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.] [CBSE 2012]

Answer 1

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$$\begin{gathered} =\frac{\mathrm{\theta }}{360°}\mathrm{\pi }{\left(\mathrm{OA}\right)}^2-\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB} \\ =\frac{90°}{360°}\times 3.14{\left(10\right)}^2-\frac{1}{2}\times 10\times 10 \\ =78.5-50 \\ =28.5{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of minor segment is 28.5 cm²