# Free Perimeter and Area 03 Practice Test - 7th grade

The diameter of a circular park is 14 m. Find the perimeter of the park. Assume (π= 227).

A.

4.4 km

B.

44 m

C.

440 mm

D.

4400 cm

#### SOLUTION

Solution : B and D

Given, the diameter of the circular park = 14 m.

Hence, radius = 142 = 7 m.

Circumference of the circular park =2×π×r=2×227×7=44 m.

44 m = 4400 cm [Since, 1m = 100 cm]

A 2 m wide path is built along the border and inside a square garden of side 11 m. The area of the path is_____.

A.

73 m2

B.

72 m2

C.

77 m2

D.

75 m2

#### SOLUTION

Solution : B Let ABCD be the square park and the region shaded in green be the path.

Here, side length of the outer square, AB = 11 m (given)
and the path is 2 m wide.

Side length of the inner square, HG
= 11 - 2 - 2 = 7 m

We know that area of a square = side2

Area of the path = Area of the outer square - Area of the inner square
=11272=12149=72 m2

EF divides the rectangle ABCD in two equal parts of _____. A. 6 cm2
B. 5 cm2
C. 8 cm2
D. 10 cm2

#### SOLUTION

Solution : A

EF divides the rectangle ABCD into two trapeziums.

Area of trapezium ABFE
=12×(Sum of parallel sides)×(Distance between them
=12×(AE+BF)×AB

=12×(1+2)×4

=12×3×4=6 cm2

Area of trapezium CDEF =12×(FC+ED)×CD

=12×(1+2)×4

=12×3×4=6 cm2

Area of both the trapeziums are equal.

EF divides the rectangle ABCD in two equal parts and the area of both the parts are 6 cm2.

The area of the given figure is ____ cm2​.  (Both the ends are semi-circular)(All dimensions in cm and π  = 22/7) ___

#### SOLUTION

Solution :

Area of the two semi-circles = 12(π×r2)+12(π×r2)=π×r2=227×7×7=154cm2

Area of rectangular part = length × breadth = 10 ×14 = 140 cm2

Total area = Area of the two semi-circles + Area of the rectangle  = 154+140 = 294 cm2​.

Find the total area of a metal sheet which is in the form of a square of side 5 m.

A. 25 m2
B. 55 m2
C. 20 m2
D. 10 m2

#### SOLUTION

Solution : A

Given: The metal sheet in the form of a square of side 5 m.

The area of a square of side 'a' is given by a2.
So, area of the sheet  =52=25 m2

There is a square park in a society of side 7 m. What is the cost of laying bricks in the park at the rate of  10 / m2.

A.

470

B.

440

C.

490

D.

500

#### SOLUTION

Solution : C

Total cost of laying bricks in the park  = Area of the park  x  rate

Area of a the park =  (7 x 7) m2

= 49 m2

Cost = 49 m2 x ( 10) / m2

=  490

What is the radius (in cm) of the tyre in a car, if it covers 0.396 km in 300 rotations? (Use π = 227)

A.

14 cm

B.

21 cm

C.

7 cm

D.

28 cm

#### SOLUTION

Solution : B

The circumference of the tyre = 2πr (Distance travelled in each rotation)

Total Distance = Distance covered in 1 rotation x Number of rotations

0.396=2×227×r×300

r=0.396×72×22×300

Hence r = 0.00021 km or  21 cm.

The base and corresponding height of a parallelogram is 6m and 8m respectively. The area of the parallelogram is ___ m2 .

#### SOLUTION

Solution :

Area of parallelogram = b x h = 48 sq m.

Puja decides to cut a circular disc into 4 equal parts. What is the perimeter of each part? (Given that the radius of the disc = 7cm, π = 227)

A.

38 cm

B.

25 cm

C.

35 cm

D.

27 cm

#### SOLUTION

Solution : B

When a circular disc is cut into four equal parts, the perimeter of each part is more than 2πr4. This is because the length of two radii are also added to the perimeter in each part.

Perimeter of each part = 2πr4+ r + r

= πr2 + 2r

= 22×77×2 + 2r

= 11+2×7

= 11 + 14

= 25 cm

Find the area of the shaded part in the figure given below. (Take π  = 227)(All the dimensions are in cm) A.

11 cm2

B.

10.5 cm2

C.

9.5 cm2

D.

11.5cm2

#### SOLUTION

Solution : B

Here we have two figures: a square of side 7 cm and a circle of diameter 7 cm, or radius 72 cm
Area of the shaded portion = Area of the square – Area of the circle

=  a2πr2    (where a is the side of the square and r is the radius of the circle)

= 72227×72×72

= 49 – 38.5

= 10.5 cm2