Question

a tree is broken at a height of 6m from the ground and its top touches the ground at a distance of 8m from the base of the tree.find the original height of the tree.

Answer 1

${h}^{2} = {p}^{2} + {b}^{2} \\ {h}^{2} = {8}^{2} + {6}^{2} \\ {h}^{2} = 64 + 36 \\ {h}^{2} = 100 \\ h = \sqrt{100} \\ h = 10$
Original height=10+6
10m

Answer 2

Answer:

16 m

Step-by-step explanation:

We are given that a tree is broken at a height of 6 m from the ground

So, Right triangle is formed

So. Perpendicular = 6 m

Its top touches the ground at a distance of 8 m from the base of the tree.

So, Base = 8 m

$Hypotenuse^2 = Perpendicular^2+Base^2$

$Hypotenuse^2 = 6^2+8^2$

$Hypotenuse = \sqrt{6^2+8^2}$

$Hypotenuse =10$

Original height = 10+6 = 16 m

Hence the original height of tree is 16 m