Question

A tree is broken at a height of 4m from ground Its top touches the ground at a distance of 3m from the base of the tree find the original height of the tree​

Answer 1

Answer:

Step-by-step explanation:

its simple

form a rightangled triangle take sides down part of tree=4m

another side=3m

and upper part is nothing but hypotenuse

=$\sqrt{3^2+4^2}$=5m

total height=upperpart+downpart

5m+3m=8m

consider when tree is broken it is a is perpendicular

Answer 2

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= 3m and BC = 4m

Using Pythagoras theorem,
In ΔABC

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

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

(AC)^2 =16 + 9

(AC)^2 =25

AC= 5
​

AC=5m

Hence, the total height of the tree(A’ B)

=A′C + CB = 5+ 4= 9m