Question

the top of a ladder of length 15 mreaches a window 9m above the ground . what is the distance between the base of the wall and that of the ladder​

Answer 1

Given :

• The top of a ladder of length 15m reaches a window 9m above the ground.

To Find :

• The distance between the base of the wall and that of the ladder.

Solution :

✰ In this question, it is given that The top of a ladder of length 15m reaches a window 9m above the ground and we have to find the distance between the base of the wall and that of the ladder. So now the ladder here is nothing but the Hypotenuse (H) and the distance from the ground to the window is nothing but Perpendicular and the distance between the base of the wall and that of the ladder is nothing but Base . So Simply we can apply Pythagoras Theorem to find the Base. Now Pythagoras Theorem states that (Hypotenuse)² = (Perpendicular)² + (Base)² . So let's fill the given values in the formula to find the distance between the base of the wall and that of the ladder.

⠀⠀⠀

⠀⠀⠀⟼⠀⠀⠀(H)² = (P)² + (Base)²

⠀⠀⠀⟼⠀⠀⠀(15)² = (9)² + (Base)²

⠀⠀⠀⟼⠀⠀⠀225 = 81 + (Base)²

⠀⠀⠀⟼⠀⠀⠀225 - 81 = (Base)²

⠀⠀⠀⟼⠀⠀⠀144 = (Base)²

⠀⠀⠀⟼⠀⠀⠀√144 = Base

⠀⠀⠀⟼⠀⠀⠀12m = Base

⠀

Thus The distance between the base of the wall and that of the ladder is 12m

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Answer 2

❥Given:

• Top of a ladder is of length 15m reaches a window 9m above the ground.

❥To find :

• The distance between the base of the wall and that of the ladder.

❥Solution :

As,

$1) \: \boxed{\mathbf{length \: of \: ladder =hypotenuse(h) }}$

$2) \boxed{\mathbf{distance \: from \: ground \: to \: window = perpendicular(p)}}$

$3) \boxed{\mathbf{distance \: between \: the \: base \: of \: the \: wall = base(b)}}$

So in this we can use Pythagoras Theorum , to find answer ...

Formula :

$\begin{gathered}{:} \longrightarrow \color{yellow}{ \boxed{ \color{olive}\sf \: (hypotenous)^{2} \:= \: (perpendicular)^{2} \: + \: (base)^{2} }}\\ \\ \end{gathered}$

A/Q

$\begin{gathered}{:} \longrightarrow \color{magenta}{ \boxed{ \color{teal}\sf \:(h) ^{2} =(p) ^{2} + (b)^{2} \ \: }}\\ \\ \end{gathered}$

$\begin{gathered}{:} \longrightarrow \color{green}{ \boxed{ \color{aqua}\sf \: (15) ^{2} = (9)^{2} + (b) ^{2} \:}}\\ \\ \end{gathered}$

$\begin{gathered}{:} \longrightarrow \color{aqua}{ \boxed{ \color{green}\sf \:225 = \: 81 + (b) ^{2} }}\\ \\ \end{gathered}$

$\begin{gathered}{:} \longrightarrow \color{red}{\boxed{ \color{lightgreen}\sf \: 225 \: - \: 81= \: (b) ^{2} }}\\ \\ \end{gathered}$

$\begin{gathered}{:} \longrightarrow \color{pink}{ \boxed{ \color{yellow}\sf \: 144\: = \: (b) ^{2} }}\\ \\ \end{gathered}$

$\begin{gathered}{:} \longrightarrow \color{orange}{ \boxed{ \color{purple}\sf \: \sqrt{144} \: = \: base}}\\ \\ \end{gathered}$

$\begin{gathered}{:} \longrightarrow \color{green}{ \boxed{ \color{red}\sf \:12m\: = \:base}}✓\\ \\ \end{gathered}$

$\therefore {\color{pink}\pmb{the \: distance \: between \: the \: base \: of \: the \: wall \: is \: }} \boxed{\color{blue} \pmb{12m}}$

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