Question

In the given figure, ∆ABC is right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3 cm and AC = 4 cm. Find the area of the shaded region.

Answer 1

In triangle $∆$ABC, we have:
$$\begin{gathered} BC=\sqrt{A{B}^2+A{C}^2} \\ =\sqrt{9+16} \\ =\sqrt{25} \\ =5\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Ar}(\mathrm{shaded}\mathrm{part})=\mathrm{Ar}\left(∆ABC\right)+\mathrm{Ar}\left(\mathrm{semicircle}APB\right)+\mathrm{Ar}\left(\mathrm{semicircle}AQC\right)-\mathrm{Ar}\left(\mathrm{semicircle}BAC\right) \\ =\left(\frac{1}{2}\times 3\times 4\right)+\left(\frac{1}{2}\mathrm{\pi }\times 1.5\times 1.5\right)+\left(\frac{1}{2}\mathrm{\pi }\times 2\times 2\right)-\left(\frac{1}{2}\mathrm{\pi }\times 2.5\times 2.5\right) \\ =\left\{6+\frac{1}{2}\mathrm{\pi }\left(4+\frac{9}{4}-\frac{25}{4}\right)\right\} \\ =6+0 \\ =6{\mathrm{cm}}^2 \end{gathered}$$