Though the centroid of an equilateral triangle, a line parallel to the base is drawn' on this line' an arbituary point P is taken inside the triangle. Iet h denote the perpendicular distance of P from the base of the triangle. Let h₁ and h₂ perpendicular distance of P from the other two sides of the triangle. Then

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$h = \frac{h_{1} + h_{2}}{2}$

$h = \frac{2 h_{1} h_{2}}{h_{1} + h_{2}}$

$h = \sqrt{h_{1} h_{2}}$

$h = \frac{(h_{1} + h_{2}) \sqrt{3}}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} h = \frac{a}{2 \sqrt{3}} \\ \begin{aligned} A r (Δ A B C) & = Ar (Δ A P B) + Ar (Δ B P C) + Ar (Δ A P C) \\ \frac{\sqrt{3}}{4} a^{2} & = \frac{1}{2} a (h + h_{1} + h_{2}) \Rightarrow h_{1} + h_{2} = \frac{a}{\sqrt{3}} \end{aligned} \end{array}$

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