Question

Two poles of heights $6m$ and $11m$ stand on a plane ground. If the distance between their feet is $12m$, find the distance between their tops.

Answer 1

Let us draw a diagram from the given information.

Let us draw a perpendicular from $B$ on $CD$ which meet $CD$ at $P$.

It is clear that $BP=12m$ because it is given that the distance between feet of the two poles is $12m$.

After drawing the perpendicular we get a rectangle $BACP$ such that $AB=PC=6m$ and
$BP=AC=12m$.

Because of this construction, we also obtained a right angled triangle $\Delta BPD$.

Now we will use Pythagoras theorem,

$B{D}^2=B{P}^2+P{D}^2$

Let us substitute the values of $BP$ and $PD$ we get,

$(BD{)}^2={12}^2+5^2[\because PD=11-6=5m$]
$\Rightarrow (BD{)}^2=144+25$
$\Rightarrow (BD{)}^2=169$
$\Rightarrow BD=\sqrt{\left(168\right)}$
$\therefore BD=13m$

Hence, the distance between the top of the two poles is $13m$.