Question

Two poles of height $6\mathrm{m}$ and $11\mathrm{m}$ stands on a plane ground, if the distance between their feet is $12\mathrm{m}$. Find the distance between their tops.

Answer 1

Find the required distance:

Let $\mathrm{AB}$ be the pole of height $11\mathrm{m}$, $\mathrm{CD}$ be the pole of height $6\mathrm{m}$ and $\mathrm{BD}$ be the distance between their feet is $12\mathrm{m}$

Draw a perpendicular from $\mathrm{C}$ to $\mathrm{AB}$ which meets $\mathrm{AB}$ at $\mathrm{E}$.

Here, $\mathrm{CD}$$=\mathrm{EB}=6\mathrm{m}$ and $\mathrm{BD}$$=\mathrm{EC}=12\mathrm{m}$

Now from the figure, $\mathrm{AE}=\mathrm{AB}-\mathrm{EB}$

$\Rightarrow$ $\mathrm{AE}=11-6$

$\Rightarrow$ $\mathrm{AE}=5\mathrm{m}$

On using Pythagoras Theorem in $△\mathrm{AEC}$, we get;

${\left(\mathrm{AC}\right)}^2={\left(\mathrm{AE}\right)}^2+{\left(\mathrm{EC}\right)}^2$ [Square of hypotenuse is equal to sum of the square of other two sides]

$\Rightarrow$${\left(\mathrm{AC}\right)}^2={\left(5\right)}^2+{\left(12\right)}^2$

$\Rightarrow$${\left(\mathrm{AC}\right)}^2=25+144$

$\Rightarrow$${\left(\mathrm{AC}\right)}^2=169$

$\Rightarrow$ $\mathrm{AC}=13\mathrm{m}$ [Taking square root on both sides]

Hence, the distance between their tops is $13\mathrm{m}$