Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1)
Mensuration – I (Area of Circle Exercise 21.1) 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 circumference of the circle whose radius is
(i) 14cm
(ii) 10m
(iii) 4km
Solution:
(i) Given radius = 14cm
We know that the circumference of the circle = 2 p r
p value is 22/7
C = 2 × (22/7) × 14
C = 88cm
(ii) Given radius = 10m
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 10
C = 62.86 m
(iii) Given radius = 4km
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 4
C = 25.142 km
2. Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km
Solution:
(i) Given diameter = 7cm
We know that radius = diameter/2
Therefore, r = d/2
r = 7/2
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 7/2
C = 22 cm
Page 2
1. Find the circumference of the circle whose radius is
(i) 14cm
(ii) 10m
(iii) 4km
Solution:
(i) Given radius = 14cm
We know that the circumference of the circle = 2 p r
p value is 22/7
C = 2 × (22/7) × 14
C = 88cm
(ii) Given radius = 10m
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 10
C = 62.86 m
(iii) Given radius = 4km
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 4
C = 25.142 km
2. Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km
Solution:
(i) Given diameter = 7cm
We know that radius = diameter/2
Therefore, r = d/2
r = 7/2
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 7/2
C = 22 cm
(ii) Given diameter = 4.2 cm
We know that radius = diameter/2
Therefore, r = 4.2/2
r = 2.1
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 2.1
C = 13.2 cm
(iii) Given diameter = 11.2 km
We know that radius = diameter/2
Therefore, r = 11.2/2
r = 5.6
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 5.6
C = 35.2 km
3. Find the radius of a circle whose circumference is
(i) 52.8 cm
(ii) 42 cm
(iii) 6.6 km
Solution:
(i) Given circumference, C = 52.8 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (52.8 × 7)/ (2 × 22)
r = 369.6/ 44
r = 8.4 cm
(ii) Given circumference, C = 42 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (42 × 7)/ (2 × 22)
r = 294/ 44
r = 6.68 cm
(iii) Given circumference, C = 6.6 km
Page 3
1. Find the circumference of the circle whose radius is
(i) 14cm
(ii) 10m
(iii) 4km
Solution:
(i) Given radius = 14cm
We know that the circumference of the circle = 2 p r
p value is 22/7
C = 2 × (22/7) × 14
C = 88cm
(ii) Given radius = 10m
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 10
C = 62.86 m
(iii) Given radius = 4km
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 4
C = 25.142 km
2. Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km
Solution:
(i) Given diameter = 7cm
We know that radius = diameter/2
Therefore, r = d/2
r = 7/2
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 7/2
C = 22 cm
(ii) Given diameter = 4.2 cm
We know that radius = diameter/2
Therefore, r = 4.2/2
r = 2.1
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 2.1
C = 13.2 cm
(iii) Given diameter = 11.2 km
We know that radius = diameter/2
Therefore, r = 11.2/2
r = 5.6
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 5.6
C = 35.2 km
3. Find the radius of a circle whose circumference is
(i) 52.8 cm
(ii) 42 cm
(iii) 6.6 km
Solution:
(i) Given circumference, C = 52.8 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (52.8 × 7)/ (2 × 22)
r = 369.6/ 44
r = 8.4 cm
(ii) Given circumference, C = 42 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (42 × 7)/ (2 × 22)
r = 294/ 44
r = 6.68 cm
(iii) Given circumference, C = 6.6 km
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (6.6 × 7)/ (2 × 22)
r = 46.2/ 44
r = 1.05 km
4. Find the diameter of a circle whose circumference is
(i) 12.56 cm
(ii) 88 m
(iii) 11.0 km
Solution:
(i) Given circumference, C = 12.56 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (12.56 × 7)/ (2 × 22)
r = 87.92/ 44
r = 1.99 cm
But diameter = 2r
= 2 × 1.99 = 3.99 cm
(ii) Given circumference, C = 88 m
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (88 × 7)/ (2 × 22)
r = 616/ 44
r = 14 m
But diameter = 2r
= 2 × 14 = 28 m
(iii) Given circumference, C = 11.0 km
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (11 × 7)/ (2 × 22)
r = 77/ 44
r = 1.75 km
But diameter = 2r
Page 4
1. Find the circumference of the circle whose radius is
(i) 14cm
(ii) 10m
(iii) 4km
Solution:
(i) Given radius = 14cm
We know that the circumference of the circle = 2 p r
p value is 22/7
C = 2 × (22/7) × 14
C = 88cm
(ii) Given radius = 10m
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 10
C = 62.86 m
(iii) Given radius = 4km
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 4
C = 25.142 km
2. Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km
Solution:
(i) Given diameter = 7cm
We know that radius = diameter/2
Therefore, r = d/2
r = 7/2
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 7/2
C = 22 cm
(ii) Given diameter = 4.2 cm
We know that radius = diameter/2
Therefore, r = 4.2/2
r = 2.1
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 2.1
C = 13.2 cm
(iii) Given diameter = 11.2 km
We know that radius = diameter/2
Therefore, r = 11.2/2
r = 5.6
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 5.6
C = 35.2 km
3. Find the radius of a circle whose circumference is
(i) 52.8 cm
(ii) 42 cm
(iii) 6.6 km
Solution:
(i) Given circumference, C = 52.8 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (52.8 × 7)/ (2 × 22)
r = 369.6/ 44
r = 8.4 cm
(ii) Given circumference, C = 42 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (42 × 7)/ (2 × 22)
r = 294/ 44
r = 6.68 cm
(iii) Given circumference, C = 6.6 km
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (6.6 × 7)/ (2 × 22)
r = 46.2/ 44
r = 1.05 km
4. Find the diameter of a circle whose circumference is
(i) 12.56 cm
(ii) 88 m
(iii) 11.0 km
Solution:
(i) Given circumference, C = 12.56 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (12.56 × 7)/ (2 × 22)
r = 87.92/ 44
r = 1.99 cm
But diameter = 2r
= 2 × 1.99 = 3.99 cm
(ii) Given circumference, C = 88 m
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (88 × 7)/ (2 × 22)
r = 616/ 44
r = 14 m
But diameter = 2r
= 2 × 14 = 28 m
(iii) Given circumference, C = 11.0 km
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (11 × 7)/ (2 × 22)
r = 77/ 44
r = 1.75 km
But diameter = 2r
= 2 × 1.75 = 3.5 km
5. The ratio of the radii of two circles is 3: 2. What is the ratio of their circumferences?
Solution:
Given that the ratio of the radii = 3: 2
So, let the radii of the two circles be 3r and 2r respectively.
And let C
1
and C
2
be the circumference of the two circles of radii 3r and 2r respectively.
C
1
= 2 p × 3r = 6 p r … (i)
Now C
2
= 2 × 2 p r = 4 p r … (ii)
Consider, C
1
/C
2
= (6 p r)/ 4 p r = 6/4 = 3/2
C
1
: C
2
= 3: 2
6. A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and
bent into the form of a circle. Find the radius of the circle so formed.
Solution:
Given length of rectangular wire = 18.7 cm
Breadth of rectangular wire = 14.3 cm
According to the question length of wire = perimeter of the rectangle
= 2 (l + b) = 2 × (18.7 + 14.3)
= 2 (33)
= 66 cm
Let the wire bent in the form of circle of radius r cm then we have
Circumference = 66cm
2 p r = 66
2 × (22/7) × r = 66
(44/7) r = 66
r = (66 × 7)/44
r = 462/44
= 10.5 cm
7. A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. It
is re-bent to form a circular ring. What is the diameter of the ring?
Solution:
Given side of equilateral triangle = 6.6 cm
Page 5
1. Find the circumference of the circle whose radius is
(i) 14cm
(ii) 10m
(iii) 4km
Solution:
(i) Given radius = 14cm
We know that the circumference of the circle = 2 p r
p value is 22/7
C = 2 × (22/7) × 14
C = 88cm
(ii) Given radius = 10m
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 10
C = 62.86 m
(iii) Given radius = 4km
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 4
C = 25.142 km
2. Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km
Solution:
(i) Given diameter = 7cm
We know that radius = diameter/2
Therefore, r = d/2
r = 7/2
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 7/2
C = 22 cm
(ii) Given diameter = 4.2 cm
We know that radius = diameter/2
Therefore, r = 4.2/2
r = 2.1
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 2.1
C = 13.2 cm
(iii) Given diameter = 11.2 km
We know that radius = diameter/2
Therefore, r = 11.2/2
r = 5.6
We know that the circumference of the circle = 2 p r
C = 2 × (22/7) × 5.6
C = 35.2 km
3. Find the radius of a circle whose circumference is
(i) 52.8 cm
(ii) 42 cm
(iii) 6.6 km
Solution:
(i) Given circumference, C = 52.8 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (52.8 × 7)/ (2 × 22)
r = 369.6/ 44
r = 8.4 cm
(ii) Given circumference, C = 42 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (42 × 7)/ (2 × 22)
r = 294/ 44
r = 6.68 cm
(iii) Given circumference, C = 6.6 km
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (6.6 × 7)/ (2 × 22)
r = 46.2/ 44
r = 1.05 km
4. Find the diameter of a circle whose circumference is
(i) 12.56 cm
(ii) 88 m
(iii) 11.0 km
Solution:
(i) Given circumference, C = 12.56 cm
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (12.56 × 7)/ (2 × 22)
r = 87.92/ 44
r = 1.99 cm
But diameter = 2r
= 2 × 1.99 = 3.99 cm
(ii) Given circumference, C = 88 m
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (88 × 7)/ (2 × 22)
r = 616/ 44
r = 14 m
But diameter = 2r
= 2 × 14 = 28 m
(iii) Given circumference, C = 11.0 km
We know that the circumference of the circle = 2 p r
Therefore radius, r = C/2 p
r = (11 × 7)/ (2 × 22)
r = 77/ 44
r = 1.75 km
But diameter = 2r
= 2 × 1.75 = 3.5 km
5. The ratio of the radii of two circles is 3: 2. What is the ratio of their circumferences?
Solution:
Given that the ratio of the radii = 3: 2
So, let the radii of the two circles be 3r and 2r respectively.
And let C
1
and C
2
be the circumference of the two circles of radii 3r and 2r respectively.
C
1
= 2 p × 3r = 6 p r … (i)
Now C
2
= 2 × 2 p r = 4 p r … (ii)
Consider, C
1
/C
2
= (6 p r)/ 4 p r = 6/4 = 3/2
C
1
: C
2
= 3: 2
6. A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and
bent into the form of a circle. Find the radius of the circle so formed.
Solution:
Given length of rectangular wire = 18.7 cm
Breadth of rectangular wire = 14.3 cm
According to the question length of wire = perimeter of the rectangle
= 2 (l + b) = 2 × (18.7 + 14.3)
= 2 (33)
= 66 cm
Let the wire bent in the form of circle of radius r cm then we have
Circumference = 66cm
2 p r = 66
2 × (22/7) × r = 66
(44/7) r = 66
r = (66 × 7)/44
r = 462/44
= 10.5 cm
7. A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. It
is re-bent to form a circular ring. What is the diameter of the ring?
Solution:
Given side of equilateral triangle = 6.6 cm
Length of the wire = the perimeter of equilateral triangle
Perimeter of equilateral triangle = 3 × side
= 3 × 6.6 = 19.8 cm
Therefore circumference = 19.8 cm
C = 2 p r
19.8 = 2 × (22/7) × r
19.8 = (44/7) r
r = (19.8 × 7)/44
r = 138.6/44
= 3.15 cm
Diameter = 2r
Therefore diameter of ring = 2 × 3.15 = 6.3 cm
8. The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car
during the period, the wheel makes 1000 revolutions.
Solution:
It may be noted that in one revolution, the cycle covers a distance equal to the
circumference of the wheel.
Given the diameter of the wheel = 63 cm
We know that circumference of the wheel = p d
= 22/7 x 63
= 198 cm.
Thus, the cycle covers 198 cm in one revolution.
Therefore the distance covered by the cycle in 1000 revolutions = (198 x 1000)
= 198000 cm
= 1980 m.
9. The diameter of a wheel of a car is 98 cm. How many revolutions will it make to
travel 6160 meters.
Solution:
In one revolution of the wheel, the car travels a distance equal to the circumference of
the wheel.
Given diameter of the wheel of a car = 98 cm
Circumference of the wheel of the car = p d
= 22/7 x 98
Read More
|
76 videos|345 docs|39 tests
|
FAQs on Mensuration – I (Area of Circle Exercise 21.1) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. What is the formula for finding the area of a circle? |  |
Ans. The formula for finding the area of a circle is A = πr^2, where A represents the area and r represents the radius of the circle.
| 2. How do I calculate the radius of a circle if I know the area? | |
Ans. To calculate the radius of a circle if you know the area, you can use the formula r = √(A/π), where r represents the radius and A represents the area of the circle.
| 3. Can I use the diameter to find the area of a circle? | |
Ans. Yes, you can use the diameter to find the area of a circle. The formula for finding the area of a circle using the diameter is A = (π/4) × d^2, where A represents the area and d represents the diameter of the circle.
| 4. How can I find the circumference of a circle if I only know the area? | |
Ans. If you only know the area of a circle and want to find the circumference, you can use the formula C = 2√(πA), where C represents the circumference and A represents the area of the circle.
| 5. What is the relationship between the radius and diameter of a circle? | |
Ans. The radius of a circle is half the length of its diameter. In other words, the diameter is twice the length of the radius. So, if you know the radius, you can find the diameter by multiplying it by 2, and vice versa.
Related Exams
About this Document
|
4.76/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) covers topics
like and RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1).
Introduction of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Mensuration – I (Area of Circle Exercise 21.1) 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 21.1)
In this doc you can find the meaning of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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
Mensuration – I (Area of Circle Exercise 21.1) RD Sharma Solutions | Mathematics (Maths) Class 7
,ppt
,Mensuration – I (Area of Circle Exercise 21.1) RD Sharma Solutions | Mathematics (Maths) Class 7
,Free
,Mensuration – I (Area of Circle Exercise 21.1) RD Sharma Solutions | Mathematics (Maths) Class 7
,Important questions
,past year papers
,MCQs
,Objective type Questions
,Sample Paper
,shortcuts and tricks
,Exam
,Semester Notes
,Summary
,mock tests for examination
,video lectures
,Previous Year Questions with Solutions
,Extra Questions
,practice quizzes
,study material
,Viva Questions
;
Additional Information about RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) for Class 7 Preparation
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) Free PDF Download
The RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1)
The importance of RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1) Notes
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1).
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 21.1) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) Class 7 Questions
The "RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1) on the App
Students of Class 7 can study RD Sharma Solutions: Mensuration – I (Area of Circle Exercise 21.1) 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 21.1),
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 21.1) 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