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# CAT 2020 Quant Question [Slot 2] with Solution 24

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

1. $\frac{{\sqrt 3 {s^2}}}{2}$
2. $\frac{{{s^2}}}{{\sqrt 3 }}$
3. $\frac{{2{s^2}}}{{\sqrt 3 }}$
4. $\frac{{{s^2}}}{{2\sqrt 3 }}$
Option: 2
Solution:

PD + PE + PF = s

Area of

$= \frac{1}{2} \times AB \times PE + \frac{1}{2} \times BC \times PD + \frac{1}{2} \times AC \times PF$

As $AB = BC = CA,$ we've

$= \frac{1}{2} \times AB(PD + PE + PF) = \frac{1}{2}AB \times s - (1)$

Now $\frac{{\sqrt 3 }}{4}A{B^2} = \frac{1}{2}AB \times s$

$\Rightarrow AB = \frac{2}{{\sqrt 3 }}s$

Required value $= \frac{1}{2} \times \frac{2}{{\sqrt 3 }} \times {s^2} = \frac{{{s^2}}}{{\sqrt 3 }}$

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