Question

A tree is broken at a height of 5 𝑚 from the ground and its top touches the ground at distance of 12 𝑚 from the base of the tree. Find the original height of the tree.

Answer 1

The original length of the tree is equal to the sum of the length of BC and AB.

In ∆𝐴𝐵𝐶

Hypotenuse = AB = ?

Base = AC = 12 m

Perpendicular = BC = 5 cm

Using Pythagoras theorem,

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

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

= 25 + 144

= 169

Hence, AB = $\sqrt{169}$

= 13 m

The original height of the tree = BC + AB

= 5 m + 13 m = 18 m

Therefore, the height of the tree is equal to 18 m.