Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > NCERT Exemplar Solutions: Perimeter & Area
NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Page 2
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
Page 3
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
We have,
Side of square = 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of
the wire is same.
Perimeter of square = Circumference of circle
4 × side = 2 × p × r
4 × 22 = 2 ×
22
7
× r
r = 14 cm
Therefore, radius of circle is 14 cm.
5. Area of the circle obtained in Question 4 is
(a) 196 cm^2 (b) 212 cm^2 (c) 616 cm^2 (d) 644 cm^2
Solution:-
(c) 616 cm
2
Area of circle = pr
2
=
22
7
× 14 × 14
= 22 × 14 × 2
= 616 cm
2
6. Area of a rectangle and the area of a circle are equal. If the dimensions
of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
We have,
length = 14 cm,
breadth = 11 cm
Also,
area of rectangle = area of circle
length × breadth = pr
2
14 × 11 =
22
7
× r
2
r
2
= 49
r = 7 cm
Page 4
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
We have,
Side of square = 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of
the wire is same.
Perimeter of square = Circumference of circle
4 × side = 2 × p × r
4 × 22 = 2 ×
22
7
× r
r = 14 cm
Therefore, radius of circle is 14 cm.
5. Area of the circle obtained in Question 4 is
(a) 196 cm^2 (b) 212 cm^2 (c) 616 cm^2 (d) 644 cm^2
Solution:-
(c) 616 cm
2
Area of circle = pr
2
=
22
7
× 14 × 14
= 22 × 14 × 2
= 616 cm
2
6. Area of a rectangle and the area of a circle are equal. If the dimensions
of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
We have,
length = 14 cm,
breadth = 11 cm
Also,
area of rectangle = area of circle
length × breadth = pr
2
14 × 11 =
22
7
× r
2
r
2
= 49
r = 7 cm
7. Area of shaded portion in Fig. 9.15 is
(a) 25 cm^2 (b) 15 cm^2 (c) 14 cm^2 (d) 10 cm^2
Solution:-
(d) 10 cm
2
We have,
length of rectangle = 5 cm,
breadth of the rectangle = 3 cm + 1 cm
= 4 cm
So,
Area of rectangle = length × breadth
= 5 × 4
= 20 cm
2
By observing the figure, the shaded part covered exactly half of the rectangle,
Therefore,
Area of shaded part is =
2
Area of rectangle
=
20
2
= 10 cm
2
8. Area of parallelogram ABCD (Fig. 9.16) is not equal to
(a) DE × DC (b) BE × AD (c) BF × DC (d) BE × BC
Page 5
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
We have,
Side of square = 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of
the wire is same.
Perimeter of square = Circumference of circle
4 × side = 2 × p × r
4 × 22 = 2 ×
22
7
× r
r = 14 cm
Therefore, radius of circle is 14 cm.
5. Area of the circle obtained in Question 4 is
(a) 196 cm^2 (b) 212 cm^2 (c) 616 cm^2 (d) 644 cm^2
Solution:-
(c) 616 cm
2
Area of circle = pr
2
=
22
7
× 14 × 14
= 22 × 14 × 2
= 616 cm
2
6. Area of a rectangle and the area of a circle are equal. If the dimensions
of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
We have,
length = 14 cm,
breadth = 11 cm
Also,
area of rectangle = area of circle
length × breadth = pr
2
14 × 11 =
22
7
× r
2
r
2
= 49
r = 7 cm
7. Area of shaded portion in Fig. 9.15 is
(a) 25 cm^2 (b) 15 cm^2 (c) 14 cm^2 (d) 10 cm^2
Solution:-
(d) 10 cm
2
We have,
length of rectangle = 5 cm,
breadth of the rectangle = 3 cm + 1 cm
= 4 cm
So,
Area of rectangle = length × breadth
= 5 × 4
= 20 cm
2
By observing the figure, the shaded part covered exactly half of the rectangle,
Therefore,
Area of shaded part is =
2
Area of rectangle
=
20
2
= 10 cm
2
8. Area of parallelogram ABCD (Fig. 9.16) is not equal to
(a) DE × DC (b) BE × AD (c) BF × DC (d) BE × BC
Solution:-
(a) DE × DC
We have,
Area of parallelogram = base × corresponding height
So,
Area of parallelogram ABCD = DC × BF
AD × BE = BC × BE [as, AD = BC]
9. Area of triangle MNO of Fig. 9.17 is
(a) 1/2 MN × NO (b) 1/2 NO × MO (c) 1/2 MN × OQ (d) 1/2 NO
×OQ
Solution:-
(d)
1
2
NO × OQ
MNO is a triangle.
We have,
Area of triangle =
1
2
(base × height)
=
1
2
× NO × OQ
10. Ratio of area of ?MNO to the area of parallelogram MNOP in the same
figure 9.17 is
(a) 2 : 3 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1
Solution:-
(c) 1 : 2
From figure, we have,
Area of ?MNO =
1
2
× base × height
=
1
2
× NO × OQ
Area of parallelogram MNOP = base × corresponding height
Read More
|
76 videos|345 docs|39 tests
|
About this Document
|
4.96/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: NCERT Exemplar Solutions: Perimeter & Area for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for NCERT Exemplar Solutions: Perimeter & Area have been prepared according to the Class 7 exam syllabus. Information about NCERT Exemplar Solutions: Perimeter & Area covers topics
like and NCERT Exemplar Solutions: Perimeter & Area Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Exemplar Solutions: Perimeter & Area.
Introduction of NCERT Exemplar Solutions: Perimeter & Area in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & NCERT Exemplar Solutions: Perimeter & Area 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: NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for NCERT Exemplar Solutions: Perimeter & Area | 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 NCERT Exemplar Solutions: Perimeter & Area
In this doc you can find the meaning of NCERT Exemplar Solutions: Perimeter & Area defined & explained in the simplest way possible. Besides explaining types of
NCERT Exemplar Solutions: Perimeter & Area theory, EduRev gives you an ample number of questions to practice NCERT Exemplar Solutions: Perimeter & Area 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
Important questions
,Viva Questions
,Exam
,NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7
,video lectures
,MCQs
,shortcuts and tricks
,Objective type Questions
,mock tests for examination
,study material
,Extra Questions
,NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7
,practice quizzes
,Previous Year Questions with Solutions
,Semester Notes
,Summary
,ppt
,past year papers
,Sample Paper
,Free
,NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7
;
Additional Information about NCERT Exemplar Solutions: Perimeter & Area for Class 7 Preparation
NCERT Exemplar Solutions: Perimeter & Area Free PDF Download
The NCERT Exemplar Solutions: Perimeter & Area 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 NCERT Exemplar Solutions: Perimeter & Area now and kickstart your journey towards success in the Class 7 exam.
Importance of NCERT Exemplar Solutions: Perimeter & Area
The importance of NCERT Exemplar Solutions: Perimeter & Area 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.
NCERT Exemplar Solutions: Perimeter & Area Notes
NCERT Exemplar Solutions: Perimeter & Area Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to NCERT Exemplar Solutions: Perimeter & Area.
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, NCERT Exemplar Solutions: Perimeter & Area Notes on EduRev are your ultimate resource for success.
NCERT Exemplar Solutions: Perimeter & Area Class 7 Questions
The "NCERT Exemplar Solutions: Perimeter & Area 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 NCERT Exemplar Solutions: Perimeter & Area on the App
Students of Class 7 can study NCERT Exemplar Solutions: Perimeter & Area alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the NCERT Exemplar Solutions: Perimeter & Area,
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 NCERT Exemplar Solutions: Perimeter & Area 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