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If the base of an Isosceles triangle is of length 2p and the length of altitude dropped to the base is ‘q’ then the distance from the midpoint of the base to the side of the triangle is

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$\frac{2 p q}{\sqrt{p^{2} + q^{2}}}$

$\frac{4 p q}{\sqrt{p^{2} + q^{2}}}$

$\frac{p q}{\sqrt{p^{2} + q^{2}}}$

$\frac{3 p q}{\sqrt{p^{2} + q^{2}}}$

answer is A.

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Detailed Solution

$\sin θ = \frac{x}{p} \Rightarrow x = p . \sin θ$

Area of $Δ A B C = \frac{1}{2} (2 p) q$

$p q = \frac{1}{2} (2 p) \sqrt{p^{2} + q^{2}} \sin θ$

$p q = \sqrt{p^{2} + q^{2}} . x$

$x = \frac{p q}{\sqrt{p^{2} + q^{2}}}$

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