Word Problems on Proportion

Word Problems on Proportion | Free Printable Proportion Word Problems with Solutions PDF

The word problems on the proportion aid you in a mathematical comparison between two numbers. The proportion gets majorly based on ratio and fractions. This article on Proportion Word Problems Worksheet with Answers PDF gives you a various number of proportion problems. At the end of this page, you can get clear knowledge of the concept of proportion. It says when two ratios are equivalent they are in proportion.

Proportion encourages solving many real-life problems. The ratio and proportion are key foundations to grasp the various concepts in mathematics. Proportions are denoted by the symbol “::”, “=”. This free printable Worksheet on Word Problems on Proportion provides various types of questions with answers to make you understand clearly how to solve the proportions problems in exams. By solving proportion examples with answers pdf, you can also improve your problem-solving skills & math skills.

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Proportion Word Problems with Answers

Example 1:
If the numbers are 4, 15, 7, and 20. What number should be added to make the numbers proportional?
Solution:
Given numbers are 4, 15, 7, and 20.
Let the needed number be k.
Now, we write the numbers according to the problem
4+k, 15+k, 7+k, and 20+k are proportional numbers.
Here,
\(\frac{4+k}{15+k}\) = \(\frac{7+k}{20+k}\)
⇒ (4+k) × (20+k) = (7+k) × (15+k)
⇒ 80+ 4k +20k +k² = 105+ 7k+ 15k+ k²
⇒ 24k+80 = 22k+42
⇒ 24k- 22k = 105-80
⇒ 2k = 25
⇒ k = \(\frac{25}{2}\)
⇒ k = 12.5
Thus, the required number is 12.5.

Example 2:
Priya enlarged the size of a photo to a height of 20 inches. What is the new width if it was originally 4 inches tall and 2 inches in width?
Solution:
The height of a photo is 20 inches.
Now, to find the new width.
The ratio to calculate is 4: 2 = 20: w
⇒ \(\frac{4}{2}\) = \(\frac{20}{w}\)
⇒ 4w = 20×2
⇒ 4w = 40
⇒ w = \(\frac{40}{4}\)
⇒ w = 10.
Thus, the new width of the photo is 10inches.

Example 3:
If P is directly proportional to n when P is 4 and n is 6. Find the value of P when n is 8.
Solution:
Given P is proportional to n ( P ∝ n ).
Now, convert to an equation multiplied by k the constant of variation.
⇒ P = nk
To find k use the given condition
P = 4 when n = 6
P = nk
⇒ k = \(\frac{n}{P}\) = \(\frac{6}{4}\)
Now, the equation is P = \(\frac{6}{4}\) × n = \(\frac{6n}{4}\)
When n = 8, then
P= \(\frac{6n}{4}\) = \(\frac{6×8}{4}\) = \(\frac{48}{4}\) = 12.
Hence, the value of P is 12 when n is 8.

Example 4:
Find the third proportion of 14 and 22?
Solution:
The proportions given are 14 and 22.
Let the third proportion be x.
According to the problem, 14 and 22 are in continued proportion.
Now,
\(\frac{14}{22}\) = \(\frac{22}{x}\)
⇒ 14× x = 22 × 22
⇒ 14x = 484
⇒ x = 484/14 = 34.57
Therefore, the third proportion is 34.57.

Example 5:
Ram, Prudhvi, and Mahesh have $15, $22, and $25 respectively with them. Father asks them to give him an equal amount so that the money held by them now is in continued proportion. Find the amount taken from each of them?
Solution:
Let the amount taken from each of them be $m.
Now,
15-m, 22-m, and 25-m are in continued proportion.
Thus, \(\frac{15-m}{22-m}\) = \(\frac{22-m}{25-m}\)
⇒ (15-m)(25-m) = (22-m)(22-m)
⇒ 375 – 15m -25m +m² = 484 – 44m +m²
⇒ 375 – 40m = 484 -44m
⇒ 375 – 484 = -44m +40m
⇒ -109 = -4m
⇒ m = 109/4
⇒ m = 27.25
Therefore, the required amount is $27.25.

Example 6:
Keerti ran 150meters in 25seconds. How long did she take to run 3meters?
Solution:
Given that
keerti ran 150metres in 25seconds.
Now, to find the time she takes to run 3mts.
Let k be the time required.
\(\frac{25}{150}\) = \(\frac{k}{3}\)
⇒ k = \(\frac{25}{150}\) × 3
⇒ k = \(\frac{75}{150}\)
⇒ k = 0.5
Thus, keerti took 0.5seconds to complete 3meters.

Example 7:
Check whether the following numbers form a proportion or not.
(i) 4.5, 3.5, 6.6, and 8.8
(ii) 2\(\frac{2}{4}\), 1\(\frac{3}{2}\), 2.2, and 5.5
Solution:
(i) Given 4.5: 3.5 = \(\frac{4.5}{3.5}\) = \(\frac{45}{35}\) = \(\frac{9}{7}\)
6.6: 8.8 = \(\frac{6.6}{8.8}\) = \(\frac{66}{88}\) = \(\frac{6}{8}\)
Therefore, \(\frac{4.5}{3.5}\) ≠ \(\frac{6.6}{8.8}\)
Thus, 4.5, 3.5, 6.6, and 8.8 are not in proportion.

(ii) Given 2\(\frac{2}{4}\), 1\(\frac{3}{2}\), 2.2, and 5.5
2\(\frac{2}{4}\): 1\(\frac{3}{2}\)
= \(\frac{10}{4}\): \(\frac{5}{2}\)
= \(\frac{10}{4}\) × 4: \(\frac{5}{2}\) × 4
= 10: 10 = 1: 1
2.2: 5.5 = \(\frac{2.2}{5.5}\) = \(\frac{22}{55}\) = \(\frac{2}{5}\) = 2: 5
Thus, the given numbers 2\(\frac{2}{4}\), 1\(\frac{3}{2}\), 2.2, and 5.5 are not in proportion.

Example 8:
Find the fourth proportional of numbers 3, 6, and 12?
Solution:
Given three proportional numbers are 5, 6, and 12.
Let the fourth proportional be x.
Now, 3, 6,12, and x be the proportionality numbers.
Thus,
\(\frac{3}{6}\) = \(\frac{12}{x}\)
⇒ 3x = 12×6
⇒ 3x = 72
⇒ x = \(\frac{72}{3}\)
⇒ x = 24.
Hence, the fourth proportional number is 24.

Example 9:
If the mean proportion is 16 and the third proportional is 64 then find the two numbers?
Solution:
Given mean proportion is 16 and the third proportion is 64.
Let the required numbers be a and b.
According to the given problem,
√ab = 16
⇒ ab = 16²
⇒ ab = 256
Now, \(\frac{b²}{a}\) = 64
⇒ b² = 64a
⇒ a = \(\frac{b²}{64}\)
Substitute, a = \(\frac{b²}{64}\) in ab = 256
⇒ \(\frac{b²}{64}\) × b = 256
⇒ \(\frac{b³}{64}\) = 256
⇒ b³ = 256 × 64
⇒ b³ = 28 × 26
⇒ b³ = 212 × 2²
⇒ b = 24 ×2²
⇒ b = 26
⇒ b = 64
So, from the equation a = \(\frac{b²}{64}\), we get
a = \(\frac{64²}{64}\)
⇒ a = \(\frac{4096}{64}\)
⇒ a = 64
Therefore, the required two numbers are 64 and 64.

Word Problems on Ratio

Word Problems on Ratio | Free Printable Ratio Word Problems with Answers

Are you searching for the word problems on ratio, this page helps you to get answers to your questions. Here, students or teachers can get many types of ratios word problems which helps to understand the concept thoroughly. The word ‘Ratio’ is a term used to compare two or more numbers or quantities.

By using ratio it is very simple to solve the problems and gives the result in the simplest form. We use ‘:‘ to denote the ratio while comparing the quantities and read as ” is to “. Let us practice some examples from this ratio word problems worksheet with answers pdf and get an idea of how to solve the problems on ratio.

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Ratio Word Problems with Solutions

Example 1: 
Find the quantity if it is divided in the ratio 6: 4, the smaller part is 48.
Solution: 
Let the quantity be ‘m’.
Then the ratio of two parts is written as \(\frac{6m}{6+4}\) and \(\frac{4m}{6+4}\).
Here, the smaller part is 48, we get
\(\frac{4m}{6+4}\) = 48
⇒ \(\frac{4m}{10}\) = 48
⇒ 4m = 48×10
⇒ 4m = 480
⇒ m = \(\frac{480}{4}\)
⇒ m = 120
Therefore, the quantity is 120.

Example 2:
A herd of 50 cows and buffaloes has 14 cows and some buffaloes. What is the ratio of buffaloes to cows?
Solution:
Given a total of 50 cows and buffaloes.
There are 14 cows given.
To get the count of buffaloes = 50 – 14 = 36.
The ratio of buffaloes to cows is 36: 14 = 18: 7.
Hence, the simplest form of ratio is 18: 7.

Example 3:
Mr. Ram’s class has 40 students, of which 18 are girls. Find the ratio of girls to boys?
Solution:
Total students in a class = 40.
No. of girls in a class = 18 (given)
No. of boys in a class = Total students – No. of girls = 40 – 18 = 22.
There are 22 boys in a class.
The ratio of girls to boys is 18: 22.
Here, there is a common factor 2. So, we can write the ratio in the simplest form by dividing it with the factor 2.
The simplest form of ratio is 9: 11.

Example 4:
If the ratio of p: q = 4: 3, then find (2p- 3q) : (5p+q)?
Solution:
Given p: q = 5: 3, then p = 5m and q = 3m (m ≠ 0 is a common multiplier).
Now, (2p- 3q) : (5p+q) = \(\frac{2p-3q}{5p+q}\)
= \(\frac{(2×5m)-(3×3m)}{(5×4m)+3m}\)
= \(\frac{10m-9m}{20m+3m}\)
= \(\frac{1m}{23m}\)
= \(\frac{1}{23}\)
= 1: 23.
Therefore, the ratio (2p- 3q) : (5p+q) = 1: 23.

Example 5:
Two numbers are in the ratio 3: 4. If 3 is added to the first number and 8 is added to the second number, they are in the ratio of 3: 5. Find the numbers?
Solution:
Given the ratio of two numbers is 3: 4.
Let the numbers be 3x and 4x.
According to the problem given,
\(\frac{3x+3}{4x+8}\) = \(\frac{3}{5}\)
⇒ 5(3x+3) = 3(4x+8)
⇒ 15x+15 = 12x+24
⇒ 15x-12x = 24-15
⇒ 3x = 9
⇒ x = \(\frac{9}{3}\)
⇒ x = 3.
Hence, the original numbers are 3x = 3×3 = 9 and 4x = 4×3 = 12.
Therefore, the numbers are 9 and 12.

Example 6:
Find the ratio of a: c from the quantities a: b = 4: 5, b: c = 2: 6?
Solution:
Given the ratios a: b = 4: 5, b: c = 2: 6.
a: b = 4: 5 ⇒ \(\frac{a}{b}\) = \(\frac{4}{5}\) ——(i)
b: c = 2: 6 ⇒ \(\frac{b}{c}\) = \(\frac{2}{6}\) ——(ii)
Now, multiply the equations (i) and (ii), we get
\(\frac{a}{b}\) × \(\frac{b}{c}\) = \(\frac{4}{5}\) × \(\frac{2}{6}\)
⇒ \(\frac{a}{c}\) = \(\frac{8}{30}\)
The ratio a: c = 8: 30 = 4: 15.
Thus, the ratio a: c = 4: 15.

Example 7:
If we have the ratio of tomatoes to apples is 2: 4. If there are 18 tomatoes. How many apples are there?
Solution:
The ratio of tomatoes to apples is 2: 4 means that for every 2 tomatoes, you have 4 apples.
Here, you have 18 tomatoes, or we can say 9 times as much.
So, you need to multiply the apples by 9.
⇒ The apples we have is 4 = 4 × 9 = 36.
Thus, there are 36 apples.

Example 8:
If the equation (2x+5y): (6x-4y) = 8: 5. Find the ratio of x: y?
Solution:
Given, (2x+5y): (6x-4y) = 8: 5
Now,
\(\frac{2x+5y}{6x-4y}\) = \(\frac{8}{5}\)
⇒ 5(2x+5y) = 8(6x-4y)
⇒ 10x+25y = 48x-32y
⇒ 10x – 48x = -32y – 25y
⇒ -38x = -57y
⇒ 38x = 57y
⇒ \(\frac{x}{y}\) = \(\frac{57}{38}\)
⇒ x: y = 57: 38.
Hence, that ratio of x: y is 57: 38.

Example 9:
Manoj leaves $ 2461600 behind. Manoj’s wish was the money is to be divided between his son and daughter in the ratio of 3: 2. Find the money received by his son and daughter?
Solution:
Manoj has money of $ 2461600 and is to be shared with his son and daughter in the ratio of 3: 2.
We know if a quantity x is divided in the ratio of a: b, then the two parts are look alike. i.e., \(\frac{ax}{a+b}\) and \(\frac{bx}{a+b}\).
Now, the money received by his son = \(\frac{3}{3+2}\) × $ 2461600
= \(\frac{3}{5}\) × $ 2461600
= 3 × $ 492320
= $ 1476960
Next, the money received by his daughter = \(\frac{2}{3+2}\) × $ 2461600
= \(\frac{2}{5}\) × $ 2461600
= 2 × $ 492320
= $ 984640
Thus, the money received by Manoj’s son is $ 1476960 and by his daughter is $ 984640.

Worksheet on Comparison on Ratios

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

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

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

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: b2
Now, 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 Ratios in Lowest Term

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

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.

Do Check:

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

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

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.

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.

Collinear Points Theorem Statement

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.

Collinear Points Theorem Construction

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

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

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

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

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)
 sq. cm
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

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

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.

Right-angled Triangle

 

 

 

 

 

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.

Midpoint Theorem on Right-Angled Triangle Statement & Proof

Prove that in a right-angled triangle the length of the median drawn to the hypotenuse is half the hypotenuse in length.

Right-angled Triangle with Median

Given:

In ∆ABC, ∠B = 90°, and BD is the median drawn to hypotenuse AB.

To Prove:

We have to prove that in a right-angled triangle ABC,

BD = ½ AC.

Construction:

From the above diagram ∆ABC, construct a right-angled triangle to prove the theorem statement.

Draw a line E and DE ∥ BC such that DE cuts AB at E.

Constructed right-angled triangle

Proof:

In ∆ABC, AD = ½ AC ( D is the midpoint of AC )

In ∆ABC,

(i) D is the midpoint of AC (by given)

(ii) DE ∥ BC (by construction)

Therefore, E is the midpoint of AB (by the converse of the midpoint theorem) —- (1)

ED ⊥ AB (since ED ∥ BC and BC ⊥ AB) —- (2)

In ∆AED and ∆BED,

(i) AE = EB (from (1))

(ii) ED = ED (common side)

(iii) ∠AED = ∠BED = 90° (from (2))

Therefore, ∆AED ≅ ∆BED (by SAS congruency theorem)

AD = BD (by c.p.c.t) —- (3)

Therefore, BD = ½ AC (using (3) in AD = ½ AC).

Hence, the theorem is proved.

Do Check:

Examples on Midpoint of Hypotenuse of Right-angled Triangles

Example 1:
In ∆PQR, QS is the median of ∆PQR. If the hypotenuse of PR= 12cm. Find the length of the median QS.

Example1 for Right-angled Triangle
Solution:
We know in a right-angled triangle, the length of the median of the hypotenuse is half the length of the hypotenuse.
Given PR= 12cm
Now, have to find the length of the median QS.
The formula used for a right-angled triangle is QS= ½PR.
QS= ½×12
⇒ QS= 6cm.
Therefore, the length of the median QS of ∆PQR is 6cm.

Example 2:
In ∆ABC, BD is the median of ∆ABC. If the length of the median BD is 14cm. Find the length of the hypotenuse AC.
Example2 for Right-angled Triangle
Solution: 
The statement used for a right-angled triangle is the median drawn to the hypotenuse is half the hypotenuse in length.
Given the length of the median BD= 14cm.
Now, find the length of the hypotenuse BD.
The formula we use is BD = ½AC.
14= ½ × AC
⇒ 14 × 2 = AC
⇒ AC= 28cm.
Therefore, the length of the hypotenuse AC of ∆ABC is 28cm.

FAQ’s on Midpoint of Right-Angled Triangle

1. What is the midpoint of the hypotenuse of a right triangle called?
The midpoint of the hypotenuse of a right-angled triangle or a right triangle is known as the circumcenter.

2. Does the median of a triangle form a right angle?
The median of a triangle does not always form a right angle to the side on which it is falling. In case, if we have an equilateral triangle or an isosceles triangle one median falls on the non-equal side of an isosceles triangle.

3. What is the midpoint of a right triangle?

The midpoint of the hypotenuse of a right triangle is the circumcenter of the triangle. Let assume any of the three points on the circle like A(a,0), B(b,0), and C(b,c) thus, the midpoint of the hypotenuse is equal to the center of the circle.

4. Why is the median half the hypotenuse?
Normally, the median of a triangle is drawn from one vertex to the midpoint of the opposite side of the vertex. In a right-angled triangle, the median to the hypotenuse has its property, i.e., the length of the hypotenuse is equal to half the length of the hypotenuse.

Problems on Ratio

Problems on Ratio | Download Free Ratio Problems with Answers

Worked out Problems on Ratio are available for your understanding. Learn how to solve various Ratio Problems by going through the step-by-step solutions provided. Sample Ratio Questions and Answers made available will assist you to learn the concept of ratios easily. Learn the formulas, tips & tricks for solving the ratio problems. All the Ratio Questions and Answers available here are given in a simple and easy-to-understand manner so that you can proceed and enhance your math skills. Become proficient in the concept by having enough practice as it is a must-have skill.

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Ratio Problems with Solutions PDF

I. If X: Y=5: 6, find the value of 3x+2y: 5x+y.

Solution:

Given that,
X: Y=5: 6 then X=5k, Y=6k where k is the common multiplier.
Therefore, 3x+2y: 5x+y=3x+2y/5x+y=3(5k)+2(6k)/5(5k)+6k
=15k+12k/25k+6k
=27k/31k
=27:31
Therefore, 3x+2y: 5x+y=27:31.


II. Divide 1500 into A, B, C such that A is 3/4th of B and B: C =2:5?

Solution:

Given that,
A is 3/4th of B –>A=3/4B–>A/B=3/4
B/C=2/5
For finding A:B: C we have to multiply A/B with 2 and B/C with 4.
A/B=6/8 and B/C=8/20
So A: B: C=6: 8: 20
6x+8x+20x=1500
34x=1500
x=1500/34
=44.11
6x=6(44.11)=264.66
8x=8(44.11)=352.88
20x=20(44.11)=882.2
Therefore, A,B and C will get Rs 264.66,Rs 352.88 and Rs 882.2.


III. If (2x + 3y) : (7x – 5y) = 5 : 3 then find the ratio x : y?

Solution:

Given that,
(2x + 3y) : (7x – 5y) = 5 : 3
2x+3y/7x-5y=5/3
3(2x+3y)=5(7x-5y)
6x+9y=35x-25y
6x-35x=-25y-9y
-29x=-34y
29x=34y
x/y=34/29
X: Y=34/29
Therefore, X: y=34:29.


IV. If a : b = 3 : 8, b : c = 7 : 2 and c : d = 6 : 14 , what is a : d?

Solution:

Given that,
a : b = 3 : 8-> a/b=3/8 –>(1)
b: c=7:2-> b/c=7/2 —->(2)
c: d=6:14->c/d=6/14 —>(3)
Multiplying (1),(2), and (3) we get,
a/b × b/c × c/d=3/8 × 7/2 × 6/14
a/d=9/16
Therefore, a:d=9:16.


V. If 3A=5B=7C, find A:B: C.

Solution:

let 3A=5B=7C=k
3A=k–> A=k/3
5B=k–>B=k/5
7c=k–>C=k/7
A:B:C=?
k/3:K/5:K/7
L.C.M of (3,5,7) is 105.
35k:21k:15k
=35:21:15
Therefore, A:B:C=35:21:15.


VI. If x : y=2 : 3,y : z=4 : 7 then find x : z.

Solution:

Given that,
x : y=2 :3,y : z=4 :7
x : y=2 :3 => x/y=2/3
y : z=4 : 7=>y/z=4/7
x/y× y/z=2/3× 4/7
Therefore,x/z=8/21
Hence, x : z=8 : 21


VII. A School has 520 students. The ratio of the number of boys to the number of girls is 4:1. This ratio changes to 5:3 after the admission of 80 new students. Find the number of newly admitted girls?

Solution:

Given that,
Number of students in the school=520
The ratio of the number of boys to the number of girls is 4:1.
Let the number of boys=4x
Let the number of girls=x
Therefore, 4x+x=520
5x=520
x=520/5=104
Boys=4x=4(104)=416
Girls=x=104
Also given no. of students newly admitted=80
i.e. Total no. of students in the school=520+80=600
The ratio of Boys to Girls=5:3
Let the number of boys be 5x.
Let the number o girls be 3x.
5x+3x=600
8x=600
x=600/8=75
B=5x=5(75)=375
G=3x=3(75)=225
The number of newly admitted Girls=225-104=121
Therefore, the number of newly admitted Girls is 121.


VIII. A milkman has a can of 180 liters of milk and water, the ratio of milk and water is 7:2. How much water must be added to the can to make this ratio 4:2?

Solution:

Given that,
The mixture of milk and water=180 lit
The ratio of milk and water is=7/2
Therefore, Milk/Mix=7/9 (Since mix=milk+water=7+2=9)
Milk=7/9×Mix
Milk=7/9×180=7(20)=140 liters
Water=180-140=40 liters
Let the water added to the mixture be x.
milk/water+x=4/2
140/40+x=4/2
2(140)=4(40+x)
280=160+4x
4x=120
x=120/4=30 liters
Therefore, 30 liters of water should be added to the can.


IX. The monthly pocket money of Raju and Sandeep is in the ratio 7:9. Their expenditures are in the ratio 5:7. If each saves Rs 60 every month, find their monthly pocket money?

Solution:

The monthly pocket money of Raju and Sandeep is in the ratio 7:9.
Let the Raju pocket money be 7x.
Let the Sandeep pocket money b 9x.
Their expenditures are in the ratio 5:7.
Let the Raju expenditure be 5y.
Let the Sandeep expenditure be 7y.
Each saves Rs 60 every month.
i.e. 7x-5y=60–>(1)*7
9x-7y=60 –.(2)*5
49x-35y=480
45x-35y=300
–   +        –
—————
4x       =180
—————-
x=180/4=45
subs x=45 in eq(1)
7(45)-5y=60
315-5y=60
5y=255
y=255/5=51
Monthly pocket money of Raju=7x=7(45)=Rs 315
Monthly pocket money of Sandeep=9x=9(45)=Rs 405
Therefore, The monthly pocket money of Raju and Sandeep are Rs 315 and Rs 405.


X. A profit of Rs 8400 is to be shared among three partners in the ratio 3: 4: 5. How much does each partner get?

Solution:

Given that,
Profit=Rs 8400
Profit of Rs 8400 has to be shared among three partners in the ratio 3: 4: 5.
Let the first partner share be 3x.
Let the second partner share be 4x.
Let the third partner share be 5x.
3x+4x+5x=8400
12x=8400
x=8400/12=700
So, 3x=3(700)=2100
4x=4(700)=2800
5x=5(700)=3500
Therefore, the shares of the three partners are Rs 2100, Rs 2800, Rs 3500.