Question

In $△ABC$, 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

The correct option is A right angled triangle

Given that $8R^2=a^2+b^2+c^2$

$\Rightarrow \frac{a^2}{4R^2}+\frac{b^2}{4R^2}+\frac{c^2}{4R^2}=2$

Using sine rule, we get

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$

Put $C=\pi -(A+B)$

${\sin }^{2}A+{\sin }^{2}B+{\sin }^2(A+B)=2$.

Now expand $\sin (A+B)$

$\Rightarrow 1-{\sin }^{2}A+1-{\sin }^{2}B=(\sin A\cos B+\sin B\cos A{)}^2$.

$\Rightarrow {\cos }^{2}A+co{s}^{2}B={\sin }^{2}A{\cos }^{2}B+2\sin A\sin B\cos A\cos B+{\sin }^{2}B{\cos }^{2}A$

$\Rightarrow {\cos }^{2}A-{\cos }^{2}A{\sin }^{2}B+{\cos }^{2}B-{\cos }^{2}B{\sin }^{2}A=2\sin A\sin B\cos A\cos B$

$\Rightarrow {\cos }^{2}A(1-{\sin }^{2}B)+{\cos }^{2}B(1-{\sin }^{2}A)=2\sin A\sin B\cos A\cos B$

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

$\Rightarrow 2{\cos }^{2}A{\cos }^{2}B=2\sin A\sin B\cos A\cos B$

$\Rightarrow \cos A\cos B(1-\sin A\sin B)=0$

$\cos A=0$ or $\cos B=0$

$A=\frac{\pi }{2}$ or $B=\frac{\pi }{2}$

$\therefore △ABC$ is a right-angled triangle.