Question

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

Answer 1

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     = 60o

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

hope this helps

Answer 2

Given :  ABCD is a square of side 6cm.

To Find :  the area of shaded region.  

ABCD is a square with side = 6

Area of ABCD = 6 * 6 = 36 cm²

Radius of Arc = 6 cm

Area of BAC arc = (90/360) π (6)² =  (1/4) (3.14) * 36 = 28.26 cm²

Area of ABD arc = (90/360) π (6)² =  (1/4) (3.14) * 36 = 28.26 cm²

Let say intersection point of arc = P

Lets draw two straight line AP & BP = 6 cm =  radius

Area of Arc APB = (60/360) * 3.14 * 6² = 18.84 cm²

Area of Arc BAP = (60/360) * 3.14 * 6² = 18.84 cm²

Area of Δ APB = (√3 / 4 ) * 6² = 15.58 cm²

Portion x   is APB  

Area of Portion x = Area of Arc APB + Area of Arc BAP - Area of ΔAPB

Area of Portion x = 18.84 + 18.84 - 15.58 =  22.1 cm²

Area of Shaded region = Area of BAC arc+ Area of ABD arc - 2*Area of Portion x

=> Area of Shaded region = 28.26 + 28.26 - 2*(22.1)

=> Area of Shaded region = 56.52 - 44.2

=> Area of Shaded region = 12.32 cm²  

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