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In any triangle $\mathrm{ABC}$, if $\mathrm{a}=\sqrt{2} \mathrm{~cm}, \mathrm{~b}=\sqrt{3} \mathrm{~cm}$ and $\mathrm{c}=\sqrt{5} \mathrm{~cm}$, find area of triangle.
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Verified Answer
The correct answer is:
$\sqrt{6} / 2$
We know that, $\cos C=\frac{a^2+b^2-c^2}{2 a b}=0$
then, $\quad \angle \mathrm{C}=\frac{\pi}{2}$
so, $\quad \Delta=\frac{1}{2} a b \sin C$ $[\because \sin C=1]$
$\Delta=\frac{1}{2} \times(\sqrt{2})(\sqrt{3})(1)$
$\Delta=\frac{\sqrt{6}}{2}$ sq. cm
then, $\quad \angle \mathrm{C}=\frac{\pi}{2}$
so, $\quad \Delta=\frac{1}{2} a b \sin C$ $[\because \sin C=1]$
$\Delta=\frac{1}{2} \times(\sqrt{2})(\sqrt{3})(1)$
$\Delta=\frac{\sqrt{6}}{2}$ sq. cm
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