Two poles of heights $6\text{m}$ and $11\text{m}$ stand vertically on a plane ground. If the distance between their feet is $12\text{m}$, what is the distance between their tops?

Let the two vertical poles be $AB=6\text{m}$ and $CD=11\text{m}$. The distance between their feet is $BD=12\text{m}$.
Draw a line parallel to the ground from the top of the shorter pole ($A$) to meet the taller pole ($CD$) at point $E$.

This forms a right-angled triangle $\Delta AEC$ at $E$, where:

* The base $AE=BD=12\text{m}$
* The height $CE=CD-ED=CD-AB=11-6=5\text{m}$

Using the Pythagoras theorem to find the distance between their tops ($AC$):

$$A{C}^2=A{E}^2+C{E}^2$$

$$A{C}^2=12^2+5^2=144+25=169$$

$$AC=\sqrt{169}=13\text{m}$$