Q.5 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.

Let AB be the original height of the tree and broken at C touching the ground at D such that
AC = 5 m
and AD = 12 m
In right triangle \(\triangle \)CAD,
\(AD^2 + AC^2 = CD^2 \) (By Pythagoras property)
Or, \( (12)^2 + (5)^2 = CD^2 \)
Or, 144 + 25 =\( CD^2 \) Or, 169 = \( CD^2\)
\(\therefore \)CD =\(\sqrt{169} \) = 13 m
But CD = BC
AC + CB = AB
5 m + 13 m = AB
\(\therefore \) AB = 18 m .
Thus, the original height of the tree = 18 m.