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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