Question

In the given figure, $△ABC$ is right-angled at $A$. Semi circles are drawn on $AB,AC$ and $BC$ as diameters. It is given that $AB=3cm$ and $AC=4cm$. Find the area of the shaded region in $c{m}^2$.

Answer 1

In $△ABC,\angle A={90}^o$

By Pythagoras theorem,

$B{C}^2=A{B}^2+A{C}^2$

$B{C}^2=3^2+4^2$

$9+16=25$

$\therefore BC=5cm$

$A\left(\text{shadded region}\right)=A(△$ $\text{ABC)+A (semicircle AB) + (semicircle AC) - A(semicircle BAC)}$

$=\frac{1}{2}\times 3\times 4+\frac{1}{2}\times \pi {\left(\frac{3}{2}\right)}^2+\frac{1}{2}\times \pi {\left(\frac{4}{2}\right)}^2-\frac{1}{2}\times \pi {\left(\frac{5}{2}\right)}^2$

$=6+\frac{9\pi }{8}+2\pi -\frac{25\pi }{8}$

$=6+2\pi -2\pi$

$=6c{m}^2$