Question

In the figure above, $AB=6$ and $BC=8$. Calculate the area of triangle $ABC$.

• A. $12\sqrt{2}$
• B. $12\sqrt{3}$
• C. $24\sqrt{2}$
• D. $24\sqrt{3}$
• E. $36\sqrt{3}$

Answer 1

The correct option is B $12\sqrt{3}$

Draw a perpendicular line from point $A$ to the line $BC$ turning $AB$ as hypotenuse and $BC$ as base.
$\sin {60}^0=\frac{AP}{AB}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{AP}{6}$
$\Rightarrow AP=3\sqrt{3}$
So, if the base of the triangle is segment $BC$ and height $AP$, then the area(A) of triangle $ABC$;
$A=\frac{1}{2}\times BC\times AP$
As, $AP=3\sqrt{3}$
$A=\frac{1}{2}\times 8\times 3\sqrt{3}=12\sqrt{3}$
Hence, option B is correct.