# Area and Perimeter of a Semicircle

Here we will discuss about the area and perimeter of a semicircle with some example problems.

Area of a semicircle = $$\frac{1}{2}$$ πr$$^{2}$$

Perimeter of a semicircle = (π + 2)r

Solved example problems on finding the area and perimeter of a semicircle:

1. Find the area and perimeter of a semicircle of radius 7 cm. (Use π = $$\frac{22}{7}$$).

Solution:

Given, radius = r = 7 cm.

Then, area of semicircle = $$\frac{1}{2}$$ πr$$^{2}$$

= $$\frac{1}{2}$$ × $$\frac{22}{7}$$ × 7$$^{2}$$ cm$$^{2}$$

= 11 × 7 cm$$^{2}$$

= 77 cm$$^{2}$$

Perimeter of a semicircle = (π + 2)r

= ($$\frac{22}{7}$$ + 2) × 7 cm

= $$\frac{36}{7}$$ × 7 cm

= 36 cm

2. Find the area and perimeter of the figure in which PQRS is a square of side 28 cm and STR is a semicircle. (Use π = $$\frac{22}{7}$$).

Solution:

The required area = Area of the square PQRS + Area of the semicircle STR

= a$$^{2}$$ + $$\frac{1}{2}$$ πr$$^{2}$$

= 28$$^{2}$$ cm$$^{2}$$ + $$\frac{1}{2}$$ × π × ($$\frac{1}{2}$$ × 28)$$^{2}$$ cm$$^{2}$$

= (28$$^{2}$$ + $$\frac{1}{2}$$ × $$\frac{22}{7}$$ × 14$$^{2}$$) cm$$^{2}$$

= (28$$^{2}$$ + $$\frac{1}{2}$$ × $$\frac{22}{7}$$ × 14 × 14) cm$$^{2}$$

= (28$$^{2}$$ + 11 × 2 × 14) cm$$^{2}$$

= (28$$^{2}$$ + 11 × 28) cm$$^{2}$$

= 28(28 + 11) cm$$^{2}$$

= 28 × 39 cm$$^{2}$$

= 1092 cm$$^{2}$$

The required perimeter = PQ + PS + QR + semicircular arc STR

= 28 cm + 28 cm + 28 cm + π × ($$\frac{1}{2}$$ SR)

= 84 cm + $$\frac{22}{7}$$ × $$\frac{1}{2}$$ × 28 cm

= 84 cm + 11 × 4 cm

= 84 cm + 44 cm

= 128 cm

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