Question:
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.]
Solution:
Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$=\frac{\theta}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$
$=\frac{90^{\circ}}{360^{\circ}} \times 3.14(10)^{2}-\frac{1}{2} \times 10 \times 10$
$=78.5-50$
$=28.5 \mathrm{~cm}^{2}$
Hence, the area of minor segment is 28.5 cm2