Question

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.

Answer 1

Let $A^{\prime }CB$ represents the tree before it broke at the point $C$ and let the top $A^{\prime }$ touches the ground at $A$ after it broke.
Then $\Delta ABC$ is a right-angled triangle, right angled at $B.$
$AB=12m$ and $BC=5m$

Using Pythagoras theorem, In $\Delta ABC$

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

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

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

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

$\Rightarrow AC=13m$

And, $A^{\prime }C=AC$

Hence, the total height of the tree $=AC+CB=13+5=18m.$