Question

If a circle is inscribed in a right angled triangle having right angle at B, then diameter of the circle is

• A. $AB+BC-AC$
• B. $BC+CA-AB$
• C. $AC+AB-BC$
• D. $AB+BC+CA$

Answer 1

Answer: (A)

$AB+BC-AC$

As we know, $△$$ABC$ is a right angle triangle,

so, area of $△$$ABC$ $=$ $\frac{1}{2}$$AB$$\times$$BC$

Also we can see in the figure that, area of $△$$ABC$ $=$ area of $△$$AOB$ $+$ area of $△$$BOC$ $+$ area of $△$$AOC$

so, area of $△$$ABC$ $=$ = $\frac{1}{2}$$r$$\times$$AB$ $+$ $\frac{1}{2}$$r$$\times$$BC$ $+$ $\frac{1}{2}$$r$$\times$$CA$

$\therefore$, area of $△$$ABC$ $=$ = $\frac{1}{2}$$r$$\times$$(AB+BC+CA)$

Equating both the equations, we get

$AB$$\times$$BC$ $=$ $r$$\times$$(AB+BC+CA)$

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

$2r=$ $\frac{(AB+BC{)}^2-(A{B}^2+B{C}^2)}{AB+BC+CA}$

Also, we know by pythagoras theorem, $A{C}^2$ $=$ $A{B}^2$ $+$ $B{C}^2$

$2r=$ $\frac{(AB+BC{)}^2-A{C}^2}{AB+BC+AC}$

$2r=$ $\frac{(AB+BC-AC)\times (AB+BC+AC)}{AB+BC+AC}$

$\therefore$$2r=$ $AB+BC-AC$

So Option A is correct