In the given figure $O$ is the centre of the circle with radius $28 c m$ . Find the area of the minor segment $A S B$ .

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Solution

Area of shaded region= ar. sector- area of $△ A O B$
In $△ A O B, O M ⊥ A B, ∴ ∠ A O M = ∠ M O B = 22.5 °$
& $A M = M B = \frac{1}{2} A B$
Also, $△ A O B = 2 \times a r △ A O M$
Now, $cos 22.5 ° = \frac{O M}{O A} \Rightarrow O M = 28 \times cos 22.5 ° = 25.90 ㎝$
$A M = O A sin 22.5 ° = 10.72 ㎝$
$a r △ A M O = \frac{1}{2} A M \times O M = \frac{1}{2} \times 10.72 \times 25.90 = 138.824 ㎠$
$∴ a r △ A O B = 2 \times 138.824 = 277.144 ㎠$
Required area $= \frac{45 °}{360 °} \times \frac{22}{7} \times 28 \times 28 - 277.144 = 308 - 277.144 = 30.35 ㎠$

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In the given figure O is the centre of the circle with radius 28 cm. Find the area of the minor segment ASB.

In the given figure $O$ is the centre of the circle with radius $28 c m$ . Find the area of the minor segment $A S B$ .