Question

Two poles of height 7m and 12m stand on a plane ground . If the distance between the feet of a poles be 12m . Then find the distance between their tops

Answer 1

notes on the figure
1. the left pole equals to 12 m(5+7)
2. the right pols equal to 7m
3. the lower distance between them is given that is 12m
4. a triangle is formed above the rectangle formed by them

the triangle ABC is a right-angled triangle

by Pythagoras 
(ac)^2=(ab)^2+(bc)^2
ac^2=5^2+12^2
ac^2=25+144
ac^2=169
ac=root 169
ac=13m

their distance is 13m

sorry for this horrible drawing but I had no choice on my PC.

Answer 2

T
Let the he height of first tower =12m
height of second tower =5m
the distance between the two tower =12m
then, the length of first tower=(7+5)m the length of second tower. The right triangleABC can be constructed using the length of the two tower.
AB =Base of the two tower =12m
BC=height of the first tower w.r.t second tower AC=distance between the top of the two tower in the form of hypotenuse of ABC
using pythaguras theorem
AC^2=AB^2+BC^2
AC^2=5x5+12x12
AC =square root of (169)
AC =13 metres
Hence the distance between the two tower =13m