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

(A) 4

(B) 3

(C) 2

(D) 1

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#### Solution

Correct answer: C

It is given that AB = 5 and BC = 12

Using Pythagoras theorem

AC^{2}=AB^{2}+BC^{2}

5^{2}+12^{2}

=169

Thus 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.

Thus 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.**

Concept: Right-angled Triangles and Pythagoras Property

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