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.

Make a $∆$ABC and B is a point where the tree is broken from the height of 5 m from the ground.

The top of the tree touches the ground at 12 m from the base of the tree,

By Pythagoras property,

$B{C}^2 = A{B}^2 + A{C}^2$

$B{C}^2 = 5^2 + {12}^2$

$B{C}^2 = 25 + 144$

$B{C}^2 = 169$

$BC =\sqrt{169}$

$BC = 13 m$

So, the original height of the tree $= AB + BC$

$= 5 + 13$

$= 18 m$