Question

If the area of $\Delta$ $ABC$ is $8\sqrt{3}$ sq. units, then the length of $AB$ is

• A. $4$ units
• B. $6$ units
• C. $7$ units
• D. $8$ units

Answer 1

The correct option is D $8$ units

Let the length of $AB$ be $x$.

Correspondingly, lengths of $AC$ and $BC$ would be $x\sin \left({60}^o\right)$ and $x\cos \left({60}^o\right)$ respectively.

i.e. $AC=\frac{x\sqrt{3}}{2}$ and $AB=\frac{x}{2}$

The area of the triangle $=\frac{1}{2}\times AC\times BC$

$8\sqrt{3}=\frac{\left(\frac{x\sqrt{3}}{2}\right)\times \left(\frac{x}{2}\right)}{2}$

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

Since the area is given to be $8\sqrt{3}$, we get

$x^2=64$

$\therefore x=8$