Question

a tree is broken by the wind but doesn't separate.if the point from where it breaks is 9 m. above the ground and its top touches the ground at a distance of 12 m from its foot. find out the total height of the tree before it broke. with diagram???? don't try to copy from gogle copied answer will be deleted ​

Answer 1

Answer:

The total height is 24 cm

Step-by-step explanation:

by phithogoras uppopado:

√{(standing part^2)+( distance from foot^2)

=√{(9^2)+(12^2)

=15

now,

total height= broken part + standing part

= 15 + 9 = 24

Answer 2

Correct Question-:

• A tree is broken by the wind but doesn't separate. If the point from where it breaks is 9 m. Above the ground and its top touches the ground at a distance of 12 m from its foot. Find out the total height of the tree before it broke.

AnswEr-:

• $\underline{\boxed{\star{\sf{\blue{ Total \:Height \:of \:Tree\: = 24 m.}}}}}$

Explanation:

$\frak{By\:Analysing\: Figure-:} \begin{cases} \sf{The\:broken \:tree\:forms\:a\:Right \:Angled \:Triangle \: ABC} & \\\\ \sf{AB \: = \: Height \: of\:the\:point \:from\:where\:the\:tree\:brokes \: }& \\\\ \sf{or\:Perpendicular_{(Triangle)}= \frak{9 \ m}} & \\\\ \sf{ BC = \: Distance \: of\:root\:from\:where\: the\: tree \: touches \: ground \: to\: broken \: part\:or\:Base_{(Triangle)}\:= \frak{12m}} & \\\\ \sf{ AC\: =Hypotenuse \:of\:Right \:Angled\:Triangle \: or\: the\:broken\:part\:of\:tree\:} \end{cases}$

$\frak{To\:\: Find-:} \begin{cases} \sf{The\:total\:Height\:of\:tree.} \end{cases}$

Now ,

By Applying Pythagoras Theorem-:

$\underline{\boxed{\star{\sf{\blue{ Pythagoras\: Theorem\: = \: Perpendicular² + Base² = Hypotenuse ².}}}}}$

• $\frak{Here\:\: -:} \begin{cases} \sf{AB = \:Perpendicular.} & \\\\ \sf{BC= Base}& \\\\ \sf{AC=Hypotenuse}\end{cases}$

Now ,

• $\underline{\boxed{\star{\sf{\blue{ Pythagoras\: Theorem\: = \: AB² + BC² = AC ².}}}}}$

• $\frak{Here\:\: -:} \begin{cases} \sf{AB = \:Perpendicular.=\: 9 \:m} & \\\\ \sf{BC= Base\:=\:12m}& \\\\ \sf{AC=Hypotenuse=??}\end{cases}$

Now ,

• $\implies{\sf{\large { AB²\: =\: 12² + 9²}}}$
• $\implies{\sf{\large { AB²\: =\: 144 + 81}}}$
• $\implies{\sf{\large { AB²\: =\: 225m}}}$
• $\implies{\sf{\large { AB\: =\: \sqrt{225}}}}$
• $\implies{\sf{\large { AB\: =\: 15m}}}$

Therefore,

• $\underline{\boxed{\star{\sf{\blue{ Hypotenuse\:or\:AB \: = 15 m.}}}}}$

Now ,

• $\underline{\boxed{\star{\sf{\blue{ Total\: Height \:of\:Tree\: = \: AB + AC \:or\:Perpendicular\:+\:Hypotenuse.}}}}}$
• $\frak{Here\:\: -:} \begin{cases} \sf{AB = \:Perpendicular.=\: 9 \:m} & \\\\ \sf{AC=Hypotenuse=15m}\end{cases}$

Now ,

• $\implies{\sf{\large { 15 + 9 =\: 24m}}}$

Hence ,

• $\underline{\boxed{\star{\sf{\blue{ Total \:Height \:of \:Tree\: = 24 m.}}}}}$

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