Question

Two poles of height a metre and b metre , are p metre apart. Prove that the height of the point of intersection of lines joining the top of each pole to the foot of the opposite pole is given by ab/a+b metres.

Answer 1

hence proved .please mark as brain list

Answer 2

According to the question, the arrangement of the two poles is given in the image below:

The height of pole AB is a

The height of pole DC is b

The distance between two poles is p

The height to be determined is taken as h

Solution:

In $\Delta A B C$, $A B \| E F$,

So, from the diagram, $\triangle A B C$ is similar to $\triangle E F C$,  

$\Rightarrow \frac{A B}{B C}=\frac{E F}{F C}$

$\Rightarrow \frac{A B}{E F}=\frac{B C}{F C}$

$\Rightarrow \frac{a}{h}=\frac{p}{F C}$

$\therefore F C=\frac{p h}{a} \rightarrow(1)$

Now, in $\Delta E F C$, $E F \| D C$,

So, from the diagram, $\triangle D C B$ is similar to $\triangle E F B$,

$\Rightarrow \frac{D C}{B C}=\frac{E F}{F B}$

$\Rightarrow \frac{\mathrm{DC}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{FB}}$

$\Rightarrow \frac{\mathrm{b}}{\mathrm{h}}=\frac{\mathrm{p}}{\mathrm{FB}}$

$\therefore \mathrm{FB}=\frac{\mathrm{ph}}{\mathrm{b}} \rightarrow(2)$

Now, on adding equation (1) and equation (2),

$\Rightarrow \mathrm{FC}+\mathrm{FB}=\frac{\mathrm{ph}}{\mathrm{a}}+\frac{\mathrm{ph}}{\mathrm{b}}$

$\because F C+F B$ is the distance between two poles.

$\because F C+F B=p$

$\Rightarrow \mathrm{p}=\frac{\mathrm{ph}}{\mathrm{a}}+\frac{\mathrm{ph}}{\mathrm{b}}$

$\Rightarrow \mathrm{p}=p h\left[\frac{1}{a}+\frac{1}{b}\right]$

$\Rightarrow \mathrm{p}=\mathrm{ph}\left[\frac{\mathrm{a}+\mathrm{b}}{\mathrm{ab}}\right]$

$\therefore \mathrm{h}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}$

Hence proved.