From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10
cm and 6 cm. Find the area of the triangle.
Thinking Process
(i) From a interior point a triangle, three triangle will be formed Determine area of each triangle by using the formula =½ (Base x Height).
(ii) Determine the area of an equilateral triangle by using the Heron’s formula i.e.,
$\sqrt{s(s-a)(s-b)(s-c)}$
(iii) Further, equate the area of triangle by Heron’s formula and the sum of all three triangle. Simplify it to get the side of D. Hence, find area of the triangle.
Let sides of an equilateral triangle be a $\mathrm{m}$.
Area of $\triangle O A B=\frac{1}{2} \times A B \times O P \quad\left[\because\right.$ area of a triangle $=\frac{1}{2}$ (base $\times$ height) $]$
$=\frac{1}{2} \times a \times 14=7 a \mathrm{~cm}^{2}$ $\ldots(i)$
Area of $\Delta O B C=\frac{1}{2} \times B C \times O Q$
$=\frac{1}{2} \times a \times 10=5 a \mathrm{~cm}^{2}$ $\ldots$ (ii)
Area of $\Delta O A C=\frac{1}{2} \times A C \times O R$
$=\frac{1}{2} \times a \times 6=3 a \mathrm{~cm}^{2}$ .....(iii)
$\therefore$ Area of an equilateral $\triangle A B C=$ Area of $(\triangle O A B+\triangle O B C+\triangle O A C)$
$=(7 a+5 a+3 a)=15 a \mathrm{~cm}^{2}$ ...(iv)
We have, semi-perimeter $s=\frac{a+a+a}{2} \Rightarrow s=\frac{3 a}{2} \mathrm{~cm}$
$\therefore$ Area of an equilateral $\triangle A B C=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]
$=\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}$
$=\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}=\frac{\sqrt{3}}{4} a^{2}$ $\ldots(v)$
From Eqs. (iv) and (v).
$\frac{\sqrt{3}}{4} a^{2}=15 a$
$\Rightarrow$ $a=\frac{15 \times 4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{60 \sqrt{3}}{3}=20 \sqrt{3} \mathrm{~cm}$
On putting $a=20 \sqrt{3}$ in Eq. (v), we get
Area of $\triangle A B C=\frac{\sqrt{3}}{4}(20 \sqrt{3})^{2}$
$=\frac{\sqrt{3}}{4} \times 400 \times 3$
$=300 \sqrt{3} \mathrm{~cm}^{2}$
Hence, the area of an equilateral triangle is $300 \sqrt{3} \mathrm{~cm}^{2}$.