Find the area of $△ A B C$ , in terms of $x$ given that $A B = x$ , $∠ A B C = 60^{o}$ and $A C ⊥ B C$

\frac{x^{2} \sqrt{3}}{8}

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\frac{x^{2} \sqrt{3}}{4}

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\frac{x^{2} \sqrt{3}}{2}

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x^{2} \sqrt{3}

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Solution

The correct option is A $\frac{x^{2} \sqrt{3}}{8}$
Since $A C ⊥ B C, ∠ A C B = 90^{o}$
$A C = x sin (60^{o}) = \frac{x \sqrt{3}}{2}$
$B C = x cos (60^{o}) = \frac{x}{2}$
Area $= \frac{1}{2} \times A C \times B C$
$= \frac{1}{2} \times \frac{x}{2} \times \frac{x \sqrt{3}}{2}$
$= \frac{x^{2} \sqrt{3}}{8}$

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