Math, asked by azar129th, 7 months ago

A
ladder
is placed against
wall such
that its foot is at a distance of 8 m from the wall
And its top reaches a window 6 m above the ground
Find the length of the ladder

Answers

Answered by ItzAditt007
55

Answer:-

The Length Of The Ladder Is 10 m.

Explanation:-

Given:-

  • Distance between the foot of the ladder and the wall (AB) = 8 m.

  • Height of the window (AC) = 6 m.

ToFind:-

  • The length of ladder.

Theorem Used:-

  • Pythagoras Theorem:- In a right angle triangle Square of Hypotenuse is equal to the sum of the square of its base and height.

\\ \large\bf\longrightarrow H^2 = P^2 + B^2.

Where,

  • H = Hypotenuse.
  • P = Perpendicular I.e. height.
  • B = Base.

So Here,

  • H = Length Of ladder = ?? [To Find].

  • P = AC = 6 m.

  • B = AB = 8 m.

Now,

By Applying Pythagoras Theorem,

\\ \bf\mapsto H^2 = P^2 + B^2.

\\ \tt\mapsto H {}^{2}  = (6 \: m) {}^{2} + (8 \: m) {}^{2}  .

\\ \tt\mapsto H =  \sqrt{36 \: m {}^{2}  + 64 \: m {}^{2} } .

\\ \tt\mapsto H =  \sqrt{100 \: m {}^{2} }

\\ \tt\mapsto H =  \sqrt{10 \times 10}  \:  \: m .

\\  \large\bf\mapsto H = 10 \: m.

Therefore The Length Of The Ladder Is H = 10m.

Attachments:
Answered by Anonymous
43

AnsWer :

\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{P}}\put(7.7,1){\large\sf{Q}}\put(10.6,1){\large\sf{R}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.1,2){\sf{\large{8 m}}}\put(9,0.7){\sf{\large{6 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}

  • PR is the ladder.

  • QR is the distance between base and ladder.

  • PQ is height of the window.

\bigstar \: \underline{\sf Now,by  \: using \:  Phythagoras \:  theorem  \: we \:  get : } \\

:\implies \sf (Hypotenuse)^{2}  = (Perpendicular)^{2}   +  (Base)^{2}  \\  \\

:\implies \sf (PR)^2 = (PQ)^2  +  (QR)^2 \\  \\

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

:\implies \sf PR^2 = 36  +  64 \\  \\

:\implies \sf PR=  \sqrt{36 + 64} \\  \\

:\implies \sf PR= \sqrt{100}\:m^2  \\  \\

:\implies \underline{ \boxed{ \sf PR= 10 \: meter}} \\

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

Attachments:
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