Question 105
A chord AB of length 3$$\sqrt{2}$$ unit subtends a right angle at the centre 0 of a circle. Area of the sector AOB (in sq. units) is
Solution
From $$\triangle$$AOB, $$\angle$$ AOB = 90
=> $$OA^2 + OB^2 = AB^2$$
=> $$2r^2 = (3\sqrt{2})^2 = 18$$
=> $$r^2 = 9$$ => $$r = 3$$ units
$$\therefore$$ Area of sector AOB
= $$\frac{1}{4} \pi r^2 = \frac{1}{4} \pi * 9$$
= $$\frac{9 \pi}{4}$$ sq. units
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