Question

Two poles of heights 3m and 2m are erected upright on the ground and ropes are stretched from the top of each to the foot of the other. At what height above the ground do the ropes cross each other

Answer 1

Solution:-
Let the hight of the two poles be 'a' and 'b' meters respectively and let they be 'p' meters apart.
Height of the pole AB = a meter
Height of the pole CD = b meter
Distance between the poles = p meters
Let the point of intersection of lines joining the top of the poles be 'E' and its height be 'h' meters. Suppose BF = x meter.
Draw EF ⊥ BC
In Δ ABC and Δ EFC,
∠ ACB = ∠ ECF      (Common)
∠ ABC = ∠ EFC      (90°)
∴ Δ ABC ~ Δ EFC   (AA similarity)
⇒ AB/EF = AC/EC = BC/FC  (Corresponding sides are proportional)
⇒ a/h = p/(p - x)
⇒ ap - ax = ph  .....(1)
Similarly,  Δ DCB ~ Δ EFB
⇒ b/h = p/x
⇒ x = ph/b  ......(2)
From (1) and (2), we get
ap - aph/b = ph
⇒ h(1 + a/b) = a
⇒ h{(a + b)/b} = a
⇒ h = ab/(a + b)
Or, h = pq/(p + q)
Hence proved.


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