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. $\frac{\sqrt{3}}{2}\sqrt{bc+ca+ab}$

Answer 1

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

It is given that, $l=\sqrt{b^2+c^2}$; $m=\sqrt{c^2+a^2}$; $n=\sqrt{a^2+b^2}$

Now from cosine rule: $\cos L=\frac{m^2+n^2-l^2}{2mn}=\frac{(c^2+a^2)+(a^2+b^2)-b^2-c^2}{2\sqrt{(c^2+a^2)(a^2+b^2)}}$

$\Rightarrow \cos L=\frac{a^2}{\sqrt{(c^2+a^2)(a^2+b^2)}}$

$\Rightarrow \sin L=\sqrt{1-{\cos }^{2}L}=\frac{\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}}{\sqrt{(c^2+a^2)(a^2+b^2)}}$

Therefore, $△=\frac{mn\sin L}{2}=\frac{\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}}{2}$