Question : From any point inside an equilateral triangle, the lengths of perpendiculars on the sides are $a$ cm, $b$ cm, and $c$ cm. Its area (in cm2) is:
Option 1: $\frac{\sqrt{2}}{3}(a+b+c)$
Option 2: $\frac{\sqrt{3}}{3}(a+b+c)^{2}$
Option 3: $\frac{\sqrt{3}}{3}(a+b+c)$
Option 4: $\frac{\sqrt{2}}{3}(a+b+c)^{2}$
Correct Answer: $\frac{\sqrt{3}}{3}(a+b+c)^{2}$
Solution :
OD = $a$ cm, OE = $b$ cm, and OF = $c$ cm
BC = AC = AB
Area of $\triangle ABC$ = Area of $(\triangle BOC + \triangle COE + \triangle BOA)$
= $\frac{1}{2}\times BC \times a+\frac{1}{2}\times AC\times b+\frac{1}{2}\times AB\times c$
= $\frac{1}{2}\times BC \times (a+b+c)$.....................(1)
Again,
Area of $\triangle ABC=\frac{\sqrt3}{4}BC^2$...................(2)
From equation (1) and (2), we get:
⇒ $\frac{1}{2}\times BC \times (a+b+c)=\frac{\sqrt3}{4}BC^2$
⇒ $BC=\frac{2}{\sqrt3}(a+b+c)$
Required area = $\frac{\sqrt3}{4}\times BC^2$
= $\frac{\sqrt3}{4}\times (\frac{2}{\sqrt3}(a+b+c))^2$
= $\frac{\sqrt3}{4}\times \frac{4}{3}(a+b+c)^2$
= $\frac{\sqrt3}{3}(a+b+c)^2$
Hence, the correct answer is $\frac{\sqrt3}{3}(a+b+c)^2$.
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