Question

The perimeter of a triangle ABC right angled at C is $70$. and the inradius is $6$. then $\left|a-b\right|$

• A. $1$
• B. $2$
• C. $8$
• D. $9$

Answer 1

The correct option is A $1$

By Pythagoras theorem, $a^2+b^2=c^2$

Area of triangle$=\Delta =\frac{1}{2}ab$

Semiperimeter$=2s=a+b+c=70$ ......$\left(1\right)$

$\Rightarrow s=\frac{70}{2}=35$

$r=\frac{\Delta }{s}\Rightarrow \Delta =rs$

$\Rightarrow \frac{1}{2}ab=35\times 6$

$\Rightarrow ab=420$ ......$\left(2\right)$

We have $a^2+b^2=c^2$

$\Rightarrow {(a+b)}^2-2ab=c^2$

$\Rightarrow {(70-c)}^2-2\times 420=c^2$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow {(70-c)}^2-840=c^2$

$\Rightarrow 4900+c^2-140c-840=c^2$

$\Rightarrow 4900-840=140c$

$\Rightarrow 140c=4060$

$\Rightarrow c=29$

$\therefore |a-b|=\sqrt{{(a+b)}^2-4ab}$

$=\sqrt{{(70-c)}^2-4\times 420}$

$=\sqrt{{(70-29)}^2-1680}$

$=\sqrt{{\left(41\right)}^2-1680}$

$=\sqrt{1681-1680}=\sqrt{1}=1$unit