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Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is
A. 5
B. 4
C. 7
D. 6

Last updated date: 17th Jul 2024
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Hint: Use the basic proportionality theorem to show that if two triangles are similar, their ratio of opposite sides will also be comparable.In order to solve this first we have considered the figure as shown below then we have equated their alternate side ratio of the triangles formed in order to find the value of a as represented in the figure.

 Complete step-by-step solution:

Let us consider two triangles ABC and ADC intersecting at P. And the perpendicular from P to AC is drawn. Let the Distance between AC is x and the perpendicular divides it in ratio ${x_1}:{x_2}$
Avoid mixing up the comparable triangle with the congruent triangle.
It is said that two or more figures are comparable if they have the same shape but not necessarily the same size. However, two or more figures are considered to be congruent if they have the same size and shape.
According to initial data we have AB = 15 m and CD = 10 m
From theory of basic proportionality, we have
$\tan {\theta _1} = \dfrac{{10}}{x} = \dfrac{H}{{{x_1}}}$
$ \Rightarrow {x_1} = \dfrac{{Hx}}{{10}}$
$\tan {\theta _2} = \dfrac{{15}}{x} = \dfrac{H}{{{x_2}}}$
$ \Rightarrow {x_2} = \dfrac{{Hx}}{{15}}$
Since ${x_1} + {x_2} = x$
$ \Rightarrow {x_1} + {x_2} = x = \dfrac{{Hx}}{{10}} + \dfrac{{Hx}}{{15}}$
$ \Rightarrow 1 = H(\dfrac{1}{{10}} + \dfrac{1}{{15}})$
$ \Rightarrow H = \dfrac{{10 \times 15}}{{10 + 15}}$
$ \Rightarrow H = \dfrac{{150}}{{25}} = 6m$
 Hence option D is correct.

 Note: For the height and distance questions just stick with the basic trigonometric formulas and similar triangle properties. Make sure you make a correct diagram and try to find the solution in the diagram itself.