**Chapter 11 Perimeter and Area Exercise 11.3**

Question 5.

From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Solution:

Radius of the circular sheet = 4 cm

∴ Area = πr^{2} = π × 4 × 4 = 16π cm^{2}

Radius of the circle to be removed = 3 cm

∴ Area of sheet removed = πr^{2} = 9π cm^{2}

Area of the remaining sheet

= (16π – 9π) cm2 = 7π cm^{2}

= 7 × 3.14 cm^{2} = 21.98 cm^{2}

Hence, the required area = 21.98 cm^{2}.

Question 7.

Find the perimeter of the given figure, which is a semicircle including its diameter.

Solution:

Given: Diameter = 10 cm

Hence, the required perimeter

= 25.7 cm. (approx.)

Question 11.

A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)

Solution:

Side os the square sheet = 6 m

∴ Area of the sheet = (Side)^{2} = (6)^{2} = 36 cm^{2}

Radius of the circle = 2 cm

∴ Area of the circle to be cut out = πr^{2}

Area of the left over sheet

Question 12.

The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)

Solution:

Circumference of the circle = 31.4 cm

2πr = 31.4

Area of the circle = 7πr^{2} = 3.14 × 5 × 5 = 78.5 cm^{2}

Hence, the required radius = 5 cm and area = 78.5 cm^{2}.

Question 14.

A circular flower garden has an area of 314 m^{2}. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler can water the entire garden?

[Take π = 3.14]

Solution:

Area of the flower garden = 314 m^{2}

Radius of the circular portion covered by the sprinkler = 12 m

∴ Area = 7πr^{2} = 3.14 × 12 × 12

= 3.14 × 144 m^{2} = 452.16 m^{2}

Since 452.16 m^{2} > 314 m^{2}

Yes, the sprinkler will water the entire garden.

Question 15.

Find the circumference of the inner and the outer circles, shown in the given figure. (Take π = 3.14)

Solution:

Radius of the outer circle = 19 m

∴ Circumference of the outer circle = 2πr

= 2 × 3.14 × 19 = 3.14 × 38 m

= 119.32 m

Radius of the inner circle

= 19m – 10m = 9m

∴ Circumference = 2πr = 2 × 3.14 × 9

= 56.52 m

Here the required circumferences are 56.52 m and 119.32 m.

Question 16.

Question 17.

The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)

Solution:

Length of minute hand = 15 cm

∴ Radius = 15 cm

Circumference = 2πr

= 2 × 3.14 × 15 cm = 94.2 cm

Since the minute hand covers the distance in 1 hour equal to the circumference of the circle. Here the required distance covered by the minute hand = 94.2 cm.