Question

$\Delta ABC,AB=5,BC=12$ and $AC=13$, then the radius of the circle touching all the three sides is ______.

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is A $2$

Given, $AB=5,BC=12$ and $AC=13$,
$A{B}^2+B{C}^2=(5{)}^2+(12{)}^2=25+144=169={13}^2=A{C}^2$
$\therefore A{B}^2+B{C}^2=A{C}^2$
$\therefore \Delta ABC$ is a right angle triangle and $\angle B$ is a right angle.
We know that, the radius of the circle touching all the three sides is $\frac{AB+BC-AC}{2}$.
$\therefore$ The required radius of a circle
$=\frac{AB+BC-AC}{2}$
$=\frac{5+12-13}{2}$
$=\frac{4}{2}$
$=2$

Hence, the answer is $2$.