Question

Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.

Answer 1

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$$\begin{gathered} =\frac{90°}{360°}\times \mathrm{\pi }{\left(\mathrm{OA}\right)}^2-\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB} \\ =\frac{1}{4}\times \frac{22}{7}\times {\left(35\right)}^2-\frac{1}{2}\times 35\times 35 \\ =\frac{1}{4}\times \frac{22}{7}\times {\left(35\right)}^2-\frac{1}{2}\times 35\times 35 \\ =962.5-612.5 \\ =350{\mathrm{cm}}^2 \end{gathered}$$
Area of major segment APB = Area of circle − Area of minor segment
$$\begin{gathered} =\mathrm{\pi }{\left(\mathrm{OA}\right)}^2-350 \\ =\frac{22}{7}\times {\left(35\right)}^2-350 \\ =3850-350 \\ =3500{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of major segment is 3500 cm²