Question

a ladder is placed against a wall such that its foot is at a distance of 2.5 metre from the world and its top reaches a window 6 metre above the ground find the length of the ladder.​

Answer 1

Answer:

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Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• A ladder is placed against a wall such that its foot is at a distance of 2.5 metre from the world and its top reaches a window 6 metre above the ground find the length of the ladder.

$\large\underline{ \underline{ \sf \maltese\red{ \: Diagram:- }}}$

• Diagram is in the attachment

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

• Perpendicular = 6m
• Base = 2.5 m
• Hypotenuse or Ladder = ?

$\large\underline{ \underline{ \sf \maltese\green{ \: Solution:- }}}$

Let AB be the ladder and CA be the wall with the window at A .

• BC = 2.5m
• CA = 6m

$\underbrace\red{\boxed{ \sf \blue{ By \: using \: Pythagoras \: Theorem \: }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: (Hypotenuse)^{2} \: = \: (Perpendicular)^{2} \: + \: (base) ^{2} }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ (AB)^{2} \: = \: (BC)^{2} \: + \: (CA) ^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ (AB)^{2} \: = \: (2.5)^{2} \: + \: (6) ^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ (AB)^{2} \: = \: 6.25 \: + \: 36}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ (AB)^{2} \: = \: 42.25}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ (AB) \: = \: \sqrt{42.25} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ (AB) \: = \: 6.5 }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{ \:AB\: = \: 6.5m }}}$

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$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Length\: of \: Ladder = \underline {\underline{ 6.5m}}}\\\end{gathered}\end{gathered}$

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