Let p, q and r be the altitudes of a triangle | vedclass

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(B) Let the sides of the triangle be $a, b, c$. The area $S$ of the triangle is given by:$S = \frac{1}{2} a p = \frac{1}{2} b q = \frac{1}{2} c r$From

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$\frac{t}{S}$

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(B) Let the sides of the triangle be $a, b, c$. The area $S$ of the triangle is given by:$S = \frac{1}{2} a p = \frac{1}{2} b q = \frac{1}{2} c r$From this, we can express the reciprocals of the altitudes as:$\frac{1}{p} = \frac{a}{2S}, \frac{1}{q} = \frac{b}{2S}, \frac{1}{r} = \frac{c}{2S}$Adding these expressions, we get:$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{a+b+c}{2S}$Since the perimeter is $2t = a+b+c$, we substitute this into the equation:$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{2t}{2S} = \frac{t}{S}$

Let $p, q$ and $r$ be the altitudes of a triangle with area $S$ and perimeter $2t$. Then, the value of $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ is

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