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A tree is broken at a height of $5\ m$ from the ground and its top touches the ground at a distance of $12\ m$ from the base of the tree. Find the original height of the tree.
Given: A tree is broken at a height of $5\ m$ from the ground and its top touches the ground at a distance of $12\ m$ from the base of the tree.
To do: To find the original height of the tree.
Solution:
Let $ACB$ represent the tree before it breaks at the point $C$, and let the top $A$ touch and the ground at $A$ after it before.
Then $\Delta ABC$ is a right-angle triangle, right angles at $B$.
$AB=12\ m$
$BC=5\ m$
On using Pythagoras theorem,
$(AC)^2=(AB)^2+(BC)^2$
$\Rightarrow (AC)^2=(12)^2+(15)^2$
$\Rightarrow (AC)^{2\ }=144+25$
$\Rightarrow (AC)^2=169$
$AC=13\ m$
Thus, the total height of the tree
$=AC+CB$
$=13\ m+5\ m=18\ m$
Updated on: 10-Oct-2022
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