**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) 43.36cm^{2}

B) 32.07^{2}

C) 18.21^{2}

D) 6.93^{2}

The correct answer is C.

### Explanation:

Area of \(\bigtriangleup\) RSO

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

= 25.14cm^{2}; 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

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

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

= \(\sqrt{18}\) = 6.93cm^{2}

Area of shaded region

= 25.14 - 6.93

= 18.21cm^{2}

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Area of shaded portion = Area of semicircle

Area of \(\bigtriangleup\) RSO

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

= 25.14cm

^{2}; 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

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

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

= \(\sqrt{18}\) = 6.93cm

^{2}Area of shaded region

= 25.14 - 6.93

= 18.21cm

^{2}