In triangle ABC, B = 90 degrees. Semicircles are constructed on sides AB, AC, and BC, as shown below. Prove that the total area of the shaded region is equal to the area of triangle ABC.
+128407 Note that the area of the right triangle = (1/2)(BC) (AB)
And note that the area between the right triangle and the circle = (1/2)pi (AC/2) ^2 - (1/2) (BC)(AB) =
(1/2) [ pi ( AC)^2 / 4 - (BC)(AB) ] = (1/2) [ BC^2 + AB^2 ] / 4 - (1/2) (BC)(AB) (2)
The area of the two semi-circles = (1/2) [ (BC/2)^2 + (AB/2)^2 ] = (1/2) [ BC^2 + AB^2 ]/ 4 (3)
So....the shaded area = (3) - (2) =
(1/2) [ BC^2 + AB^2 ] /4 - [ (1/2) (BC^2 + AB^2 ] / 4 - (1/2) (BC)(AB) ] =
(1/2) (BC)(AB) which is the same as the area of the right triangle
