A triangle is drawn by joining a point on the semi-circle to the ends of the diameter. Two semi-circles are drawn with the other two sides as diameter.
Find the sum of the areas of shaded part.
Let AB=a, AC=b
∴ BC = √a2+b2 (Δ ABC is a right triangle)
ΔBAC=90∘ (angle in semi-circle)
Now, area of Δ ABC = 12ab
Now, area of semi-circle as diameter AB = 12π(a2)2
and , area of semi-circle as diameter AC = 12π(b2)2
area of semi-circle as diameter BC =12π[√a2+b22]2
Now, area of shaded part
= [12πa24+12πb24]−[12π(b2+a24)−12ab]
=π8(a2+b2)−π8(a2+b2)+12ab
= 12ab = 12(AB)(BC)