# Class 7 Maths Perimeter And Area Important Questions For Exam

IMPORTANT QUESTIONS FOR CBSE EXAMINATION | CLASS 7 MATHEMATICS
PERIMETER AND AREA – Chapter 11

1. One of the sides and the corresponding height of a parallelogram are 6 cm and 2 cm respectively. Find the area of the parallelogram.

2. Find the area of a square park whose perimeter is 480m.

3. Find the breadth of a rectangular garden; if its area is 120 square metre and the length is 20 m. Also find its perimeter?

4. The perimeter of a rectangular sheet is 120 cm. If the length is 30 cm, find its breadth. Also find the area?

5. Find the height, if the area of the parallelogram is 48 square cm and the base is 6 cm.

6. Find the base, if the area of the triangle is 48 square cm and the height is 4 cm.

7. What is the circumference of a circle of diameter 12 cm (take pi = 3.14)

8. What is the circumference of a circular disc of radius 28 cm (pi = 22/7).

9. Convert the following:
i) 50 square cm in square mm.
ii) 10 square metre in square cm.

10. A rectangular park is 40 m long and 24 m wide. A path 3m wide is constructed outside the park. Find the area of the path?

1. Area of the parallelogram = b x h
Given b= 6 cm and h = 2 cm
Area = 6 x 2 = 12 sq. cm.

2. Given perimeter of a square = 480 m
Length of one side =480/4 = 120 m
Area = side x side = 120 x 120 = 14400 sq. m.

3. Given area = 120 sq. m and length = 20 m
b = Area/length = 120/20 = 6m.
Perimeter = 2 x (l +b) = 2 x (20 + 6) = 2 x 26 = 52 m

4. Given perimeter of a rectangular sheet = 120 cm and length = 30 cm
Let breadth be b then 2 x (30 + b) = 120
30 + b = 120/2 = 60
b = 60 – 30 = 30 cm
Area = side x side = 30 x 30 = 900 sq. cm (since it is a square)

5. Given area of the parallelogram = 48 q cm and breadth = 6cm
Area = b x h
48 = 6 x h
h = 48/6 = 8 cm.

6. Area of a triangle = 48 sq. cm and h = 4 cm
Area = ½ x b x h
48 = ½ x b x 4
96 = 4b
b = 96/4 = 24 cm

7. Circumference of a circle = Pi x d
Given d = 12 cm
C = 12 x 3.14 = 37.68cm

8. Circumference of a circle = 2 x pi x r
Given r = 28 cm
C = 2 x 22/7 x 28
= 176 cm

9. i. 50 sq. cm = 5000 sq. mm
ii. 10 sq. m = 100000 sq. cm

10. Let ABCD be the inner rectangle and PQRS be the outer rectangle.
To find the area of the path,
we need to find Area of PQRS – Area of ABCD.
We have PQ = 40 + 3 + 3 = 46m
PS = 24 + 3 + 3 = 30 m
Area of ABCD = l x b = 40 x 24 = 960 sq. m
Area of PQRS = l x b = 46 x 30 = 1380 sq. m
Area of the path = 1380 – 960 = 420 sq. m.

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