Question

Two poles of height a metres and b metres are P metres apart. Prove that the height h of the point of intersection N of the lines joining the top of each pole to the foot of the opposite pole is ab/a+b metres

Answer 1

$h = \frac{ab}{a + b}$ (Proved)

Step-by-step explanation:

See the attached diagram.

Let the two poles are AB (a meters) and CD (b meters) ate placed P distance apart i.e. AC = P meters.

Now, the lines AD and BC meet at point F and EF is the height of the point (h meters).

Now, considering Δ ABC, since AB ║ EF and Δ ABC and Δ CEF are similar, so

$\frac{EF}{AB} = \frac{EC}{AC}$

⇒ $\frac{h}{a} = \frac{y}{P}$ ............ (1)

Again, considering Δ ACD, since CD ║ EF and Δ ACD and Δ AEF are similar, so

$\frac{EF}{CD} = \frac{AE}{AC}$

⇒  $\frac{h}{b} = \frac{x}{P}$ .............. (2)

Now, adding equations (!) and (2) we get,

$(\frac{h}{a} + \frac{h}{b}) = \frac{x + y}{P}$

⇒ $\frac{h(a + b)}{ab} = 1$

⇒ $h = \frac{ab}{a + b}$ (Proved)