Question

In the adjoining figure, $ABC$ is a right angled triangle right angled at $A$. semi-circle are drawn on AB,AC and BC as diameters. find the area of the shaded region.

• A. $6.5c{m}^2$
• B. $7c{m}^2$
• C. $6c{m}^2$
• D. $4c{m}^2$

Answer 1

The correct option is C $6c{m}^2$

Given figure,

In right $△ABC$,

${BC}^2={AB}^2+{AC}^2$

$\Rightarrow {BC}^2={\sqrt{{\left(3\right)}^2+\left(4\right)}}^2$ ${\left(3\right)}^2+{\left(4\right)}^2$

$\Rightarrow {BC}^2=\sqrt{9+16}$ $9+16\dots \left[UsingPythagorastheorm\right]$

$\Rightarrow {BC}^2=\sqrt{25}$ $25$

$\Rightarrow {BC}^2=\sqrt{{\left(5\right)}^2}{\left(25\right)}^2$

$\Rightarrow BC=5cm$

Area of shaded region $=(ar\left(semicircleABP\right)+ar\left(semicircleACQ\right)+ar\left(semicircleQBC\right)-ar\left(semicircleABC\right))$

$=\frac{1}{2}\pi \left(\frac{1}{2}{AB}^2\right)+\frac{1}{2}\pi \left(\frac{1}{2}{AC}^2\right)+\frac{1}{2}\times AB\times AC-\frac{1}{2}\pi \left(\frac{1}{2}{BC}^2\right)$

$=\frac{1}{2}(\pi \times \frac{1}{4}{AB}^2+\pi \times \frac{1}{4}{AC}^2-\pi \times \frac{1}{4}{BC}^2)+\frac{1}{2}\times 3\times 4$

$=\pi \times \frac{1}{8}\times ({AB}^2+{AC}^2+{BC}^2)+6$

$=\pi \times \frac{1}{8}.(9+16-25)+6$

$=\pi \times \frac{1}{8}\times 0+6$

$=6{cm}^2$

$\therefore$ Area of the shaded region $=6{cm}^2$

$\therefore$ so, the answer is $6{cm}^2$