Question

A tree is broken at a height of 5 m from the ground and its top touches the ground at adistance of 12 m from the base of the tree .Find the original height of the tree.

Answer 1

Answer:-

The Original Height of the Tree is 18 meters.

Explanation:-

Given:-

• A tree has broken from 5 m above the ground (Let us say from point A).

• Its top touches the ground at a distance of 12 m from its base (Let us say point C).

To Find:-

• The original height of the tree.

Theorem used:-

Pythagoras Theorem,

$\large \blue{ \longrightarrow \boxed{ \red{ \bf {h}^{2} = b {}^{2} + {p}^{2}.}}}$

Where,

• h is Hypotenuse.

• b is Base.

• p is Perpendicular.

So Here,

• h = ??

• b = 12 m.

• p = 5m.

Now,

By putting the above values in theorem we get:-

$\tt\longrightarrow {h}^{2} = {b}^{2} + {p}^{2}. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\longrightarrow h = \sqrt{ {b}^{2} + {p}^{2} } . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\longrightarrow h = \sqrt{ (12m) {}^{2} + (5m) {}^{2} } . \\ \\ \tt\longrightarrow h = \sqrt{144 {m}^{2} + 25 {m}^{2} }. \: \: \: \: \\ \\ \tt\longrightarrow h = \sqrt{169m {}^{2} }. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\longrightarrow h = 13 \: m. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore The total height of the tree,

= 13m + 5m.

= 18m.

Answer 2

✔ Question :-

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 height of tree.

✔ Solution :-

By Using Pythagoras Theorem, We can find the height.

$\sf{{(h)}^{2} = {(p)}^{2} + {(b)}^{2}}$

Here,

h = Hypotenuse,

p = perpendicular,

b = base.

We can put values provided.

➝ h² = p² + b²

➝ h² = 12² + 5²

➝ h = √144 + 25

➝ h = √169

➝ h = 13m.

Now, 13 + 5 = 18m.

Hence, the original height of the tree = 18m.