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 (in cm) is?

Given: AB = 5 cm, BC = 12 cm
Using Pythagoras theorem, AC² = AB² + BC² = 5² + 12² = 25 + 144 = 169
AC = 13.
We know that two tangents drawn to a circle from the same point that is exterior to the circle are of equal lengths.
So, AM = AQ = a
Similarly MB = BP = b and PC = CQ = c
We know AB = a + b = 5
BC = b + c = 12
and AC = a + c = 13
Solving simultaneously we get a = 3, b = 2 and c = 10
We also know that the tangent is perpendicular to the radius
Thus OMBP is a square with side b.
Hence the length of the radius of the circle inscribed in the right angled triangle is 2 cm.