Question

In Fig. ABC is a right angled triangle in which $\angle A={90}^{\circ }$, AB = 21 cm and AC = 28 cm. Semi - circles are described on AB, BC and AC as diameters. Find the area if the shaded region.

Answer 1

In right ΔABC, by Pythagoras theorem:

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

$A{C}^2={28}^2+{21}^2$

$A{C}^2=35$

$Areaof△ABC=\frac{BC\times AB}{2}=\frac{21\times 28}{2}=294c{m}^2$

Semicircle's area = $\frac{1}{2}\times \frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}=481.25c{m}^2$

Quadrant's area = $\frac{1}{4}\times \frac{22}{7}\times 21\times 21=346.5c{m}^2$

Area of the shaded region
= Semicircle's area +Area of ΔABC - Quadrant's area
= 481.25 + 346.5 - 294 = 428.75 $c{m}^2$