Find the area of the shaded segment.

11

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12

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13

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14

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Solution

The correct option is A $11$
$A_{s e c t o r} = \frac{n}{360} π r^{2}$
$= \frac{60}{360} π 6^{2}$
= $18.84 c m^{2}$
$A_{t r i a n g l e} = \frac{1}{2} b h$
$b = s i n 60^{0} * 6 =$ $3 \sqrt{3}$
$h = c o s 60^{0} * 6 =$ $3$
$A_{t r i a n g l e} = \frac{1}{2} b h$
= $\frac{1}{2} 3 \sqrt{3} \times 3$
= $7.79$
$A_{s e g m e n t} = 18.84 - 7.79 \approx 11 c m^{2}$

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