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 poles to the foot of the opposite pole is,

To find : $$EF = x = ?$$

Solution : In $$\triangle$$ ABC and $$\triangle$$ EFC

$$\angle ACB = \angle ECF$$  (common)

$$\angle ABC = \angle EFC = 90$$

=> $$\triangle ABC \sim \triangle EFC$$

=> $$\frac{x}{2} = \frac{5 - d}{5}$$ ----------Eqn(I)

Similarly, $$\triangle BCD \sim \triangle BFE$$

=> $$\frac{x}{3} = \frac{d}{5}$$ --------Eqn(II)

Adding Eqns (I) & (II), we get :

=> $$\frac{x}{2} + \frac{x}{3} = \frac{5 - d}{5} + \frac{d}{5}$$

=> $$\frac{3x + 2x}{6} = \frac{5}{5}$$

=> $$5x = 6 \times 1 = 6$$ 

=> $$x = \frac{6}{5} = 1.2 m$$