3) Circle O has diameter AE with AE = 8 cm. Point C is on the circumference of the circle such that AC and CE are congruent. Also, AC is a diameter of semicircle ABC, and CE is a diameter of semicircle CDE, as shown in the figure. In square centimeters, what is the total combined area of the shaded regions?
Believe it or not, the two shaded regions have the same combined area as that of right triangle ACE
Proof :
Radius of semi-circle CDE = (1/2)CE
So...area of semi-circle CDE = (1/2)pi [(1/2)CE]*2 = (pi / 8) CE^2
And similarly.....the area of semi-circle ABC = (pi/8)AC^2
Area of right triangle ACE = (1/2)(AC)(CE)
Area of two unshaded regions between the right triangle and the circle =
(1/2)pi[(1/2)AE]^2 - (1/2)(AC)(CE) =
(1/2) pi (1/4) [ AE]^2 - (1/2)(AC)(CE) = { note that AE^2 = CE^2 + AC^2 }
(1/8) pi [ CE^2 + AC^2 ] - (1/2)(AC)(CE)
So....the area of the two shaded regions =
(pi/8) (CE^2 + AC^2] - [ (1/8)pi (CE^2 + AC^2 ) - (1/2)(AC)(CE) ] =
(1/2)(AC)(CE).....which is the same area as right triangle ACE
And the area of right triangle ACE = (1/2) (CE)(AC) = (1/2)(8/√2)(8/√2) = 64 / 4 = 16 cm^2 = area of two shaded regions