Question

In $\Delta ABC,\angle A={90}^o,AB=5$ cm and $BC=13$ cm. Find the radius of the incircle of $\Delta ABC$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

R.E.F image

Pythagoras Theorem

$A{C}^2=\sqrt{{13}^2-5^2}$

$A{C}^2=144$

$AC=12$

and $a=13,b=12,c=5$

$S=\frac{a+b+c}{2}=\frac{13+12+5}{2}=\frac{30}{2}$

$S=15$

So, $r=(S-a)\tan \frac{A}{2}$

$r=(15-13)\tan \frac{90}{2}$

$r=2\tan {45}^o$

$r=2\times 1$

$r=2cm$

So, option (B) is right.