CAT 2020 Quant Question [Slot 2] with Solution 24

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 \( = \frac{1}{2} \times \frac{2}{{\sqrt 3 }} \times {s^2} = \frac{{{s^2}}}{{\sqrt 3 }}\)