In the given figure, ΔABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and 0 is the centre of the incircle of ΔABC. [Take π = 3.14.]

Given AB = 6 cm, BC = 10 cm

Consider ∆ABC, ∠BAC = 90°

⇒ 10^{2} = 6^{2} + AC^{2} (putting given values)

⇒ 100 = 36 + AC^{2}

⇒ 100 – 36 = AC^{2}

⇒ AC^{2} = 64

∴ AC = 8 cm

Join OB, OA, OC, OE, OF, OD

Here OE = OF = OD = radius of circle = r cm

∠OEC = ∠ODB = ∠OFB = 90° (angle at the point of contact of radius & tangent)

Area ∆ABC = Area of ∆OAC + Area of ∆OCB + Area of ∆OAB → eqn1

∴ Area of ∆OAC = 4r → eqn2

∴ Area of ∆OCB = 5r → eqn3

∴ Area of ∆OAB = 3r → eqn4

⇒ Area of ∆ABC = 3×8

∴ Area of ∆ABC = 24 cm^{2}→ eqn5

Putting all the values in equation we get;

⇒ 24 = 4r + 5r + 3r

⇒ 24 = 12r

∴ r = 2 cm

Area of circle = πr^{2}

Put the value of r, we get,

⇒ Area of circle = π×2^{2}

⇒ Area of circle = 3.14×4 (putting π = 3.14)

∴ Area of circle = 12.56 cm^{2}→ eqn6

Area of shaded region = Area of triangle – Area of circle

⇒ Area of shaded region = 24 – 12.56 (from eqn5 and eqn6)

∴ Area of shaded region = 11.44 cm^{2}

Area of shaded region is 11.44 cm^{2}.

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