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# In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO...

### Question

In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4cm. Find the area of the shaded region.

### Options

A) 43.36cm2

B) 32.072

C) 18.212

D) 6.932

The correct answer is C.

### Explanation:

Area of shaded portion = Area of semicircle

Area of $$\bigtriangleup$$ RSO

Area of semicircle = $$\frac {\pi r^2}{2} = \frac{22\times 4 \times 4}{7 \times 3}$$

= 25.14cm2; Area of $$\bigtriangleup$$RSO

=$$\sqrt{s(s - 1)(s - b)(s - c)}$$; where

s = $$\frac{a + b + c}{2}$$

s = $$\frac{4 + 4 + 4}{2}$$

= 6cm

= $$\sqrt{6(6 - 4)(6 - 4) (6 - 4)}$$

= $$\sqrt{6(2) (2) (2)}$$

= $$\sqrt{18}$$ = 6.93cm2

Area of shaded region

= 25.14 - 6.93

= 18.21cm2

## Dicussion (1)

• Area of shaded portion = Area of semicircle

Area of $$\bigtriangleup$$ RSO

Area of semicircle = $$\frac {\pi r^2}{2} = \frac{22\times 4 \times 4}{7 \times 3}$$

= 25.14cm2; Area of $$\bigtriangleup$$RSO

=$$\sqrt{s(s - 1)(s - b)(s - c)}$$; where

s = $$\frac{a + b + c}{2}$$

s = $$\frac{4 + 4 + 4}{2}$$

= 6cm

= $$\sqrt{6(6 - 4)(6 - 4) (6 - 4)}$$

= $$\sqrt{6(2) (2) (2)}$$

= $$\sqrt{18}$$ = 6.93cm2

Area of shaded region

= 25.14 - 6.93

= 18.21cm2

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