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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