Question

In $△ABC,8\Delta =(b+c)(bc+1)$, then circumradius of $△ABC$ is ($\Delta$ denotes area of triangle and $a,b,c$ are length of sides $BC,AC$ and $AB$ respectively)

• A. $\sqrt{\Delta }$
• B. $\frac{1}{\sqrt{2\Delta }}$
• C. $\sqrt{2\Delta }$
• D. $\frac{1}{\sqrt{\Delta }}$

Answer 1

The correct option is A $\sqrt{\Delta }$

$8\Delta =(b+c)(bc+1)\Rightarrow 8\times \frac{1}{2}bc\sin A=(b+c)(bc+1)\Rightarrow 4\sin A=(b+c)(1+\frac{1}{bc})\Rightarrow 4\sin A=b+\frac{1}{b}+c+\frac{1}{c}$
Maxmimum value of LHS is $4$ and minimum value of RHS is $4$, so both will be equal when
$\sin A=1\Rightarrow A=\frac{\pi }{2}$ and
$b=c=1$

$\Delta =\frac{1}{2}$
Circumradius $=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=\sqrt{\Delta }$