Question

A point O is taken inside an equilateral ∆ABC. If OL ⊥ BC, OM ⊥ AC and ON ⊥ AB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of ∆ABC.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{each}\mathrm{side}\mathrm{of}△ABCbex\mathrm{cm}. \\ \therefore ar(∆ABC)=ar(∆BOC)+\mathrm{ar}(∆AOC)+\mathrm{ar}(∆AOB) \\ =\frac{1}{2}\times BC\times OL+\frac{1}{2}\times AC\times OM+\frac{1}{2}\times AB\times ON \\ =\frac{1}{2}\times x\times 14+\frac{1}{2}\times x\times 10+\frac{1}{2}\times x\times 6 \\ =\frac{1}{2}x\left(14+10+6\right) \\ =\frac{1}{2}x\times 30 \\ =15x{\mathrm{cm}}^2 \\ ar(∆ABC)=15x{\mathrm{cm}}^2 \\ \mathrm{or},\frac{\sqrt{3}}{4}{x}^2=15x \\ \mathrm{or},\sqrt{3}{x}^2=60x \\ \mathrm{or},\sqrt{3}{x}^2-60x=0 \\ \mathrm{or},\sqrt{3}x(x-20\sqrt{3})=0 \\ \mathrm{or},\sqrt{3}x=0and(x-20\sqrt{3})=0 \\ \mathrm{or},x=0(\mathrm{Side}\mathrm{cannnot}\mathrm{be}0.) \\ \mathrm{and},x=20\sqrt{3} \\ \therefore \mathrm{ar}(∆ABC)=15x=15\times 20\sqrt{3}=300\sqrt{3}{\mathrm{cm}}^2 \end{gathered}$$