Match the following with the area of shaded regions in $c m^{2}$ .

\frac{250 π}{3}

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\frac{50 π}{3} - 25 \sqrt{3}

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\frac{50 π}{3}

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Solution

Here, $r = 10$ cm and $θ = 60^{\circ} .$

For the given figure:

$A r e a o f m i n o r s e c t o r$
$= \frac{θ}{360} \times π r^{2}$
$= \frac{60}{360} \times π \times 10 \times 10$
$= \frac{50 π}{3} c m^{2}$

For the given figure:

$A r e a o f m a j o r s e c t o r = A r e a o f c i r c l e - A r e a o f m i n o r s e c t o r$

$A r e a o f c i r c l e$
$= π r^{2}$
$= π \times 10^{2}$
$= 100 π c m^{2}$

$\Rightarrow A r e a o f m a j o r s e c t o r$
$= 100 π - \frac{50 π}{3}$
$= \frac{250 π}{3} c m^{2}$

For the given figure:

$A r e a o f t h e s e g m e n t = A r e a o f m i n o r s e c t o r - A r e a o f △ O A B$
$A r e a o f △ O A B$
$= \frac{\sqrt{3}}{4} \times 10 \times 10$
$= 25 \sqrt{3} c m^{2}$
[ $△ O A B$ is an equilateral triangle with side 10 cm.]
$\Rightarrow$ Area of the segment
$= \frac{50 π}{3} - 25 \sqrt{3} c m^{2}$

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