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

In the given figure, BC represents the unbroken part of the tree.
Point C represents the point where the tree broke and CA represents the broken part of the tree.
Triangle ABC, thus formed, is right-angled at B.
Applying Pythagoras theorem in Δ ABC,
$AC^2= BC^2+ AB^2$
$AC^2= (5 m)^2+ (12 m)^2$
$AC^2= 25 m^2+ 144 m^2= 169 m^2$
$AC = 13$ $m$

Thus, original height of the tree = $AC + CB = 13$ $m + 5$ $m$
= $18$ $m$