Question

In fig abc is a right angled triangle in which angle a=90, ab=21 cm and ac= 28cm. semi circles are described on ab,bc and ac as diameters. find the area of the shaded region

Answer 1

In right ΔABC, by Pythagoras theorem:
AC2 = AB2+BC2
AC2 = 282 + 212
AC2 =35
 Area of ΔABC = ½*BC*AB  
 = ½*21*28  
=294
 Semicircle's area = ½* 22/7*35/2*35/2
= 481.25 Quadrant's area 
= ¼*22/7*21*21  
=346.5 Area of the shaded region
= Semicircle's area +Area of ΔABC - Quadrant's area
= 481.25 + 346.5 - 294 = 428.75 cm²

Answer 2

Answer:428.75

Step-by-step explanation;

In right ΔABC, by Pythagoras theorem:

AC2 = AB2+BC2

AC2 = 282 + 212

AC2 =35

Area of ΔABC = ½*BC*AB  

= ½*21*28  

=294

Semicircle's area = ½* 22/7*35/2*35/2

= 481.25 Quadrant's area  

= ¼*22/7*21*21  

=346.5 Area of the shaded region

= Semicircle's area +Area of ΔABC - Quadrant's area

= 481.25 + 346.5 - 294 = 428.75 cm²