Question

5. 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.​

Answer 1

Answer:

let A'CB represents the tree before it broken at the point C and let the top A touches the ground at A After it broke.then ABC is a right angled triangle right angled at B .

AB =12m and BC=5m

using pythagoras theoram

(AC)²+(AB)²+(BC)²

(AC)² = (12)²+(5)²

(AC)²=144+25

(AC)²=169

AC =13m

hence the total height of the tree = AC+CB=13+5=18m

Answer 2

Let a triangle ABC in which

AB is hypotenus of triangle/broken part of tree=y(let)

AC is perpendicular of triangle/ height from ground=5m

AB is base of triangle/distance of bending part of tree from base of tree=12m

in triangle ABC,by pythagorus theorem

(AB)²=(AC)²+(BC)²

(y)²=(5)²+(12)²

y²=25+144=169

$y = \sqrt{169} = 13$

original height of tree=AC+AB

=5+13

=18m