Q. Find the area of shaded region . If BC = 4 cm and AB = 5 cm.

Dear Student,

$$\begin{gathered} As, \angle ABC=90° \left( angle opposite the diameter is 90°\right) \\ AB=5cm \\ BC=4cm \\ So, \\ AC=\sqrt{A{B}^2+B{C}^2} \\ =\sqrt{5^2+4^2} \\ =\sqrt{25+16} \\ =\sqrt{41} \\ Now, \\ Area of semi circle=\frac{{\mathrm{\pi r}}^2}{2} \left[r=\frac{AC}{2}=\frac{\sqrt{41}}{2}\right] \\ =\frac{22}{7}\times {\left(\frac{\sqrt{41}}{2}\right)}^2\times \frac{1}{2} \\ =\frac{22}{7}\times \frac{41}{8} \\ \approx 16.10 c{m}^2 \\ Now Area of ∆ABC \\ =\frac{1}{2}\times AB\times BC \\ =\frac{1}{2}\times 4\times 5 \\ =10 c{m}^2 \\ Area of shaded region=Area of semi circle- Area of ∆ABC \\ =16.10-10 \\ =\mathbf{6}\mathbf{.}\mathbf{10}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}\mathbf{ }\left(\mathbf{A}\mathbf{p}\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{x}\right)\mathbf{ } \end{gathered}$$
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