Question

In a triangle $ABC$ right-angled at $B$, the inradius is

• A. $AB+BC-AC$
• B. $AB+AC-AC$
• C. $\frac{AB+BC-AC}{2}$
• D. none of these.

Answer 1

Answer: (C)

$\frac{AB+BC-AC}{2}$

Let $AE=AF=\alpha$
$BD=BE=\beta$
$CD=CF=\gamma$
$\because AB+BC+CA=2(\alpha +\beta +\gamma )$
$\Rightarrow 2s=2(\alpha +\beta +\gamma )$
$\therefore \alpha +\beta +\gamma =s$
$\Rightarrow \alpha =s-a,\beta =s-b,\gamma =s-c$
In $△BID,\tan \frac{\pi }{4}=\frac{r}{\beta }$
$\Rightarrow 1=\frac{r}{\beta }$
$\Rightarrow \beta =r=s-b$
$=\frac{a+b+c}{2}-b$
$=\frac{a+c-b}{2}$
$=\frac{AB+BC-CA}{2}$