Question

The sides of a triangle are given by $\sqrt{b^2+c^2},\sqrt{c^2+a^2},\sqrt{a^2+b^2}$, where $a,b,c>0$, then the area of the triangle equals

• A. $\frac{1}{2}\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}$
• B. $\frac{1}{2}\sqrt{a^4+b^4+c^4}$
• C. $\frac{\sqrt{3}}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$
• D. $\frac{\sqrt{3}}{2}(bc+ca+ab)$

Answer 1

The correct option is A $\frac{1}{2}\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}$

Area of triangle =$\frac{1}{2}ab\sin C$
Here, Area of $△ABC=\frac{1}{2}\sqrt{b^2+c^2}\sqrt{c^2+a^2}\sin C$
Now,
$\cos C=\frac{b^2+c^2+c^2+a^2-a^2-b^2}{2\sqrt{b^2+c^2}\sqrt{c^2+a^2}}$
$\Rightarrow \cos C=\frac{c^2}{2\sqrt{b^2+c^2}\sqrt{c^2+a^2}}$
$\Rightarrow \sin C=\sqrt{1-\frac{c^4}{(b^2+c^2)(c^2+a^2)}}$
$\Rightarrow \sin C=\frac{\sqrt{b^2{c}^2+b^2{a}^2+c^2{a}^2}}{\sqrt{b^2+c^2}\sqrt{c^2+a^2}}$
So, area $=\frac{1}{2}\sqrt{b^2{c}^2+b^2{a}^2+c^2{a}^2}$