Question

two poles of height am and b m are p m apart. prove that the height of point of intersection of the line joining the top of each pole to the foot of the opposote pole is given by. ab/a+b m

Answer 1

Answer:

Proven that height of point of intersection of the line joining the top of each pole to the foot of the opposite pole is given by $\frac{ab}{a+b} m$

Step-by-step explanation:

In the diagram attached:

BC = a meters,  DE = b meters, CE = p meters

Need to prove that FG $=\frac{ab}{a+b}$ meters

ΔEBC and ΔEFG are right angled triangles and ∠E is common.

∴ They are similar triangles.

$\frac{FG}{a} = \frac{GE}{p}$  ................ equation (i)

Similarly, ΔCDE and ΔCFG are right angled triangles and ∠C is common.

∴ They are similar triangles.

$\frac{FG}{b} = \frac{CG}{p}$$=\frac{p - GE}{p}$$=1 - \frac{GE}{p}$ $=1-\frac{FG}{a}$ [from equation (i)]

=>$FG(\frac{1}{b} + \frac{1}{a} ) = 1$

=>$FG = \frac{1}{\frac{a + b}{ab} }$$=\frac{ab}{a+b}$

Proven that height of point of intersection of the line joining the top of each pole to the foot of the opposite pole is given by $\frac{ab}{a+b} m$