Question

3. 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

Step-by-step explanation:

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 ΔABC is a right angled triangle, right angled at B.

AB=12m and BC=5m

Using Pythagoras theorem, In ΔABC

(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

Answer:

18 m

Step-by-step explanation:

let the broken point be A

let the point where its top touches the ground at 12 m be B

let the base of tree be C

AB=?

BC=12 m

AC =5 m

By pythagorus theorm

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

= 5^2 + 12^2

= 25+144

=169

So AB =13 m

Now total length of tree is the length from base to broken part+ broken part to the point it touches the ground

i.e .- AB+ AC=13+5=18m

Thanks