Question

The perimeter of a sector of a circle with central angle 90° is 25 cm. The area of the minor segment of the circle is
(a) 14 cm²
(b) 16 cm²
(c) 18 cm²
(d) 24 cm²

Answer 1

(a) 14 cm²
Let r be the radius of the circle and $\theta$ be the angle.
Now,
Perimeter of the sector$=\left(2r+\frac{2\mathrm{\pi }r\theta }{360}\right)$
$$\begin{gathered} =2r+2\times \frac{22}{7}\times r\times \frac{90}{360} \\ =\left(2r+\frac{11r}{7}\right) \\ =\frac{25r}{7} \end{gathered}$$
Also,

$$\begin{gathered} \frac{25r}{7}=25 \\ \Rightarrow r=\left(25\times \frac{7}{25}\right) \\ \Rightarrow r=7\mathrm{cm} \end{gathered}$$
Area of the minor segment$=\left(\frac{\mathrm{\pi }{r}^2\theta }{360}-\frac{1}{2}r\right)$²sin θ cm²
$$\begin{gathered} =\left|\left(\frac{22}{7}\times 7\times 7\times \frac{90}{360}\right)-\left(\frac{1}{2}\times 7\times 7\times \sin 90°\right)\right|{\mathrm{cm}}^2 \\ =\left(\frac{77}{2}-\frac{49}{2}\right){\mathrm{cm}}^2 \\ =\frac{28}{2}{\mathrm{cm}}^2 \\ =14{\mathrm{cm}}^2 \end{gathered}$$