Question

From a point $p$, inside of an equilateral triangle $ABC$, the perpendicular distances of the three sides are $6\sqrt{3}cm,9\sqrt{3}cm$ and $12\sqrt{2}cm$, respectively. Find the semi perimeter of the triangle.

• A. $81cm$
• B. $54cm$
• C. $108cm$
• D. $162cm$
• E. $noneofthese$

Answer 1

The correct option is A $81cm$

Given $6\sqrt{3}cm,9\sqrt{3}cm,12\sqrt{3}cm$ are the three perpendiculars from the point $O$ which lies inside the equilateral triangle.

Let $P_1=6\sqrt{3}cm,P_2=9\sqrt{3}cm$; and $P_3=12\sqrt{3}cm$.

If we consider the triangles $APB,APC,BPC$, we can see that the 3 triangles will be equal to total area of triangle $ABC$
Area $\left(△ABC\right)=$Area$\left(△APB\right)$+Area$\left(△APC\right)$+Area$\left(△BPC\right)$
$\Rightarrow \frac{\sqrt{3}}{4}\cdot a^2=\frac{1}{2}(9\sqrt{3}+12\sqrt{3}+6\sqrt{3})\times a$

$\Rightarrow \frac{\sqrt{3}}{2}.a=(6+9+12)\sqrt{3}cm$

$\Rightarrow \sqrt{3}a=27\sqrt{3}\times 2$

$\Rightarrow a=54cm$.

Now perimeter of equilateral triangle.

$=3a$

$=3\times 54$

$=162cm$

$\therefore$ In this case

Semi-perimeter of equilateral triangle.

$=\frac{1}{2}\times 162$

$=81cm$