Two vertical poles AB = 15 m and CD = 10 m | vedclass

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(D) Let the horizontal distance between the poles be $x$. Let $h$ be the height of the intersection point $P$ above the ground $AC$.Let the foot of th

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$6$

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(D) Let the horizontal distance between the poles be $x$. Let $h$ be the height of the intersection point $P$ above the ground $AC$.Let the foot of the perpendicular from $P$ to $AC$ be $M$. Let $AM = x_2$ and $MC = x_1$,so $x_1 + x_2 = x$.In $	riangle AMC$ and $	riangle BCD$,we have $	riangle PMC \sim 	riangle ABC$ and $	riangle PMA \sim 	riangle ADC$.From similarity,$\frac{h}{15} = \frac{x_1}{x}$ and $\frac{h}{10} = \frac{x_2}{x}$.Adding these two equations: $\frac{h}{15} + \frac{h}{10} = \frac{x_1 + x_2}{x} = \frac{x}{x} = 1$.$\frac{2h + 3h}{30} = 1$ $\Rightarrow \frac{5h}{30} = 1$ $\Rightarrow \frac{h}{6} = 1$.Therefore,$h = 6 \ m$.

Two vertical poles $AB = 15 \ m$ and $CD = 10 \ m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $BC$ and $AD$,then the height of $P$ (in $m$) above the line $AC$ is:

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