In the figure given below, ABC and CDE are two identical semi-circles of radius 2 units. B and D are the mid points of the arc ABC and CDE respectively. What is the area of the shaded region? Select
A) 4π - 1	B) 3π - 1	C) 2π - 4	D) ½(3π - 1)	E) 2π - 2
Solution   P and Q are the centers of the two semicircles.   Draw BP perpendicular to AC.   BP is radius to the semi-circle. So are AP and PC.   Therefore, BP = AP = PC = 2 units.   In semicircle ABC, area of the shaded portion is the difference between the area of half the semicircle PBC and the area of the triangle PBC.   Triangle PBC is a right triangle because PB is perpendicular to PC. PB and PC are radii to the circle and are equal. So, triangle PBC is an isosceles triangle.   Therefore, triangle PBC is a right isosceles triangle.   Area of half the semicircle - Area of region PBC   Area of the semicircle ABC = ½ area of the circle of radius 2.   So, area of half the semicircle, PBC = ¼ area of the circle of radius 2.   Area of half the semicircle, PB2π - 4 = ¼ * π * 22   Area of half the semicircle, PB2π - 4 = π sq units   Area of right isosceles triangle PB2π - 4   Area of right triangle PB2π - 4 = ½ P2π - 4 * PB   Area of triangle PB2π - 4 = ½ * 2 * 2 = 2 sq units   Area of shaded region   Area of shaded region in one of the semi circles AB2π - 4 = (π - 2) sq units   Therefore, area of the overall shaded region = 2(π - 2) sq units   2π - 4hoice 2π - 4
<result>C</result>
In the figure given below, ABC and CDE are two identical semi-circles of radius 2 units. B and D are the mid points of the arc ABC and CDE respectively. What is the area of the shaded region?