question_answer

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:

Let ACB be the tree before it broken at the point C, and let its top A touch the ground at A? after it broke. Then, ∆ A?BC is a right triangle, right-angled at B such that A?B = 12 m, BC = 5 m. By Pythagoras theorem, we have \[{{\left( AC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}\] \[{{\left( AC \right)}^{2}}={{12}^{2}}+{{5}^{2}}\] = 144 + 25 = 169 \[AC=\sqrt{169}=13\] AC = A?C = 13 m \[\left[ \because AC=AC \right]\] \[\therefore \]      AB = AC + BC = (13 + 5) m = 18 m