In the given figure, LABC is right - angled at B such that BC = 6 cm and AB = 8 cm. A circle with center O has been inscribed inside the triangle. OP ⊥AB, OQ ⊥BC and OR ⊥AC. If OP = OQ = OR = x cm then x = ?

As ABC is a right triangle,


By Pythagoras Theorem


[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]


(AC)2 = (BC)2 + (AB)2


(AC)2 = (6)2 + (8)2


(AC)2 = 36 + 64 = 100


AC = 10 cm


Now,


ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)


As we know,


Area of triangle is 1/2 × Base × Height(Altitude)


Now,


OP ⏊ AB [Given]


∴ OP is the altitude in △AOB


OQ ⏊ BC [Given]


∴ OQ is the altitude in △BOC


OR ⏊ AC [Given]


∴ OR is the altitude in △AOC


So, we have


1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)


6(8) = 8(x) + 6(x) + 10(x)


[∵ OP = OQ = OR = x, Given]


48 = 24x


x = 2 cm

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