Question

A ladder 10cm long reaches a window 8cm above the ground. Find the distance of the foot of the ladder from base of the wall.​
(plz. answer stepwise)​

Answer 1

Hope this helps you
I think it is stepwise

Answer 2

Given:-

• A ladder 10cm long reaches a window 8cm above the ground.

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To Find:-

• Find the distance of the foot of the ladder from base of the wall.

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Concept:-

• Let's go through the concept first. Concept used here is Pythagoras Theorem.

• It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

• Substitute the values in the equation and solve it.Let's do it up.

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Formulae Applied:-

$\Large\boxed{\bold{Pythagoras \: Theorem \: = AC^{2} \: = \: (AB)^{2} + (BC)^{2}}}$

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Solution:-

Given that,

A ladder 10cm long reaches a window 8cm above the ground.

Let AC is the foot of the ladder.

Let AB is the distance of window above the ground

Such that,

AC = 10cm

AB = 8cm

BC = ??

According to the question we have,

$\Large\boxed{\bold{Pythagoras \: Theorem \: = AC^{2} \: = \: (AB)^{2} + (BC)^{2}}}$

${\longrightarrow\bold{AC^{2} \: = \:(AB)^{2} + (BC)^{2}}}$

${\longrightarrow\bold{10^{2} \: = \:(8)^{2} + (BC)^{2}}}$

${\longrightarrow\bold{BC^{2} \: = \:(8)^{2} - (10)^{2}}}$

${\longrightarrow\bold{BC^{2} \: = \: {64 - 100}}}$

${\longrightarrow\bold{BC^{2} \: = \: {36}}}$

${\longrightarrow\bold{BC = \: { \sqrt{36}}}}$

${\longrightarrow\bold{BC = \: {6}}}$

${\longrightarrow\bold{BC = \: {6cm}}}$

Hence,

The distance of the foot of the ladder from the base of the wall is 6cm.

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Verification:-

$\longrightarrow\bold{BC^{2} \: = \:(8)^{2} - (10)^{2}}$

$\longrightarrow\bold{BC = 6cm}$

${\longrightarrow\bold{6^{2} \: = \:(8)^{2} - (10)^{2}}}$

${\longrightarrow\bold{36 \: = \:64 - 100}}$

${\longrightarrow\bold{36 \: = \:36}}$

${\longrightarrow\bold{LHS = RHS}}$

Hence,

It is verified.

[Note:- Triangle's perimeter/area can't be expressed in negative terms.]