Question

In a triangle $ABC,$ $R(b+c)=a\sqrt{bc},$ where $R$ is the circumradius and $a,b,c$ are the sides of triangle $ABC.$ Then the triangle is

• A. isosceles but not right-angled
• B. right-angled but not isosceles
• C. right-angled and isosceles
• D. equilateral

Answer 1

The correct option is C right-angled and isosceles

$R(b+c)=a\sqrt{bc}$
We know that $R=\frac{a}{2\sin A}$
$\therefore \frac{a}{2\sin A}\cdot (b+c)=a\sqrt{bc}$
$\Rightarrow \sin A=\frac{b+c}{2\sqrt{bc}}$
Applying A.M. $\ge$ G.M. on $b$ and $c,$
$\frac{b+c}{2}\ge \sqrt{bc}$
$\Rightarrow \frac{b+c}{2\sqrt{bc}}\ge 1$
$\Rightarrow \sin A\ge 1$
$\therefore \sin A=1$ ($\because \sin A>1$ is not possible)
$\Rightarrow A={90}^{\circ }$

Now, $b+c=2\sqrt{bc}$
$\Rightarrow (b+c{)}^2=4bc$
$\Rightarrow b=c$
$\therefore A={90}^{\circ }$ and $b=c$
Hence, triangle is isosceles and right-angled.