Question

A tree is broken at a height of $5m$ from the ground and its top touches the ground at a distance of $12m$ from the base of the tree. Find the original height of the tree.

Answer 1

Let $A^{\prime }CB$ represent the tree before it broke at 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 $\Delta ABC$
$(AC{)}^2=(AB{)}^2+(BC{)}^2$
$\Rightarrow (AC{)}^2=(12{)}^2+(5{)}^2$
$\Rightarrow (AC{)}^2=144+25$
$\Rightarrow (AC{)}^2=169$

$\Rightarrow (AC{)}^2=\sqrt{169}$
$\Rightarrow AC=A^{\prime }C=13m$ [Height cannot be negative so we will take only positive value of square root]
Hence, the total height of the tree $=A^{\prime }C+CB=13+5=18m.$