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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


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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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