Question

In $∆ABC,\mathrm{if}a=\sqrt{2},b=\sqrt{3}\mathrm{and}c=\sqrt{5}$, show that its area is $\frac{1}{2}\sqrt{6}\mathrm{sq}.$ units.

Answer 1

$$\begin{gathered} \text{Given:}a=\sqrt{2},b=\sqrt{3},c=\sqrt{5} \\ \because \cos C=\frac{a^2+b^2-c^2}{2ab} \\ \Rightarrow \cos C=\frac{2+3-5}{2\times \sqrt{6}}=0 \\ \Rightarrow \cos C=0 \\ \Rightarrow \cos C=\cos 90° \\ \Rightarrow C=90° \\ \mathrm{Thus},\sin C=\sin 90°=1 \\ \mathrm{Hence},\mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}ab\sin C=\frac{1}{2}\sqrt{6}\times 1=\frac{\sqrt{6}}{2}\mathrm{sq}.\mathrm{units}. \end{gathered}$$