In a right triangle ABC, right - angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is

Let AB, BC and AC touch the circle at points P, Q and R respectively.

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} = (12)^{2} + (5)^{2}

(AC)^{2} = 144 + 25 = 169

AC = 13 cm

Let O be the center of circle, Join OP, OQ and PR

Let the radius of circle be r,

We have

r = OP = OQ = OR

[radii of same circle] [1]

Now,

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

As we know,

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

Now,

OP ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OP is the altitude in △AOB

OQ ⏊ BC

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OQ is the altitude in △BOC

OR ⏊ AC

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ 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)

12(5) = 5(r) + 12(r) + 13(r) [Using 1]

60 = 30r

r = 2 cm

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