Question

A tree is broken at a height of 9 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

Let A’CB be the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then ΔABC is a right angled triangle, at B.

AB = 12 m and BC = 9 m

Using Pythagoras theorem,

In ΔABC

(AC)²=(AB)²+(BC)²

(AC)²=(12)²+(9)²

(AC)²=144+81

(AC)²=225

AC = √225

AC= 15 m

Hence, the total height of the tree(A’B) = A’C + CB = 15 + 9 = 24 m.