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

Answer 1

Correct option is (b) $2$ cm

Since, $\Delta ABC$ is a right angle, therefore, by Pythagoras Theorem,

$\Rightarrow AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{25+144}=\sqrt{{13}^2}=13cm$

Now, $OQ,OP$ and $OR$ are radius of the circle therefore, $OQ=OP=OR=x$ (say) and

are also perpendicular to respective sides $BC,AB$ and $AC$. [Radius $\perp$ Tangent]

Also, join $OA,OB$ and $OC$.

$\therefore ar\Delta ABC=ar\Delta AOB+ar\Delta BOC+ar\Delta AOC$

$\Rightarrow \frac{1}{2}\times AB\times BC=\frac{1}{2}\times AB\times PO+\frac{1}{2}\times QO\times BC+\frac{1}{2}\times AC\times RO$

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

$\Rightarrow 60=(5+12+13)x$

$\therefore x=\frac{60}{30}=2cm$

Hence, radius of incircle is $2cm$.