48
A chord of a circle of radius $14$ cm makes a right angle at the centre of the circle. Find the area of the minor segment.
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Area of sector $= \frac{\theta}{360^\circ} \pi r^2 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ cm$^2$.
Area of triangle $= \frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times 14^2 \times \sin 90^\circ = \frac{1}{2} \times 196 \times 1 = 98$ cm$^2$.
Area of segment = Area of sector - Area of triangle
Area of segment $= \frac{22}{7} \times 14 \times 14 \times \frac{90}{360} - \frac{1}{2} \times 14 \times 14$
$= 154 - 98 = 56$
Hence area of segment = $56$ cm$^2$
Area of triangle $= \frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times 14^2 \times \sin 90^\circ = \frac{1}{2} \times 196 \times 1 = 98$ cm$^2$.
Area of segment = Area of sector - Area of triangle
Area of segment $= \frac{22}{7} \times 14 \times 14 \times \frac{90}{360} - \frac{1}{2} \times 14 \times 14$
$= 154 - 98 = 56$
Hence area of segment = $56$ cm$^2$