In the adjoining figure, ABCD is a square of side 6cm.  Find the area of the shaded region.

Ans:    From P draw PQ ⊥ AB

AQ = QB = 3cm

(Ans: 34.428 sq cm)

Join PB. Since arc APC is described by a circle with center B,

so BA = BP=BC =6cm.

In ? PQB Cos θ  = QB/PB  = 1/2

∴θ     = 60^o

Area of sector BPA =60/360 Π (6²) = 18.84cm

Area of  Δ BPQ = 1/2(QB) (PQ) = 1/2 (3)( 6 Sin 60) = 7.794Sq.cm

Area of portion APQ = Area of sector BPA - Area of  Δ BPQ

= 18.84 - 7.794 = 11.046 Sq.cm

Area of shaded portion = 2 x Area of Quadrant ABC - 2 Area APQ

= [2 x π/4(6)²  - 2 x 11.046]

=34.428 Sq.cm