Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Given:
Two poles of heights 6 m and 11 m stand on a plain ground.
The distance between their feet is 12 m.
To do:
We have to find the distance between their tops.
Solution:
Let $DE$ and $AC$ be the poles of heights $6\ m$ and $11\ m$ on the ground.
The distance between their feet $EC= 12\ m$
This implies,
$DB=12\ m$
In right-angled triangle $ABD$,
By Pythagoras theorem,
$AB^2 + BD^2 = AD^2$
$5^2 + 12^2 = AD^2$ ($AB=AC-BC=11-6=5\ m$)
$AD^2 = 144 + 25$
$AD^2= 169$
$AD = \sqrt{169}=13$
Therefore, the distance between their tops is $13\ m$.
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