Question

Question 3
From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14cm, 10cm and 6cm.Find the area of the triangle.

Answer 1

Let ABC be an equilateral triangle, O be the interior point and AQ, BR and CP are the perpendicular drawn from point O.
Let the each side of an equilateral triangle be m.
$Areaof\Delta OAB=\frac{1}{2}\times AB\times OP[\because areaofatriangle=\frac{1}{2}(base\times height\left)\right]$
$=\frac{1}{2}\times a\times 14=7ac{m}^2\cdots \left(i\right)$
$Areaof\Delta OBC=\frac{1}{2}\times BC\times OQ$
$=\frac{1}{2}\times a\times 10=5ac{m}^2\cdots \left(ii\right)$
$Areaof\Delta OAC=\frac{1}{2}\times AC\times OR$
$=\frac{1}{2}\times a\times 6=3ac{m}^2\cdots \left(iii\right)$

$\therefore$ Area of an equilateral $\Delta ABC,=Areaof(\Delta OAB+\Delta OBA+\Delta OAC)$
$=(7a+5a+3a)=15ac{m}^2\cdots \left(iv\right)$
We have, semi-perimeter $s=\frac{a+a+a}{2}\Rightarrow s=\frac{3a}{2}cm$
$\therefore$ Area of an equilateral $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron’s formula]
$=\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$
$=\sqrt{\frac{3a}{2}\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}}=\frac{\sqrt{3}}{4}{a}^2\cdots \left(v\right)$
From Equations (iv) and (v),
$\frac{\sqrt{3}}{4}{a}^2=15a$
$\Rightarrow a=\frac{15\times 4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{60\sqrt{3}}{3}=20\sqrt{3}cm$
putting $a=20\sqrt{3}$ in Equation (v), we get
$Areaof\Delta ABC=\frac{\sqrt{3}}{4}(20\sqrt{3}{)}^2$
$=\frac{\sqrt{3}}{4}\times 400\times 3$
$=300\sqrt{3}c{m}^2$
Hence, the area of an equilateral triangle is $300\sqrt{3}c{m}^2$