Question

In the given figure, ∆ABC is a right-angled triangle with ∠B = 90°, AB = 48 cm and BC = 14 cm. With AC as diameter a semicircle is drawn and with BC as radius, A quadrant of a circle is drawn. Find the area of the shaded region.

Answer 1

Consider the triangle ABC.

$$\begin{gathered} AC=\sqrt{A{B}^2+B{C}^2} \\ =\sqrt{{48}^2+{14}^2} \\ =\sqrt{2304+196} \\ =\sqrt{2500} \\ =50\mathrm{cm} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{part}=\mathrm{Area}\left(∆\mathrm{ABC}\right)+\mathrm{Area}\left(\mathrm{semicircle}\mathrm{having}\mathrm{diameter}\mathrm{AC}\right)-\mathrm{Area}\left(\mathrm{sector}\mathrm{BCD}\right) \\ =\left[\frac{1}{2}\times 14\times 48\right]+\left[\frac{1}{2}\times \mathrm{\pi }\times 25\times 25\right]-\left[\frac{90}{360}\times \mathrm{\pi }\times 14\times 14\right] \\ =336+\frac{\mathrm{\pi }}{2}\left[\left(25\times 25\right)-\frac{\left(14\times 14\right)}{2}\right] \\ =336+\frac{22}{14}\left[625-98\right] \\ =1164.14{\mathrm{cm}}^2 \end{gathered}$$