Question

From a point in the interior of an equilateral triangle perpendicular drawn to three sides are 8 cm, 10 cm and 11 cm respectively. Find the area of triangle the nearest cm.

Answer 1

We know area of equilateral triangle
$=\frac{\sqrt{3}}{4}{a}^2$ (a is side of triangle)
Also $ar\left(\Delta ABC\right)$
$=ar\left(\Delta OAB\right)+ar\left(\Delta OBC\right)+ar\left(\Delta OAC\right)$
$=\frac{1}{2}a[8+10+11]=\frac{29}{2}a$
Equating area
$\therefore \frac{\sqrt{3}}{\text{/}{4}_2}{a}^2=\frac{29}{\text{/}2}a$
$\Rightarrow \frac{\sqrt{3}}{2}{a}^2-29a=0$
$\Rightarrow a(\frac{\sqrt{3}}{2}a-29)=0\Rightarrow a=0$
$a=\frac{58}{\sqrt{3}}$
$a\ne 0\therefore a=\frac{58}{\sqrt{3}}$
Area $=\frac{\text{/}\sqrt{3}}{\text{/}{4}_{\text{/}2}}\times \frac{\text{/}{58}^{29}\times \text{/}{58}^{29}}{\text{/}3\sqrt{3}}=\frac{641}{\sqrt{3}}c{m}^2$