Question

The inradius of the triangle whose sides are $3,5,6$ is

• A. $\sqrt{\frac{8}{7}}$
• B. $\sqrt{8}$
• C. $\sqrt{7}$
• D. $\sqrt{\frac{7}{8}}$

Answer 1

Answer: (A)

$\sqrt{\frac{8}{7}}$

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ and $s=\frac{a+b+c}{2}$
where $a=3,b=5,c=6$
$\Rightarrow \Delta =\sqrt{7\times 4\times 2\times 1}=2\sqrt{14}$
Inradius $r=\frac{\Delta }{s}=\frac{2\sqrt{14}}{7}=\sqrt{\frac{8}{7}}$