A $ 8 \mathrm{~m} $ high bamboo tree standing erect on the ground breaks at the height of $ 3 \mathrm{~m} $ from the ground. Its broken part remains attached to the main part. Find distance between the top of the tree and the base of the tree on the ground.
Given:
A \( 8 \mathrm{~m} \) high bamboo tree standing erect on the ground breaks at the height of \( 3 \mathrm{~m} \) from the ground. Its broken part remains attached to the main part.
To do:
We have to find the distance between the top of the tree and the base of the tree on the ground.
Solution:
Let $AB$ be the bamboo and $C$ be the point from where it broke.
This implies,
$CD$ is the broken part.
From the figure,
$\mathrm{CD}=5 \mathrm{~m}, \mathrm{AC}=3 \mathrm{~m}$
In triangle ACD, by Pythagoras theorem,
$CD^2=AC^2+AD^2$
$5^2=3^2+AD^2$
$25-9=AD^2$
$AD=\sqrt{16}$
$AD=4\ m$
Therefore, the distance between the top of the tree and the base of the tree on the ground is $4\ m$.
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