102
A chord PQ of a circle of diameter $28$ cm subtends an angle of $90^\circ$ at the centre O. Find the area of the sector OPCQ, where C is a point on minor arc PQ. Also, find the area of segment PCQ.
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Solution:
Area of sector OPCQ = $\frac{90}{360} \times \frac{22}{7} \times 14 \times 14$ (1 Mark)
$= 154$ cm$^2$ (0.5 Mark)
Area of the segment PCQ = Area of sector OPCQ – Area of $\triangle OPQ$
$= 154 - \frac{1}{2} \times 14 \times 14$ (1 Mark)
$= 56$ cm$^2$ (0.5 Mark)
Area of sector OPCQ = $\frac{90}{360} \times \frac{22}{7} \times 14 \times 14$ (1 Mark)
$= 154$ cm$^2$ (0.5 Mark)
Area of the segment PCQ = Area of sector OPCQ – Area of $\triangle OPQ$
$= 154 - \frac{1}{2} \times 14 \times 14$ (1 Mark)
$= 56$ cm$^2$ (0.5 Mark)