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Question
In \[∆ ABC, if a = \sqrt{2}, b = \sqrt{3} \text{ and } c = \sqrt{5}\] show that its area is \[\frac{1}{2}\sqrt{6} sq .\] units.
Solution
\[\text{ Given }: a = \sqrt{2}, b = \sqrt{3}, c = \sqrt{5}\]
\[ \because \cos C = \frac{a^2 + b^2 - c^2}{2ab}\]
\[ \Rightarrow \cos C = \frac{2 + 3 - 5}{2 \times \sqrt{6}} = 0\]
\[ \Rightarrow \cos C = 0\]
\[ \Rightarrow \cos C = \cos90°\]
\[ \Rightarrow C = 90°\]
\[Thus, \sin C = \sin90°= 1\]
\[\text{ Hence, Area of } ∆ ABC = \frac{1}{2}ab\sin C = \frac{1}{2}\sqrt{6} \times 1 = \frac{\sqrt{6}}{2}sq . \text{ units } . \]
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