Solution :
Let AB and CD be two poles.
AB = 11 m, CD = 6 m. Draw a line CE || to BD.
In \(\triangle\) AEC,
\({AC}^2\) = \({CE}^2\) + \({AE}^2\) = \((12)^2\) + \((11 – 6)^2\) = 144 + 25 = 169
\(\implies\) AC = 13 m
Hence, distance between their tops is 13 m.