Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2)
Mensuration – I (Area of Circle Exercise 21.2) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
1. Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km
Solution:
(i) Given radius = 7 cm
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 cm
2
(ii) Given radius = 2.1 m
We know that area of circle = p r
2
A = (22/7) × (2.1)
2
A = 22/7 × 4.41
A = 13.86 m
2
(iii) Given radius = 7 km
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 km
2
2. Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km
Solution:
(i) Given diameter = 8.4 cm
Therefore radius, r = d/2 = 8.4/2 = 4.2
We know that area of circle = p r
2
A = (22/7) × (4.2)
2
Page 2
1. Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km
Solution:
(i) Given radius = 7 cm
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 cm
2
(ii) Given radius = 2.1 m
We know that area of circle = p r
2
A = (22/7) × (2.1)
2
A = 22/7 × 4.41
A = 13.86 m
2
(iii) Given radius = 7 km
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 km
2
2. Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km
Solution:
(i) Given diameter = 8.4 cm
Therefore radius, r = d/2 = 8.4/2 = 4.2
We know that area of circle = p r
2
A = (22/7) × (4.2)
2
A = 55.44 cm
2
(ii) Given diameter = 5.6 cm
Therefore radius, r = d/2 = 5.6/2 = 2.8
We know that area of circle = p r
2
A = (22/7) × (2.8)
2
A = 24.64 cm
2
(iii) Given diameter = 7 km
Therefore radius, r = d/2 = 7/2 = 3.5
We know that area of circle = p r
2
A = (22/7) × (3.5)
2
A = 38.5 km
2
3. The area of a circle is 154 cm
2
. Find the radius of the circle.
Solution:
Given area of the circle = 154 cm
2
A = p r
2
154 = 22/7 r
2
r
2
= (154 × 7)/22
r
2
= 49
r = 7 cm
4. Find the radius of a circle, if its area is
(i) 4 p cm
2
(ii) 55.44 m
2
(iii) 1.54 km
2
Solution:
(i) We know that area of circle = p r
2
Given area = 4 p cm
2
A = p r
2
4 p = p r
2
r
2
= 4
Therefore r = 2 cm
Page 3
1. Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km
Solution:
(i) Given radius = 7 cm
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 cm
2
(ii) Given radius = 2.1 m
We know that area of circle = p r
2
A = (22/7) × (2.1)
2
A = 22/7 × 4.41
A = 13.86 m
2
(iii) Given radius = 7 km
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 km
2
2. Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km
Solution:
(i) Given diameter = 8.4 cm
Therefore radius, r = d/2 = 8.4/2 = 4.2
We know that area of circle = p r
2
A = (22/7) × (4.2)
2
A = 55.44 cm
2
(ii) Given diameter = 5.6 cm
Therefore radius, r = d/2 = 5.6/2 = 2.8
We know that area of circle = p r
2
A = (22/7) × (2.8)
2
A = 24.64 cm
2
(iii) Given diameter = 7 km
Therefore radius, r = d/2 = 7/2 = 3.5
We know that area of circle = p r
2
A = (22/7) × (3.5)
2
A = 38.5 km
2
3. The area of a circle is 154 cm
2
. Find the radius of the circle.
Solution:
Given area of the circle = 154 cm
2
A = p r
2
154 = 22/7 r
2
r
2
= (154 × 7)/22
r
2
= 49
r = 7 cm
4. Find the radius of a circle, if its area is
(i) 4 p cm
2
(ii) 55.44 m
2
(iii) 1.54 km
2
Solution:
(i) We know that area of circle = p r
2
Given area = 4 p cm
2
A = p r
2
4 p = p r
2
r
2
= 4
Therefore r = 2 cm
(ii) We know that area of circle = p r
2
Given area = 55.44 cm
2
A = p r
2
55.44 = p r
2
r
2
= (55.44 × 7)/22
r
2
= 17.64 m
r = 4.2 m
(iii) We know that area of circle = p r
2
Given area = 1.54 km
2
A = p r
2
1.54 = p r
2
r
2
= (1.54 × 7)/22
r
2
= 0.49 km
r = 0.7 km
r = 700m
5. The circumference of a circle is 3.14 m, find its area.
Solution:
Given circumference = 3.14m
We know that circumference of circle = 2 p r
3.14 = 2 × 3.14 × r
r = 3.14/ (2 × 3.14)
r = 0.5
We know that area of circle = p r
2
A = (22/7) × (0.5)
2
A = 0.785 m
2
6. If the area of a circle is 50.24 m
2
, find its circumference.
Solution:
Given area of a circle is 50.24 m
2
We know that area of circle = p r
2
50.24 = (22/7) × r
2
r
2
= (50.24 × 7)/22
r
2
= 15.985
Page 4
1. Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km
Solution:
(i) Given radius = 7 cm
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 cm
2
(ii) Given radius = 2.1 m
We know that area of circle = p r
2
A = (22/7) × (2.1)
2
A = 22/7 × 4.41
A = 13.86 m
2
(iii) Given radius = 7 km
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 km
2
2. Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km
Solution:
(i) Given diameter = 8.4 cm
Therefore radius, r = d/2 = 8.4/2 = 4.2
We know that area of circle = p r
2
A = (22/7) × (4.2)
2
A = 55.44 cm
2
(ii) Given diameter = 5.6 cm
Therefore radius, r = d/2 = 5.6/2 = 2.8
We know that area of circle = p r
2
A = (22/7) × (2.8)
2
A = 24.64 cm
2
(iii) Given diameter = 7 km
Therefore radius, r = d/2 = 7/2 = 3.5
We know that area of circle = p r
2
A = (22/7) × (3.5)
2
A = 38.5 km
2
3. The area of a circle is 154 cm
2
. Find the radius of the circle.
Solution:
Given area of the circle = 154 cm
2
A = p r
2
154 = 22/7 r
2
r
2
= (154 × 7)/22
r
2
= 49
r = 7 cm
4. Find the radius of a circle, if its area is
(i) 4 p cm
2
(ii) 55.44 m
2
(iii) 1.54 km
2
Solution:
(i) We know that area of circle = p r
2
Given area = 4 p cm
2
A = p r
2
4 p = p r
2
r
2
= 4
Therefore r = 2 cm
(ii) We know that area of circle = p r
2
Given area = 55.44 cm
2
A = p r
2
55.44 = p r
2
r
2
= (55.44 × 7)/22
r
2
= 17.64 m
r = 4.2 m
(iii) We know that area of circle = p r
2
Given area = 1.54 km
2
A = p r
2
1.54 = p r
2
r
2
= (1.54 × 7)/22
r
2
= 0.49 km
r = 0.7 km
r = 700m
5. The circumference of a circle is 3.14 m, find its area.
Solution:
Given circumference = 3.14m
We know that circumference of circle = 2 p r
3.14 = 2 × 3.14 × r
r = 3.14/ (2 × 3.14)
r = 0.5
We know that area of circle = p r
2
A = (22/7) × (0.5)
2
A = 0.785 m
2
6. If the area of a circle is 50.24 m
2
, find its circumference.
Solution:
Given area of a circle is 50.24 m
2
We know that area of circle = p r
2
50.24 = (22/7) × r
2
r
2
= (50.24 × 7)/22
r
2
= 15.985
r = 3.998 m
We know that circumference of circle = 2 p r
C = 2 × (22/7) × 3.998
C = 25.12 m
7. A horse is tied to a pole with 28 m long string. Find the area where the horse can
graze. (Take p = 22 / 7).
Solution:
Given the length of the string = 28m
The area over which the horse can graze is the same as the area of circle of radius 28 m
Hence required area = p r
2
A = (22/7) × (28)
2
A = 2464 m
2
8. A steel wire when bent in the form of a square encloses an area of 121 cm
2
. If the
same wire is bent in the form of a circle, find the area of the circle.
Solution:
Given area of the square = 121 cm
2
(Side)
2
= 121
Therefore side = 11 cm
We know that the perimeter of the square = 4 (side)
= 4 (11)
= 44 cm
According to the question circumference of the circle = perimeter of the square
So let r be the radius of the circle
2 p r = 44
2 × (22/7) × r = 44
Therefore r = 7 cm
We know that area of the circle = p r
2
A = (22/7) × (7)
2
A = 154 cm
2
9. A road which is 7 m wide surrounds a circular park whose circumference is 352 m.
Find the area of road.
Page 5
1. Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km
Solution:
(i) Given radius = 7 cm
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 cm
2
(ii) Given radius = 2.1 m
We know that area of circle = p r
2
A = (22/7) × (2.1)
2
A = 22/7 × 4.41
A = 13.86 m
2
(iii) Given radius = 7 km
We know that area of circle = p r
2
A = (22/7) × 7
2
A = 22 × 7
A = 154 km
2
2. Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km
Solution:
(i) Given diameter = 8.4 cm
Therefore radius, r = d/2 = 8.4/2 = 4.2
We know that area of circle = p r
2
A = (22/7) × (4.2)
2
A = 55.44 cm
2
(ii) Given diameter = 5.6 cm
Therefore radius, r = d/2 = 5.6/2 = 2.8
We know that area of circle = p r
2
A = (22/7) × (2.8)
2
A = 24.64 cm
2
(iii) Given diameter = 7 km
Therefore radius, r = d/2 = 7/2 = 3.5
We know that area of circle = p r
2
A = (22/7) × (3.5)
2
A = 38.5 km
2
3. The area of a circle is 154 cm
2
. Find the radius of the circle.
Solution:
Given area of the circle = 154 cm
2
A = p r
2
154 = 22/7 r
2
r
2
= (154 × 7)/22
r
2
= 49
r = 7 cm
4. Find the radius of a circle, if its area is
(i) 4 p cm
2
(ii) 55.44 m
2
(iii) 1.54 km
2
Solution:
(i) We know that area of circle = p r
2
Given area = 4 p cm
2
A = p r
2
4 p = p r
2
r
2
= 4
Therefore r = 2 cm
(ii) We know that area of circle = p r
2
Given area = 55.44 cm
2
A = p r
2
55.44 = p r
2
r
2
= (55.44 × 7)/22
r
2
= 17.64 m
r = 4.2 m
(iii) We know that area of circle = p r
2
Given area = 1.54 km
2
A = p r
2
1.54 = p r
2
r
2
= (1.54 × 7)/22
r
2
= 0.49 km
r = 0.7 km
r = 700m
5. The circumference of a circle is 3.14 m, find its area.
Solution:
Given circumference = 3.14m
We know that circumference of circle = 2 p r
3.14 = 2 × 3.14 × r
r = 3.14/ (2 × 3.14)
r = 0.5
We know that area of circle = p r
2
A = (22/7) × (0.5)
2
A = 0.785 m
2
6. If the area of a circle is 50.24 m
2
, find its circumference.
Solution:
Given area of a circle is 50.24 m
2
We know that area of circle = p r
2
50.24 = (22/7) × r
2
r
2
= (50.24 × 7)/22
r
2
= 15.985
r = 3.998 m
We know that circumference of circle = 2 p r
C = 2 × (22/7) × 3.998
C = 25.12 m
7. A horse is tied to a pole with 28 m long string. Find the area where the horse can
graze. (Take p = 22 / 7).
Solution:
Given the length of the string = 28m
The area over which the horse can graze is the same as the area of circle of radius 28 m
Hence required area = p r
2
A = (22/7) × (28)
2
A = 2464 m
2
8. A steel wire when bent in the form of a square encloses an area of 121 cm
2
. If the
same wire is bent in the form of a circle, find the area of the circle.
Solution:
Given area of the square = 121 cm
2
(Side)
2
= 121
Therefore side = 11 cm
We know that the perimeter of the square = 4 (side)
= 4 (11)
= 44 cm
According to the question circumference of the circle = perimeter of the square
So let r be the radius of the circle
2 p r = 44
2 × (22/7) × r = 44
Therefore r = 7 cm
We know that area of the circle = p r
2
A = (22/7) × (7)
2
A = 154 cm
2
9. A road which is 7 m wide surrounds a circular park whose circumference is 352 m.
Find the area of road.
Solution:
Given circumference of park = 352 m
But we know that circumference of circle = 2 p r = 352 m
2 × (22/7) × r = 352
r = (352 × 7)/ 44
r = 56 m
Radius of the path including the 7m wide road = (r + 7) = 56 + 7 = 63m
Therefore area of the road = p × (63)
2
- p × (56)
2
A = 22/7 × 63 × 63 – (22/7) × 56 × 56
A = 22 (9 × 63 – 8 × 56)
A = 22 (567 – 448)
A = 2618 m
2
10. Prove that the area of a circular path of uniform width h surrounding a circular
region of radius r is ph (2r + h).
Solution:
Let radius of circular region = r
Radius of circular path of uniform width h surrounding the circular region of radius,
r = r + h
Therefore area of path = p (r + h)
2
– pr
2
= p r
2
+ p h
2
+ 2 p r h – p r
2
= p h (2r + h)
Hence the proof.
11. The perimeter of a circle is 4pr cm. What is the area of the circle?
Solution:
Given perimeter of circle = 4pr cm
Which can be written as = 2 p (2r)
We know that the perimeter of a circle = 2 p r
Therefore we have radius = 2r
We also know that area of circle = p r
2
= p (2r)
2
= 4 p r
2
cm
2
12. A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find
Read More
|
76 videos|345 docs|39 tests
|
Related Exams
About this Document
|
4.81/5 Rating |
|
Oct 24, 2024 Last updated |
Document Description: RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) covers topics
like and RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2).
Introduction of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Mensuration – I (Area of Circle Exercise 21.2) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Mensuration – I (Area of Circle Exercise 21.2) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2)
In this doc you can find the meaning of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
practice quizzes
,Mensuration – I (Area of Circle Exercise 21.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,mock tests for examination
,Semester Notes
,Previous Year Questions with Solutions
,ppt
,Mensuration – I (Area of Circle Exercise 21.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Extra Questions
,shortcuts and tricks
,Free
,Exam
,video lectures
,Mensuration – I (Area of Circle Exercise 21.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Sample Paper
,Viva Questions
,Summary
,Important questions
,study material
,MCQs
,past year papers
,Objective type Questions
;
Additional Information about RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) for Class 7 Preparation
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Free PDF Download
The RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2)
The importance of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Notes
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Class 7 Questions
The "RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) on the App
Students of Class 7 can study RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.2) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup