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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

$=AC+CB$

$=13\ m+5\ m=18\ m$

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