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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