Question

A 15 m long ladder reached the window 12 m high from the ground against the wall a. Find the distance of the foot of the ladder from the wall ​

Answer 1

Given:

• A 15 m long ladder reached the window 12 m high from the ground against the a wall.

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To find:

• Distance of the foot of the ladder from the wall?

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Solution:

☯ Let AB be the wall and AC be the ladder and wall.

B is the position of window.

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$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put( - .2,2.5){\large\bf 12 m}\put(3.5,2.8){\large\bf 15 m}\put(2.3,.3){\large\bf distance}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf B}\put(.8,.4){\large\bf C}\put(5.7,.4){\large\bf A}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \theta }\end{picture}$

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Therefore,

• Length of AB = 15 m

• Length of BC = 12 m

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$\underline{\sf{\bigstar\;Using\; Pythagoras\; theorem\;in\;\triangle\;ABC\;:}}$

$:\implies\sf (AB)^2 = (BC)^2 + (AC)^2$

$:\implies\sf (15)^2 = (12)^2 + (AC)^2$

$:\implies\sf (AC)^2 = (15)^2 + (12)^2$

$:\implies\sf (AC)^2 = 225 - 144$

$:\implies\sf (AC)^2 = 81$

$:\implies\sf \sqrt{AC^2} = \sqrt{81}$

$:\implies{\boxed{\frak{\purple{AC = 9\;m}}}}\;\bigstar$

We know that,

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1 meter = 3.2808 foot

9 meter = 9 × 3.2808 = 29.53 foot

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$\therefore$ The distance of ladder from the wall would be 29.53 foot or 9 meter.