In an adjoining figure ABCD is a square of side 6 cm.
Find the area of shaded region.
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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 SO IT WILL HELP YOU
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 SO IT WILL HELP YOU
anmolsidhu4:
hey
plz mark me brainliest answer
Ok
But your answer is incorrect
It is not matching with answers given in my book
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