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

From the above figure, we observe that the height of the remaining part of tree$=5m$
i.e., $AB=5m$
The distance between the base at the point B of the tree and the point A where the top of the tree fallen is $=12m$
i.e., $AB=12m$
In $\Delta ABC$ is a right angled triangle.
Using Pythagoras theorem,
$\Rightarrow (AC{)}^2=(AB{)}^2+(BC{)}^2$
$\Rightarrow (AC{)}^2=5^2+{12}^2$
$=25+144$
$\Rightarrow AC=\sqrt{169}$
$\therefore AC=13$
Hence, the original height of the tree is $=A^{\prime }C+CB=13+5=18m$