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Let ABC be the right angled triangle such that ∠B = 90° , BC = 6 cm, AB = 8 cm. Let O be the centre and r be the radius of the in circle.
AB, BC and CA are tangents to the circle at P, N and M.
∴ OP = ON = OM = r (radius of the circle)
By Pythagoras theorem,
CA^{2} = AB^{2} + BC^{2}
⇒ CA^{2} = 8^{2} + 6^{2}
⇒ CA^{2} = 100
⇒ CA = 10 cm
Area of ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA
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