight line is said to be collinear.

You can see many real-world examples of the collinearity name as a group of people or students standing in a straight line, a bunch of oranges kept in a single row, etc. Observe the following figure to get clear knowledge of the collinear points and identify the collinear points and non-collinear points.

In the above figure, the set of collinear points are {A, E, B}, {B, F, C}, {C, G, D}, {A, H, D}, and {B, I, D}. The set of non-collinear points are {E, G}, {H, F}, and {G, F}, etc.

Collinear Points Proved by Midpoint Theorem Statement and Proof

In ∆ABC, the medians CM and BN are produced to the points P and Q respectively such that CM=MP and BN=NQ. Prove that the points P, A, and Q are collinear and A is the midpoint of PQ.

To Prove:

Consider an ∆ABC,

Given, the points M and N are the midpoints of AB and AC respectively. CM and BN are produced to P and Q respectively such that CM=MP and BN=NQ.

Now, we prove that

(i) P, A, and Q are collinear.

(ii) A is the midpoint of PQ.

Construction:

From the above triangle ABC, draw a dotted line to join the points PA, AQ, and MN.

Proof:

In ∆APC, M and N are the midpoints of PC and AC respectively (from given)

Therefore, MN ∥ AP and MN = ½AP (by the midpoint theorem) —-(1)

In ∆ABQ, M and N are the midpoints of AB and BQ respectively (from given)

Therefore, MN ∥ AQ and MN = ½AQ (by the midpoint theorem) —-(2)

Thus, AP ∥ MN and AQ ∥ MN. ( from (1) and (2))

Now, AP and AQ lie in the same straight line because both passes through the same point A and are parallel to the same straight line MN.

∴ The points P, A, and Q are collinear. [(i) proved]

Also, ½AP = ½AQ (from statements (1) and (2)) —-(3)

⇒ AP = AQ (from statement (3))

Therefore, A is the midpoint of PQ. [(ii) proved]

Hence, the given statement is proved and the points are collinear by a midpoint theorem.

Also, refer:

FAQs on How to Prove Points are Collinear using MidPoint Theorem

1. How do you prove the given points are collinear?
Three or more points are on the same straight line are collinear, if the slope of any two pairs of points is the same. For example, we have three points A, B, and C, the pairs formed are AB, BC, and CA respectively. If the slope of AB = Slope of BC = Slope of AC, then the points A, B, and C are collinear points.

2. How do you prove three points are collinear vectors?
Take the three points with position vectors a, b, and c. To prove the vectors are collinear, if and only if the vectors (a-b) and (a-c) are parallel. In other words, to prove the collinearity of the vectors, we need to show (a-b) = k(a-c), where k is a constant.

3. What is the theorem of the midpoint?
The midpoint theorem states that ” The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Area and Perimeter of Combined Figures – Definition, Facts, Examples | How do you find the Area and Perimeter of Combined Shapes?

December 18, 2021

In Mathematics, Area and Perimeter are two important properties of two-dimensional shapes which is used in our day-to-day life. The Area and Perimeter of different combined figures are explained here. Before that, you must learn about different basic shapes such as triangle, square, rectangle, circle, and sphere, etc. The Perimeter is defined as the distance of the boundary of the shape whereas the Area will explain the region occupied by it.

In this article, we will discuss the definition of Combined Figures, how to find Area and Perimeter of Combined Figures, Solved Example problems, and so on. There are various types of shapes but commonly used are square, rectangle, triangle, circle.

Also, Read :

Combined Figures – Definition

Combined Figures is defined as shapes that are composed of a combination of simpler shapes. Each shape has its own area and perimeter formula. It is applicable to any shape and size whether it is regular or irregular.

To find the perimeter, add all the outside sides of our shape, and find the area we divide our shapes into simple shapes, calculate the area of those shapes separately, and then add areas up to get total.

How to find Area and Perimeter of Combined Figures?

The following are the steps for finding the area and perimeter of combined figures:
How to find Area of Combined Shapes?
Step 1: First, divide the compound shape into a basic regular shape.
Step 2: Next, find each basic shape area separately.
Step 3: Then Add all the areas of basic shapes together.
Step 4: Now, write the final answer in square units.

How to find the Perimeter of Combined Figures?
The perimeter of the combined figure is the length of the outline of a shape. To find the perimeter of any shape like rectangle, square, and so on you have to add all the lengths of four sides. Consider x is in this case the length of the rectangle and y is the width of the rectangle.

Read Similar Shapes:

Area and Perimeter of Combined Shapes Examples

Problem 1: Find the area and perimeter of the following combined figures. The below- combined figure consists of a square and a triangle.

Solution:
As given in the question, the combined figure is given.
Now, we will find the area and perimeter of a combined figure.
The combined figure consists of a square and equilateral triangle.
First, find the area of a square and the area of an equilateral triangle.
We know, the formulas of the area of a square and the area of an equilateral triangle.
The area of a square is a² and the area of an equilateral triangle is (√3/4) × a² square units.
Substitute the value in the above formulas, we get
Area of combined figures = Area of a square + Area of an equilateral triangle.
A= a² + (√3/4) × a²
A = 8 x 8 x (1.732/4) x 8 x 8 (√3 = 1.732)
A = 64 x 1.732 x 16 = 1774 sq.units.
So, the area of combined figures is 1774 sq. units.
Now, find the perimeter. So, the perimeter of combined figures is AB+BC+CD+DE+EA
P= 8+8+8+8+8 = 40 cm.
Hence the area and perimeter of the combined figure are 1774 sq. units and 40 cm.

Problem 2: A combined Figure is as given below. Find the area and perimeter of that Figure?

Solution:
The given figure is a combination of a rectangle and two circles. ABCD is a rectangle and AEB, DFC is two semicircles both are having equal area.
The length of the rectangle is l = 4cm.
The breadth of the rectangle is, b = 2cm.
The diameter of a semicircle is 2 cm.
The radius of a semicircle is r = 22cm.
Now, we will find the area and perimeter of the combined figure.
So, the perimeter of the given combined figure is, AD+BC+AEB+DFC
P= 4+4+2×21 (circumference of a circle)
P= 8+ 2 x 21 x 2πr = 8+2 x 722 x 1
P= 8+2 x 3.14 = 14.28 cm.
So, the perimeter of the given combined figure is 14.28 cm.
Next, find the area of combined figures.
Area of a combined figure = Area of a rectangle (ABCD)+ 2 x Area of a semicircle.
After substituting the values, we will get the value that is
A = l x b + 2 x 2πr²
A= 4×2+2x7x22x1x1 = 11.14 cm²
Therefore, the area and perimeter of the given combined figure are 11.14 cm² and 14.28 cm.

Problem 3: Find the area of a below-given combined figure.

Solution:
As given in the question, the figures consist of a square and semicircle.
Now, we will find the area of a combined figure.
So, the Area of a combined figure = Area of a square + Area of a semicircle
A = sxs+1/2 π(rxr) = 14 X 14 + (1/2)(22/7)(7 x 7) = 77+196
A= 273 cm²
Thus the area of a given combined figure is 273 cm².

Problem 4: Find the perimeter of a given figure.

Solution:
In the given question consists of a figure.
Now, we will find the perimeter of that figure.
Perimeter means adding all the lengths of shapes.
Perimeter of a combined figure is,
P = 4 cm+3 cm+8 cm+5cm.
P = 20cm
Hence, the Perimeter of a given combined figure is 20cm.

Problem 5: The figure is shown in the below figure. Using figure values find the area and perimeter of a combined figure?

Solution:
The given figure consists of four semi-circles and one square.
Now, we will find the perimeter and area of the figure.
The diameter of each semicircle is 7cm.
The length of each square side is 7cm.
Using the diameter value, we will find the radius.
Radius = d/2 = 7/2 = 3.5cm.
First, find the perimeter, we have to add the four semicircles and a square. The resultant sum is Perimeter.
So, the perimeter of a given figure = 4 x Perimeter of semicircles + perimeter of a square.
P = 4 (Πr )+ 4a = (22/7)(3.5 cm) + 4(7).
P = 4(22 x 0.5 )+ 28 = 44 cm + 28 cm = 72 cm.
Thus the perimeter of the given figure is 72cm.
Next, find the area of a given figure.
Area of a combined figure = Area of 4 semicircles + Area of a square.
The Area of 1 semicircle is Πr²/2
= (1/2) x (22/7) x (7/2)²= (1/2) x (22/7) x (7/2) x (7/2)
= (1/2) x 11 x (7/2)= 77/4= 19.25 cm²
Then, the Area of 4 semi circles = 4 (19.25)= 77cm².
The area of a square is a²= (7)²= 7 x 7 = 49 cm².
Area of a combined figure = 19.25 cm² + 49 cm² = 126 cm².
Hence, the area and perimeter of a given combined figure are 72 cm and 126 cm².

Area of a Circular Ring – Definition, Formula, Examples | How do you find the Area of a Circular Ring?

December 13, 2021

In this article, you will learn about how to find the Area of a Circular Ring. A circular ring is a plane figure which is bounded by the circumference of two concentric circles of two different radii. A circle is made of multiple points arranged equidistant from a single central point and that point is called the center of the circle. The outer circle and inner circles define that the ring is concentric, that shares a common center point. The best way to think of it is a circular disk with a circular hole in it.

On this page, we will discuss the definition of the area of a circular ring, formulas, solved example problems, and so on.

Also, Read:

Area of a Circular Ring – Definition

The area of a circular ring will be subtracted from the area of a large circle to the area of a small circle. A ring-shaped object is bounded by the circumference of two concentric circles of two different radii. The dimensions of the two radii are R, r, which are the radii of the outer ring and the inner ring respectively.

Formula:

To find the area of a circular ring or annulus, to multiply the product of the sum and the difference of the two radii. It will be bounded by two concentric circles of radii R and r (R>r).
Therefore, the area of a circular ring is, area of the bigger circle – the area of the smaller circle.
= π(R + r) (R – r) = πR²- πr²
where R and r are the outer circle radius and the inner circle radius. So, it will be as R=√ r2+Aπ and r = √ R2+Aπ.

Area of a Circular Ring Examples

Problem 1:

Find the area of a flat circular ring formed by two concentric circles, circles with the same center whose radii are 8cm and 4cm?

Solution:
Given in the question, the values are
The radius (r1) of the bigger circle is 8cm.
The smaller circle radius (r2) is 4cm.
Now, we have to find the area of a circular ring.
As we see the required area is between the two circles as shown in the figure.
Using the formula, we can find the value.
The area of the shaded portion is = Area of the bigger circle – Area of the smaller circle.
A = πR²- πr²
Substitute the given values within the above formula, we get
A = π(8×8) – π(4×4) = 64π – 16π = 48π
48π = 48 x 22/7 = 6.85 x 22 = 150.7 cm²
Therefore, the area of the circular ring formed by two concentric circles is 150.7 cm².

Problem 2:

A path is 21cm wide surrounds a circular lawn with a diameter of 240cm. Find the area of the path?

Solution:
As given in the question,
A circular dawn diameter is 240cm.
So, the radius of the inner circle(r) is 120 cm.
The wide of a path is 21cm.
The radius of the outer circle(R) is 120+21 = 141cm.
Now, we have to find the area of a path.
We know the formula, Area of a path is π(R+ r) (R – r).
After the substitution of the value, we get,
Area of the path = 22/7(141+120)(141-120).
= 22/7(161)(21) = 22 x 23x 21=10616Sq.cm
Thus, the Area of the path is 10616sq.cm.

Problem 3:

The inner diameter and the outer diameter of the circular path are 628 m and 600m respectively. Find the breadth of a circular path and the area of the circular path. Consider the π value as 22/7.

Solution:
As given in the question,
The outer radius of a circular path (R) is, 628/2 = 314m.
The inner radius of a circular path (r) is, 600/2 = 300m.
Now, we have to find the breadth of a circular path and the area of a circular path.
So, the breadth of a circular path is R-r = 314m – 300m = 14m.
Next, the Area of a circular path is, π(R + r)(R – r)
Area = 22/7 (314+300)(314-300)
A= 22/7(614)(14)m² = 27016m².
Therefore, the area of the circular path is 27016m².

Problem 4:

Find the area of a circular ring formed by two concentric circles whose radii are 6.2cm and 5.8cm respectively. Take the π value as 3.14.

Solution:
As we know the radii of the outer circle and inner circles will be R and r respectively.
The concentric circle (R) value is 6.2 cm and the concentric circle (r) value is 5.8 cm.
Now, we will find the area of a circular ring value.
We know the formula, Area of a circular ring is π(R+ r) (R – r).
Substitute the given values in the above formula, we get
A=π(R+ r) (R – r) = 22/7 (6.2 +5.8)(6.2 – 5.8)
$=3.14\times 12 \times 0.4$ sq. cm
$=15.072$ sq. cm
Hence, the area of a circular ring value is 15.072 sq. cm

Problem 5:
A circular ring is 8cm wide. Find the difference between the outer circle radius and inner circle radius?

Solution:
As given in the question, the circular ring wide is 8cm.
We have to find the difference value of the outer circle and inner circle radius.

The difference between the outer circle radius and inner circle radius is R-r.
So, the value of R-r is 8 cm.
Thus, the difference value of an inner circle radius and an outer circle radius is 8cm.

Midpoint Theorem on Right Angled Triangle – Statement & Proof | How do you find the Midpoint of a Right Angled Triangle?

November 30, 2021

This article aids the children to gain more knowledge about Midpoint Theorem on Right-angled Triangle. Usually, we know a triangle is cited as a polygon that has three sides of three line segments. In other words, a triangle is just a closed figure where the sum of its angles is equal to 180 degrees. Every shape of a triangle is classified based on the angle made by the two adjacent sides of a particular triangle.

The right-angle triangle is a geometrical shape, it plays a prominent role, and is considered as a basis in trigonometry. On this page, we will be proving that in a right-angled triangle the median drawn to the hypotenuse is half the hypotenuse in length. Let us discuss more on right-angle triangles and prove the statement of the midpoint theorem on the right-angled triangles with a few examples.

Right-angled Triangle – Definition

A right-angled triangle is defined as if one of the angles of a triangle is a 90 degrees right angle, then the triangle is called a right-angled triangle or simply can call a right triangle. In a triangle, the relation between the various sides can be easily realized with the help of a Pythagoras rule. These right-angles triangles are of two types namely, isosceles right-angled triangle and a scalene right-angled triangle.

The side opposite to the right angle is called hypotenuse and it is the largest side of the right-angle triangle. In the following image, triangle ABC is a right triangle, with three sides namely, the base, the altitude, and the hypotenuse.

In the above right-angled triangle, AB is the altitude, BC is the base, and the AC is the hypotenuse. AC is the largest side and it is opposite to the right angle within a triangle.

Midpoint of Right Angled Triangle Formula

The formula is represented as the square of the hypotenuse is equal to the sum of the square of the base and the square of the altitude.

Formula: (Hypotenuse)² = (Base)² + (Altitude)².