Question

a tree is broken at a height is 5 m from the group and it's top touches the ground at a distance of 12m from the base if the tree
find the original height of tree .
did anyone know the ans plz tell me ​

Answer 1

Answer:

17m

Step-by-step explanation:

12 + 5 = 17cm

∴ The height of the orginal tree is 17cm.

Nice question...Hope it helps...

Answer 2

$\sf{\blue{\underline{\blue{\huge{Correct\:Question}}}}}$

A tree is broken at a height of 5 m from the ground and it's top touches the ground at a distance of 13 m from the base of the tree. Find the original height of the tree.

Given:-

• Height of the broken tree = 5 m
• Distance of it's top touching ground from it's base = 12 m

To find:-

Original height of the tree

Solution:-

Refer to the attachment for the figure of this question.

From the figure we can see,

Base (i.e., AB) = 12 m

Perpendicular (i.e., AC) = 5 m

We need to find

Hypotenuse (i.e., AB)

According to the Pythagoras Theorem,

$\sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}$

= $\sf{Hypotenuse = \sqrt{(Base)^2 + (Height)^2}}$

= $\sf{Hypotenuse = \sqrt{(12)^2+(5)^2}}$

= $\sf{Hypotenuse = \sqrt{144+25}}$

= $\sf{Hypotenuse = \sqrt{169}}$

= $\sf{Hypotenuse = 13\:m}$

Therefore the length of the broken part of the tree that touches the ground is 13 m

Original length of the tree:-

Height of the part of the tree standing on the ground + length of the broken part of the tree touching the ground

= 13 + 5 m

= 18 m

$\sf{\therefore}$ The original height of the tree is 18 m

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Important:-

Pythagoras theorem states that the sum of square of base and square of perpendicular equals to the square of hypotenuse.

$\sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}$

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