Question

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 [CBSE 2014]

Answer 1

Let a circle with centre O and radius r cm be inscribed in the right ∆ABC.

In right ∆ABC,

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right) \\ \Rightarrow \mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2} \\ \Rightarrow \mathrm{AC}=\sqrt{5^2+{12}^2} \\ \Rightarrow \mathrm{AC}=\sqrt{169}=13\mathrm{cm} \end{gathered}$$

Now, the sides AB, BC and CA are tangents to the circle at R, P and Q, respectively.

∴ OR ⊥ AB, OP ⊥ BC and OQ ⊥ AC

Now,

ar(∆OAB) + ar(∆OBC) + ar(∆OAC) = ar(∆ABC)

$$\begin{gathered} \Rightarrow \frac{1}{2}\times \mathrm{AB}\times \mathrm{OR}+\frac{1}{2}\times \mathrm{BC}\times \mathrm{OP}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OQ}=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB} \\ \Rightarrow \frac{1}{2}\times 5\times r+\frac{1}{2}\times 12\times r+\frac{1}{2}\times 13\times r=\frac{1}{2}\times 12\times 5 \\ \Rightarrow 5r+12r+13r=60 \\ \Rightarrow 30r=60 \\ \Rightarrow r=2 \end{gathered}$$

Thus, the radius of the circle is 2 cm.

Hence, the correct answer is option C.