Question

In the given figure, $ABC$ is a right triangle, $\angle B={90}^{\circ },AB=28cm$ and $BC=21cm$. With $AC$ as diameter a semicircle is drawn and with $BC$ as radius a quarter circle is drawn. Find the area of the shaded region.

Answer 1

Given:

$△ABC$ is a right angled triangle at $B$

Height of the triangle, $AB=28cm$

Base of the triangle, $BC=21cm$

Therefore,

Area of the triangle $ABC=\frac{1}{2}\times 21\times 28$

$=294c{m}^2$

By pythagoras theorem,

Length of $AC=\sqrt{(21{)}^2+(28{)}^2}=\sqrt{1225}=35cm$

Now,

$AC$ is the diameter of the circle.

Therefore,

Area of the semicircle $=\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{35}{2}\right)}^2=\frac{1925}{4}c{m}^2$

Again,

Radius of another circle $=BC=21cm$

Area of the quarter circle $=\frac{1}{4}\times \frac{22}{7}\times {\left(21\right)}^2=\frac{693}{2}c{m}^2$

Therefore,

Required area of the shaded region $=294+\frac{1925}{4}-\frac{693}{2}$

$=428.75c{m}^2$