Question

From a point in the interior of an equilateral triangle, perpendiculars drawn to the three sides are $8cm,10cm$ and $11cm$, respectively. Find the area of the triangle to the nearest cm. (Use $\sqrt{3}=1.73$)

Answer 1

In $\Delta ABC,OD\perp BC,OE\perp ACandOF\perp ABOD=8cm,OE=10cm,OF=11cm$
$\Delta$ABC is an equilateral triangle
therefore, let $AB=BC=CA=acm$
now, ar$\Delta ABC=ar\Delta BOC+ar\Delta COA+ar\Delta AOB\frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}\times BC\times OD+\frac{1}{2}\times AC\times OE+\frac{1}{2}\times AB\times OF\frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}\times a\times 8+\frac{1}{2}\times a\times 10+\frac{1}{2}\times a\times 11=\frac{a}{2}[8+10+11]\frac{\sqrt{3}}{4}{a}^2=\frac{29a}{2}a=\frac{29\times 4}{2\sqrt{3}}=\frac{58}{\sqrt{3}}cm\therefore ar\Delta ABC=\frac{\sqrt{3}}{4}\times {\left(\frac{58}{\sqrt{3}}\right)}^2=\frac{\sqrt{3}}{4}\times \frac{58}{3}\times 58=484.96{cm}^2\therefore ar\Delta ABC\approx 485{cm}^2$