Question

in the given figure,ABC is a right triangle right angled at A.Find the area of shaded region,if AB =6m,BC=10m and O is the center of the incricle of triangle ABC. (use π =3.14

Answer 1

AC = 8m ( Pythagoras Theorem )
Area of ∆ AOC = 1/2 * r * 8 = 4r
Area of ∆ AOB = 3r
Area of ∆ BOC = 5r
Adding these 3 equations,
12r = Area of ABC
12r = 1/2 * 6 * 8
12r = 24
r = 2
Area of circle = 3.14*2*2
= 3.14*4 = 12.56 m^2
Area of shaded region = 24-12.56
= 11.44 m^2

Answer 2

Answer:

ABC is a right angled triangle where ∠A=90∘

BC=10cm and AB=6cm

Let O be the centre and r be the radius of the in-circle.

AB,BC and CA are the tangents to the circle at P,M and N

∴IP=IM=IN=r(radius of the circle)

In △BAC,

BC2=AB2+AC2(by pythagoras theorem)

⇒102=62+AC2

⇒AC2=100−36=64

∴AC=8cm

Area of △ABC=1/2​bh=1/2×AC×AB=1/2​×8×6=24sq.cm

Area of △ABC=Area of △IAB+Area of △IBC+ Area of △ICA

⇒24=1/2​r(AB)+1/2​r(BC)+1/2​r(CA)

⇒24=1/2​r(AB+BC+CA)

⇒24=1/2​r(6+8+10)

⇒24=12r  

∴r=24/12​=2cm

Area of the circle=πr2=​ 22/7 × 2^2 =12.56sq.cm

Area of shaded region=Area of △ABC−Area of the circle.

                                    =24−12.56=11.44sq.cm

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