What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?

This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.)
Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$. 
Expand $\sin(A+B)$ and do the manipulations to get 
$2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$ 
which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\cos(A)\cos(B) = \sin(A)\sin(B) \Rightarrow \cos(A+B) = 0 \Rightarrow \cos(C) = 0$. 
Hence either $A = \pi/2$ or $B = \pi/2$ or $C = \pi/2$. 
So the triangle is a right-angled triangle.
