Question

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

Answer 1

the given information in the diagramatical form forms a right angled triangle

with base as 2.5 m and height as 6 m ladder will be like hypotenuse

hyp²= base²+height ²

hyp²= (2.5)²+ 6²
hyp² = 6.25 + 36
hyp² = 42.25
hyp= 6.5m

so the height of the ladder is 6.5 m


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Answer 2

Step-by-step explanation:

Step-by-step explanation:

Answer:

PR is the ladder.

QR is the distance between base and ladder.

PQ is height of the window.

$\begin{gathered}\bigstar \: \underline{\sf Now,by \: using \: Phythagoras \: theorem \: we \: get : } \\\end{gathered}$

★

Now,by using Phythagoras theorem we get:

$\begin{gathered}:\implies \sf (Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2} \\ \\\end{gathered}$

:⟹(Hypotenuse)

2

=(Perpendicular)

2

+(Base)

2

$\begin{gathered}:\implies \sf (PR)^2 = (PQ)^2 + (QR)^2 \\ \\\end{gathered}$

:⟹(PR) 2=(PQ) 2 +(QR) 2

$\begin{gathered}:\implies \sf PR^2 = (6 \: m)^{2} + (8 \: m)^{2} \\ \\\end{gathered}$

:⟹PR 2 =(6m) 2 +(8m) 2

$\begin{gathered}:\implies \sf PR^2 = 36 + 64 \\ \\\end{gathered}$

:⟹PR

2

=36+64

$\begin{gathered}:\implies \sf PR= \sqrt{36 + 64} \\ \\\end{gathered}$

:⟹PR=

36+64

$\begin{gathered}:\implies \sf PR= \sqrt{100 \: m} \\ \\\end{gathered}$

:⟹PR=

100m

$\begin{gathered}:\implies \underline{ \boxed{ \sf PR= 10 \: meter}} \\\end{gathered}$

:⟹

PR=10meter

$\begin{gathered}\therefore\:\underline{\textsf{Length of the ladder is \textbf{10 meter}}}. \\\end{gathered}$

∴

Length of the ladder is 10 meter

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