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|344 docs|39 tests

Top Courses for Class 7

76 videos|344 docs|39 tests
Download as PDF
Explore Courses for Class 7 exam

Top Courses for Class 7

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7

,

Previous Year Questions with Solutions

,

Sample Paper

,

past year papers

,

Objective type Questions

,

Important questions

,

Extra Questions

,

Summary

,

NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7

,

Exam

,

pdf

,

study material

,

mock tests for examination

,

NCERT Exemplar Solutions: Perimeter & Area | Mathematics (Maths) Class 7

,

MCQs

,

practice quizzes

,

shortcuts and tricks

,

Free

,

video lectures

,

Viva Questions

,

ppt

,

Semester Notes

;