Question

in the given figure ABC is right angled triangle right angled at A . semi circles are drawn on AB,AC and BC as diameters . find the area of the shaded region​

Answer 1

Answer:

Step-by-step explanation:

In △ABC,∠A=90°

By Pythagoras theorem,

$BC=\sqrt{AC^{2} AB^{2} }$

$BC=\sqrt{9+16 }$

BC=$\sqrt{25}$

∴BC=5

NOW

A(shaded region)=A(△ ABC)+A (semicircle AB) + (semicircle AC) - A(semicircle BAC)

=  $\frac{1}{2}$*3*4 +  

$\frac{1}{2}$$\pi$$\frac{3}{2} ^{2}$ +

$\frac{1}{2}$$\pi$$\frac{4}{2} ^{2}$ -

$\frac{1}{2}$$\pi$$\frac{5}{2} ^{2}$

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

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

=6 sq.cm