Question

In given figure two poles of height a metres and b metres are p metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by $\frac{ab}{a+b}$ meters.

Answer 1

Let $AB$ and $CD$ be two poles of heights $a$ metres and $b$ metres respectively such that the poles are $p$ metres apart i.e.$AC=p$ metres. Suppose the lines $AD$ and $BC$ meet at $O$ such that $OL=h$ metres.

$LetCL=xandLA=y.Then,x+y=p.$

In $△ABC$ and $△LOC$, we have

$\angle CAB=\angle CLO$ [Each equal to ${90}^{\circ }$]

$\angle C=\angle C$ [Common]

$\therefore$ $△CAB\sim △CLO$ [By AA-criterion of similarity]

$\Rightarrow$ $\frac{CA}{CL}=\frac{AB}{LO}$

$\Rightarrow$ $\frac{p}{x}=\frac{a}{h}$

$\Rightarrow$ $x=\frac{ph}{a}$...........(i)

In $△ALO$ and $△ACD$, we have

$\angle ALO=\angle ACD$ [Each equal to ${90}^{\circ }$]

$\angle A=\angle A$ [Common]

$\therefore$ $△ALO\sim △ACD$ [By AA-criterion of similarity]

$\Rightarrow$ $\frac{AL}{AC}=\frac{OL}{DC}$

$\Rightarrow$ $\frac{y}{p}=\frac{h}{b}$

$\Rightarrow$ $y=\frac{ph}{b}$ [$\because$ $AC=x+y=p$]........(ii)

From (i) and (ii), we have

$x+y=\frac{ph}{a}+\frac{ph}{b}$

$\Rightarrow$ $p=ph(\frac{1}{a}+\frac{1}{b})$ [$\because$ $x+y=p$]

$\Rightarrow$ $1=h\left(\frac{a+b}{ab}\right)$

$\Rightarrow$ $h=\frac{ab}{a+b}metres$

Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is $\frac{ab}{a+b}$ metres.