$ABCD$ is a square with $AB = 8$cm. Arcs $BC$ and $CD$ are semicircles. Express the area of the shaded region, in square centimeters, and in terms of $\pi$.
Where the two semi-circles intersect in the middle of the square, call this point E
Connect EC
Let the midpoint of DC be F
So FC and FE are radii of the top semi-circle = 4 cm
And FE is perpendicular to FC
So CFE is a right triangle ....and the area of this triangle = (1/2)FE * FC =
(1/2) (4) (4) = 8 cm ^2
And FC , FE and arc EC will form sector FEC of this top semicircle
And the area of this sector = (1/2) FE^2 ( pi/2) = (1/2) 4^2 (pi/2) = 4 pi cm^2
So...the area of the sector minus the area of the triangle will equal 1/2 of the shaded area =
[ 4pi - 8] cm^2
So...the shaded area = 2 [ 4pi - 8 ] = [ 8pi - 16 ] cm^2