Geometry-6 (XAT-2010)

Two poles of height 2 meters and 3 meters are 5 meters apart. The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is,

A. 1.2 meters

B. 1.0 meters

C. 5.0 meters

D. 3.0 meters

E. None of the above

Answer follows here:

Solution:

Refer to the above diagram,

Length of pole AB = 2 m; Length of pole CD = 3 m;

Distance between the poles = BD =5 m;

We need to find CF. Let CF = h;

Let BC=x and CD=y;

Hence, x+y = 5                       ----(1)

The only formula we use here is area of triangle = ½ (base) (height)

∆BDE = ½ (5*3) = 15/2;        ∆ABF = ½ (x*2) = x;  

∆ADB = ½ (5*2) = 5;             ∆DEF = ½ (y*3) = 3y/2;

From the diagram, it is evident that,

∆BDE+∆ADB = ∆ABF + ∆DEF + 2∆BDF

=> 15/2 + 5 = x + 3y/2 + 2∆BDF

=> 25 = 2x+3y+4∆BDF => ∆BDF = (25-2x-3y)/4    -------(2)

It is also evident that,

∆BDE-∆ADB =∆DEF - ∆ABF

=> 15/2 - 5 = 3y/2 – x => 3y-2x = 5                        ----(3)

Solving (1) and (3), we get x =2 and y=3;

Substituting x and y in (2),

∆BDF = (25-4-9)/4 = 12/4 = 3

=> ½ (5*h) = 3 =>h = 6/5 = 1.2

Answer (A)