Question

In a triangle, if $8R^2=a^2+b^2+c^2$, then the triangle is a

• A. Right angled triangle
• B. equilateral triangle
• C. Scalene triangle
• D. Obtuse angled triangle

Answer 1

Answer: (A)

Right angled triangle

Given:

$8R^2=a^2+b^2+c^2$ ...$\left(i\right)$

By Sine rule, for a triangle we know that

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin c}=2R$

Replacing $a$ with $2R\sin A$

$b=2R\sin B$ and $c=2R\sin C$.... $\left(ii\right)$

This essentially means

${\sin }^2\left(A\right)+{\sin }^2\left(B\right)+{\sin }^2\left(C\right)=2$ (from $\left(i\right)$) .... $\left(iii\right)$

Replace $C=\pi -(A+B)$ to get ${\sin }^2(A+B)$

Now,

${\cos }^2\left(A\right)+{\cos }^2\left(B\right)={\sin }^2(A+B)$ ... from $\left(iii\right)$

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Using this , we get

$2{\cos }^2\left(A\right){\cos }^2\left(B\right)=2\sin \left(A\right)\sin \left(B\right)\cos \left(A\right)\cos \left(B\right)$

$\Rightarrow \cos \left(A\right)=0$ or

$\cos \left(B\right)=0$ or

$\cos \left(A\right)\cos \left(B\right)=\sin \left(A\right)\sin \left(B\right)\Rightarrow \cos (A+B)=0\Rightarrow \cos \left(C\right)=0$

Hence either $A=\frac{\pi }{2}$ or $B=\frac{\pi }{2}$ or $C=\frac{\pi }{2}$

So, the triangle is a right-angled triangle.