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Quantitative Aptitude Question on Triangles, Circles & Quadrilaterals

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is

A

3s22\frac{\sqrt{3}s^2}{2}

B

s23\frac{s^2}{\sqrt{3}}

C

2s23\frac{2s^2}{\sqrt3}

D

s223\frac{s^2}{2\sqrt3}

Answer

s23\frac{s^2}{\sqrt{3}}

Explanation

Solution

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides

PD+PE+PF=sPD + PE + PF = s
Area =12×AB×PE+12×BC×PD+12×AC×PF\frac{1}{2}×AB×PE+\frac{1}{2}×BC×PD+\frac{1}{2}×AC×PF
As AB=BC=CAAB=BC=CA, we've
=12×AB(PD+PE+PF)=12  AB×s−(1)\frac{1}{2}×AB(PD+PE+PF)=\frac{1}{2}\;AB×s-(1)
Now 34AB2=12AB×s\frac{\sqrt{3}}{4} AB^2=\frac{1}{2}AB×s

⇒AB=23s⇒ AB=\frac{2}{\sqrt 3}s

Required value = 12×23×s2=s23\frac{1}{2}×\frac{2}{\sqrt3}×s^2 = \frac{s^2}{\sqrt3}

So, The correct option is (B): s23\frac{s^2}{\sqrt{3}}