Question

Perpendiculars are drawn on the sides of an equilateral triangle from any point within the triangle. If the lengths of these perpendiculars be $6cm,7cm$ and $9cm$, then the length of the side of the triangle is

• A. $\frac{44}{3}\sqrt{3}cm$
• B. $\frac{11}{3}\sqrt{3}cm$
• C. $\frac{22}{3}\sqrt{3}cm$
• D. $11\sqrt{3}cm$

Answer 1

The correct option is A $\frac{44}{3}\sqrt{3}cm$

Let $△ABC$ be an equilateral triangle.

Length of each side be ${}^{\prime }{a}^{\prime }cm$

$O$ is a point in the interior of the triangle .

$OX\perp BC,OY\perp CA$ and $OZ\perp AB$

Let $OZ=6cm,OX=7cm$ and $OY=9cm$

Area of $△ABC=$ Area of $△AOB$ + Area of $△BOC$+ Area of $△COA$

$\frac{\sqrt{3}}{4}{a}^2=(\frac{1}{2}\times OZ\times AB)+(\frac{1}{2}\times OX\times BC)+(\frac{1}{2}\times OY\times AC)$

$\Rightarrow \frac{\sqrt{3}}{4}{a}^2=(\frac{1}{2}\times 6\times a)+\left(\right(\frac{1}{2}\times 7\times a)+\left(\right(\frac{1}{2}\times 9\times a)$

$\Rightarrow \frac{\sqrt{3}}{4}{a}^2=3a+\frac{7}{2}a+\frac{9}{2}a$

$\Rightarrow \frac{\sqrt{3}}{4}{a}^2=11a$

$\Rightarrow a=\frac{(11\times 4)}{\sqrt{3}}$

$\Rightarrow a=\frac{44}{\sqrt{3}}cm=44\frac{\sqrt{3}}{3}cm.$