Question

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

Then the triangle is?

• A. isosceles but not right
• B. right but not isosceles
• C. right isosceles
• D. equilateral

Answer 1

The correct option is C right isosceles

Given, $R(b+c)=a\sqrt{bc}$

from sine rule: $b+c=2\sin A\sqrt{bc}$ ....(1)
As A.M$\ge$G.M

Therefore, $\frac{b+c}{2}\ge \sqrt{bc}$

Using (1), we get $\Rightarrow \sin A\ge 1$

But $-1\le \sin A\le 1$

So $\sin A=1\Rightarrow A=\frac{\pi }{2}$

Substituting in (1), we get $b+c=2\sqrt{bc}$

$\Rightarrow b^2+c^2-2bc=0$

$\Rightarrow b=c$

Therefore, the given triangle is right isosceles.