Question

In the given figure, $\Delta ABC$ is a right-angled triangle in which $\angle A$ is ${90}^{\circ }.$ Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

Answer 1

In right $\Delta BAC,$ by pythagoras theorem,

$B{C}^2=A{B}^2+B{C}^2$
$=(3{)}^2+(4{)}^2$
$=9+16=25$
$BC=\sqrt{25}=5cm$
Area of semi-circle with diameter $BC=\frac{1}{2}\pi r^2$
$=\frac{1}{2}\times \pi (\frac{5}{2}{)}^2=\frac{25}{8}\pi c{m}^2$
Area of semi-circle with diameter $AB=\frac{1}{2}\pi r^2$
$=\frac{1}{2}\times \pi (\frac{3}{2}{)}^2=\frac{9}{8}\pi c{m}^2$
Area of semi-circle with diameter $AC=\frac{1}{2}\pi r^2$
$=\frac{1}{2}\times \pi (\frac{4}{2}{)}^2=\frac{16}{8}\pi c{m}^2\pi c{m}^2$
Area of semi-circle with diameter $AC=\frac{1}{2}\pi r^2$
Area of rt $\Delta BAC=\frac{1}{2}\times AB\times AC$
$=\frac{1}{2}\times 3\times 4=6c{m}^2$
Area of dotted region $=(\frac{25}{8}\pi -6)c{m}^2$
Area of shaded region $=\frac{16}{8}\pi +\frac{9}{8}\pi -(\frac{25}{8}\pi -6)$
$=\frac{16}{8}\pi +\frac{9}{8}\pi -\frac{25}{8}\pi +6$
$=6c{m}^2$