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

This fig..

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@Dominic....fig is short form for Figure

thats is answer

given ans

It's the wrong answer @raunak mishra

Give Full answer

Answer is 34.428 Sq. Cm

In the adjoining figure, ABCD is a square of side 6cm. Find the area of the shaded region. 1190_circle33.png 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

Answer is 12.32 cm2

And the figure is....

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 = 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

In the adjoining figure, ABCD is a square of side 6cm. Find the area of the shaded region. 1190_circle33.png 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

12.336

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12.32 is the answer

This is your answer

Here