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 = \(AC + CB = 13\) \(m + 5 \) \(m\) 
= \(18\) \(m\)