Question

A tree is broken at a height of 3 m from the ground and its top touches the ground at

a distance of 4 m from the base of the tree. Find the original height of the tree.​

Answer 1

Answer:

In the given figure, BC represents the unbroken part of the tree. Point C represents the point where the tree broke and CA represents the broken part of the tree. Triangle ABC, thus formed, is right-angled at B.

Applying Pythagoras theorem in ΔABC,

AC^2 = BC^2 + AB^2

AC^2 = (3 m)^2 + (4 m)^2

AC^2 = 9 m^2 + 16 m^2 = 25 m^2

AC = 5 m

Thus, original height of the tree = AC + CB = 5 m + 3 m = 8 m

Step-by-step explanation:

hope it helps you

Answer 2

Answer:

let, AC=?, AB=3m, BC=4m

Step-by-step explanation:

by using pythagorus theorem, In ∆ABC

(AC)^2=(AB)^2+(BC)^2

(AC) ^2=(3)^2+(4)^2

(AC) ^2=9+16

(AC) ^2=25

AC=√25m

AC=5m

hence, the total height of the tree is AB+AC

3+5

8m.