In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm.
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 is

(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm

Solution:

In right triangle ABC
By using Pythagoras theorem we have
AC2 = AB2 + BC2
= 52 + 122
= 25 + 144
= 169
∴ AC2 = 169
⇒ AC = 13 cm
Now,

$\operatorname{Ar}(\triangle \mathrm{ABC})=\operatorname{Ar}(\triangle \mathrm{AOB})+\operatorname{Ar}(\triangle \mathrm{BOC})+\operatorname{Ar}(\triangle \mathrm{AOC})$

$\Rightarrow \frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}=\frac{1}{2} \times \mathrm{OP} \times \mathrm{AB}+\frac{1}{2} \times \mathrm{OQ} \times \mathrm{BC}+\frac{1}{2} \times \mathrm{OR} \times \mathrm{AC}$

$\Rightarrow 5 \times 12=x \times 5+x \times 12+x \times 13$

$\Rightarrow 60=30 x$

$\Rightarrow x=2 \mathrm{~cm}$

Hence, the correct answer is option (b).