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