Question

The heights of two vertical poles are 36m and 28m and the shortest distance btw the tops is 17m. Find how far they are apart.

Answer 1

Heights of two vertical poles = 36 m & 28m
From the shorter pole construct a line parallel to ground upto longer pole.
Now, p will be = 8 m 
and h given is 17 m
From pythagoras th. 
b^2=h^2-p^2
⇒b^2=(17m)^2 - (8m)^2
⇒b^2=289m^2 - 64m^2
⇒b^2=225 m^2
⇒b= root under (225 m^2)
⇒b=15 m
Hence, distance between two poles is 15 m. (Ans)

Answer 2

let AB be the first pole and CD be the second
AB=36m and CD=28m
let E be the point where a line parallel to the ground meets AB
we have to find CE
AC= hypotenuse of the right ΔACE=17m
and AE= 36-28=8m
in ΔACE, By Pythagoras Theorem,
$AC^{2}= AE^{2} + CE^{2}$
$CE^{2}= 17^{2} - 8^{2}$
$CE^{2}=$289-64=225
CE=√225=15m
as therefore they are separated by a distance of 225m