Question

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

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

Answer 1

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

$\Delta =\frac{1}{2}{b}^{\prime }{c}^{\prime }sin{A}^{\prime }$
$cos{A}^{\prime }=\frac{b^{\prime 2}+c^{\prime 2}-a^{\prime 2}}{2b^{\prime }{c}^{\prime }}$
or $cos{A}^{\prime }=\frac{(c^2+a^2)+(a^2+b^2)-(b^2+c^2)}{2\sqrt{\left(\right(c^2+a^2\left)\right(a^2+b^2\left)\right)}}$
$=\frac{2a^2}{2\sqrt{\left(\right(c^2+a^2\left)\right(a^2+b^2\left)\right)}}$
$\therefore si{n}^2{A}^{\prime }=1-co{s}^2{A}^{\prime }$
$=1-\frac{a^4}{c^2{b}^2+c^2{a}^2+a^2{b}^2+a^4}=\frac{\sum \left(b^2{c}^2\right)}{b^{\prime 2}{c}^{\prime 2}}$ ...(I)
$\therefore$ Area $=\frac{1}{2}{b}^{\prime }{c}^{\prime }sin{A}^{\prime }=\frac{1}{2}\sqrt{\sum b^2{c}^2}$ by (I)