Question
In a $\triangle\text{ABC},$ if $\text{a}=\sqrt{2},\text{b}=\sqrt{3}$ and $\text{c}=\sqrt{5},$ show that its area is $\frac{1}{2}\sqrt{6 }\text{ sq. units.}$
✓
Answer
The area of a triangle ABC is given by
$\triangle=\frac{1}{2}\text{ab}\sin\text{C}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
$=\frac{2+3-5}{2\sqrt{6}}$
$=0$
$\sin\text{C}=\sqrt{1-\cos^2\text{C}}$
$=1$
Therefore,
$\triangle=\frac{1}{2}\text{ab}\sin\text{C}$
$\frac{1}{2}\sqrt{6}$
$\triangle=\frac{1}{2}\text{ab}\sin\text{C}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
$=\frac{2+3-5}{2\sqrt{6}}$
$=0$
$\sin\text{C}=\sqrt{1-\cos^2\text{C}}$
$=1$
Therefore,
$\triangle=\frac{1}{2}\text{ab}\sin\text{C}$
$\frac{1}{2}\sqrt{6}$
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