Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

The triangle is right angled at $B,$ then the inradius of the triangle is:

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

Answer 1

The correct option is A $\frac{AB+BC-CA}{2}$

Area of $△ABC=\frac{1}{2}ac$
Area of $△ABC=\frac{1}{2}ra+\frac{1}{2}rb+\frac{1}{2}rc$
$\frac{1}{2}ac=\frac{1}{2}r(a+b+c)$
$\Rightarrow r=\frac{ac}{a+b+c}$
It can be written as$r=\frac{1}{2}(a+c-b)$

$r=\frac{AB+BC-CA}{2}$