Question

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. \(\frac{\sqrt{3}s^2}{2}\)
• B. \(\frac{s^2}{\sqrt{3}}\)
• C. \(\frac{2s^2}{\sqrt3}\)
• D. \(\frac{s^2}{2\sqrt3}\)

Answer 1

The Correct Option is B

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

\(⇒ AB=\frac{2}{\sqrt 3}s\)

Required value = \(\frac{1}{2}×\frac{2}{\sqrt3}×s^2 = \frac{s^2}{\sqrt3}\)

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

Answer 2

From any point on a equilateral triangle, if we draw 3 perpendiculars on other sides, those 3 perpendicular = Altitude of an equilateral triangle
We take the point to be A, and try to draw 3 perpendiculars and is same as altitude of the equilateral triangle.
So, \(s = \frac{√3}{2} a\) 
⇒ \(a = \frac{2s}{√3}\)
By substituting the value in Area of an equilateral triangle \(\frac{√3}{4}a^2\)
\(\frac{√3}{4}\times(\frac{2s}{√3})^2=\frac{s^2}{√3}\)

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