Two vertical poles of height 20 m and 24 m are standing on the opposite end of a road. Two ropes are tied to the top of each pole with the foot of the other pole.

Assuming that there is no slack in the two ropes, how high is the point of intersection of the two ropes from the ground?

Answer :

We have two poles 20 m and 24 m high . And the height of point of intersection of lines joining the top of each pole to the foot of other pole is a m .

we form our diagram , As :

Here Poles AB  =  24 m and CD  = 20 cm and Height of point of intersection FE  =  a 

And AB | |  CD  | | FE  , As AB and CD are poles attached to a horizontal plane , and FE also lies on same plane .

So, From Basic proportionality theorem , We get

In $∆$ ABC   , We know AB  | | FE , So we get

$$\begin{gathered} \frac{AC}{AB} = \frac{FC}{FE} \\ \Rightarrow \frac{AC}{24} = \frac{FC}{a} \\ \Rightarrow AC = \frac{24FC}{a} -- ( 1 ) \end{gathered}$$

And from In $∆$ CAD   , We know CD  | | FE , So we get

$$\begin{gathered} \frac{AC}{CD} = \frac{AF}{FE} \\ \Rightarrow \frac{AC}{20} = \frac{AC - FC}{a} \\ \Rightarrow a = \frac{20AC - 20FC}{AC} \\ \Rightarrow a = 20 - \frac{20FC}{AC} , Substitute value from eqiation 1 , we get \\ \Rightarrow a = 20 - \frac{20FC}{{\displaystyle \frac{24FC}{a}}} \\ \Rightarrow a = 20 - \frac{5}{{\displaystyle \frac{6}{a}}} \\ \Rightarrow a = 20 - \frac{a5}{{\displaystyle 6}} \\ \Rightarrow a + \frac{5a}{{\displaystyle 6}}= 20 \\ \Rightarrow \frac{11a}{{\displaystyle 6}}= 20 \\ \Rightarrow 11a= 120 \\ \mathbf{\Rightarrow }\mathit{a}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{10}\mathbf{.}\mathbf{9}\mathbf{ }\mathit{m}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)} \end{gathered}$$