Question

In a $△ABC,$ if $2{△}^2=\frac{a^2{b}^2{c}^2}{a^2+b^2+c^2}$ then it is

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

Answer 1

Answer: (D)

right angled

Given: $2{△}^2=\frac{a^2{b}^2{c}^2}{a^2+b^2+c^2}$
$\Rightarrow 2{△}^2(a^2+b^2+c^2)=a^2{b}^2{c}^2$
$\Rightarrow a^2+b^2+c^2={\left(\frac{abc}{△}\right)}^2.\frac{1}{2}$
$={\left(\frac{abc}{\frac{abc}{4R}}\right)}^2.\frac{1}{2}$ where $△=\frac{abc}{4R}$
$=8R^2$
Using sine rule, we have
$a^2+b^2+c^2=8R^2$
$\Rightarrow {\left(2R\sin A\right)}^2+{\left(2R\sin B\right)}^2+{\left(2R\sin C\right)}^2=8R^2$
$\Rightarrow {4R}^2[{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C]=8R^2$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$
$\Rightarrow 1-{\cos }^{2}A+1-{\cos }^{2}B+{\sin }^{2}C=2$
$\Rightarrow {\cos }^{2}A+{\cos }^{2}B-{\sin }^{2}C=0$
$\Rightarrow {\cos }^{2}A+\cos (B+C)\cos (B-C)=0$
$\Rightarrow {\cos }^{2}A+\cos (\pi -A)\cos (B-C)=0$
$\Rightarrow {\cos }^{2}A-\cos A\cos (B-C)=0$
$\Rightarrow \cos A[\cos A-\cos (B-C)]=0$
$\Rightarrow \cos A[\cos (\pi -(B+C))-\cos (B-C)]=0$
$\Rightarrow -\cos A[\cos (B+C)+\cos (B-C)]=0$
$\Rightarrow \cos A\left[2\cos B\cos C\right]=0$
$\Rightarrow \cos A\cos B\cos C=0$
$\Rightarrow A={90}^0,$ or $B={90}^0$ or $C={90}^0$