Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4)
Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
Page 2
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Page 3
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth
= 48 m x 20 m
= 960 m
2
Area of one right triangular flower bed = ½ (12 x 5) = 30 m
2
Therefore, required number of right triangular flower beds = area of the rectangular
field/ area of one right triangular flower bed.
= 960/30
Number of right triangular flower beds = 32
7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ?AC, BM ? AC,
DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm
We know that area of triangle = ½ x base x height
Area of triangle ADC = ½ (AC x DL)
= ½ (84 x 16.5)
= 693 cm
2
Area of triangle ABC = ½ (AC x BM)
= ½ (84 x 12) = 504 cm
2
Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
= (693 + 504) cm
2
= 1197 cm
2
8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD
measure 48 m and 32 m respectively and are perpendicular to each other.
Page 4
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth
= 48 m x 20 m
= 960 m
2
Area of one right triangular flower bed = ½ (12 x 5) = 30 m
2
Therefore, required number of right triangular flower beds = area of the rectangular
field/ area of one right triangular flower bed.
= 960/30
Number of right triangular flower beds = 32
7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ?AC, BM ? AC,
DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm
We know that area of triangle = ½ x base x height
Area of triangle ADC = ½ (AC x DL)
= ½ (84 x 16.5)
= 693 cm
2
Area of triangle ABC = ½ (AC x BM)
= ½ (84 x 12) = 504 cm
2
Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
= (693 + 504) cm
2
= 1197 cm
2
8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD
measure 48 m and 32 m respectively and are perpendicular to each other.
Solution:
Given diagonal AC = 48 m and diagonal BD = 32 m
Area of a quadrilateral = ½ (Product of diagonals)
= ½ (AC x BD)
= ½ (48 x 32) m
2
= 768 m
2
9. In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such
that EF ? AD and EF= 14 cm. Calculate the area of the shaded region.
Solution:
Given length of rectangle = 32m and breadth = 18m
We know that area of rectangle = length x breadth
Therefore area of the rectangle = AB x BC
= 32 m x 18 m
= 576 m
2
Page 5
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth
= 48 m x 20 m
= 960 m
2
Area of one right triangular flower bed = ½ (12 x 5) = 30 m
2
Therefore, required number of right triangular flower beds = area of the rectangular
field/ area of one right triangular flower bed.
= 960/30
Number of right triangular flower beds = 32
7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ?AC, BM ? AC,
DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm
We know that area of triangle = ½ x base x height
Area of triangle ADC = ½ (AC x DL)
= ½ (84 x 16.5)
= 693 cm
2
Area of triangle ABC = ½ (AC x BM)
= ½ (84 x 12) = 504 cm
2
Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
= (693 + 504) cm
2
= 1197 cm
2
8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD
measure 48 m and 32 m respectively and are perpendicular to each other.
Solution:
Given diagonal AC = 48 m and diagonal BD = 32 m
Area of a quadrilateral = ½ (Product of diagonals)
= ½ (AC x BD)
= ½ (48 x 32) m
2
= 768 m
2
9. In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such
that EF ? AD and EF= 14 cm. Calculate the area of the shaded region.
Solution:
Given length of rectangle = 32m and breadth = 18m
We know that area of rectangle = length x breadth
Therefore area of the rectangle = AB x BC
= 32 m x 18 m
= 576 m
2
Also given that base of triangle = 18m and height = 14m and EF ? AD
We know that area of triangle = ½ x base x height
Area of the triangle = ½ (AD x FE)
= ½ (BC x FE) [Since AD = BC]
= ½ (18 m x 14 m)
= 126 m
2
Area of the shaded region = Area of the rectangle – Area of the triangle
= (576 – 126) m
2
= 450 m
2
10. In Fig. 32, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If P, Q,
R, S be the mid-points of the sides AB, BC, CD and DA respectively, find the area of the
shaded region.
Solution:
Given ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm.
Join PR and SQ so that these two lines bisect each other at point O
Read More
|
76 videos|344 docs|39 tests
|
Related Exams
About this Document
|
4.63/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) covers topics
like and RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4).
Introduction of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 (Perimeter and Area of Rectilinear Figures Exercise 20.4)
In this doc you can find the meaning of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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
MCQs
,Summary
,Sample Paper
,Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7
,Important questions
,Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7
,Extra Questions
,Free
,Semester Notes
,video lectures
,Previous Year Questions with Solutions
,Exam
,ppt
,Viva Questions
,study material
,Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7
,Objective type Questions
,shortcuts and tricks
,past year papers
,mock tests for examination
,practice quizzes
;
Additional Information about RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) for Class 7 Preparation
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) Free PDF Download
The RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 (Perimeter and Area of Rectilinear Figures Exercise 20.4) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4)
The importance of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 (Perimeter and Area of Rectilinear Figures Exercise 20.4) Notes
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 (Perimeter and Area of Rectilinear Figures Exercise 20.4).
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 (Perimeter and Area of Rectilinear Figures Exercise 20.4) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) Class 7 Questions
The "RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) 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 (Perimeter and Area of Rectilinear Figures Exercise 20.4) on the App
Students of Class 7 can study RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4),
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 (Perimeter and Area of Rectilinear Figures Exercise 20.4) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup