Question

a tree is broken at a height of 6 metre from the ground and its top touches the ground at a distance of 8 metres from the base of the tree find the original height of the tree question of 7th standard​

Answer 1

Answer :

• Height of the tree is 16m

Given :

• A tree is broken at a height of 6 metre from the ground
• Its tops touches the ground at a distance of 8m from the base of tree

To find :

• original Height of the tree

Solution :

Given that tree is broken so,

• Let the part after breaking down be AB
• Distance of its tops from the base of tree be BC

Tree vertical is ∠b = 90⁰

Then ,

• AB = 6m
• BC = 8m

In ABC

As we know that by Pythagoras theorem

● (AC)² = (AB)² + (BC)²

● (AC)² = (6)² + (8)²

● (AC)² = 36 + 64

● (AC)² = 100

● AC = √100

● AC = 10

So Now we Have to find the original height of the tree

● AB + AC

● 6 + 10

● 16 m

Hence , original Height of the tree is 16m

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

${\underline{\underbrace{\tt{Understanding\:the\:concept}}}}$

In this question, A tree is broken at a height of 6m from the ground. It's top touches the ground at distance of 8m from the base of the tree.

Means, the base is 8m and perpendicular is 6m.

we have to find first the Hypotenuse. And then find the actual height of the tree.

◉ TO FIND:-

• Original height of tree = DB.

◉ GIVEN:-

• AC = AD

• Base = 8m = BC

• height = 6m = AB

◉ Formula used:-

• pythagoras theorem.

$\bold{ {p}^{2} + {b}^{2} = { \: \:H}^{2} }$

◉ Solution:-

In ∆ ABC,

$: \longrightarrow \bold{ {6}^{2} + {8}^{2} = { H }^{2} }$

$: \longrightarrow \bold {36 + 64 = {H }^{2}}$

$: \longrightarrow \bold{H \: = \sqrt{100} }$

$: \longrightarrow \bold \pink{ \:H = 10m} = AC$

As, AC = AD=10m

We have to find DB

So, we will add,

DB = AD + AB

$\bold{DB = 6m + 10m}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \longmapsto \bold {\tt {\purple{DB = 16m}}}$

So, The original height of tree is 16m.