Question

Two poles of height p and q metresare standing vertically on a level ground , a metre 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 pq/(p+q)

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.

Answer 2

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