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A 25-foot ladder is placed against a vertical wall of a building

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A 25-foot ladder is placed against a vertical wall of a building [#permalink]

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New post 01 Sep 2017, 07:48
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A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out

(A) 4 ft
(B) 5 ft
(C) 6 ft
(D) 7 ft
(E) 8 ft
[Reveal] Spoiler: OA

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Re: A 25-foot ladder is placed against a vertical wall of a building [#permalink]

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New post 01 Sep 2017, 08:25
The ladder when inclined against the wall forms a right triangle - the ladder being the hypotenuse, the vertical wall and ground being the other 2 sides of the right triangle.

the base is given as 7 feet. Using this info we can calculate the vertical height by applying Pythagoras theorem. Therefore height is 24 feet (easier to find this if we know the 25-24-7 right triangle). If the ladder slips by 4 feet the height is now 20 feet and the ladder length is the same (25 feet). the new base can again be calculated by Pythagoras theorem or by simply applying the 3-4-5 multiple concept of right triangle.

thus we can calculate the new base as 15 feet.
Therefore the ladder has slipped by 8 feet (15-7).

Ans is E

Please guide if there is a faster way to solve this.

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Re: A 25-foot ladder is placed against a vertical wall of a building [#permalink]

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New post 01 Sep 2017, 08:30
Bunuel wrote:
A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out

(A) 4 ft
(B) 5 ft
(C) 6 ft
(D) 7 ft
(E) 8 ft


From Pythagoras

\(AC^2=AB^2+BC^2\)

625-49=576

AB=24

From another triangle

\(25^2= 20^2+(7+x)^2\)


\(625=400 + (7+x)^2\)

\((7+x)^2= 225\)

x=8
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Solution.jpg [ 18.38 KiB | Viewed 441 times ]


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A 25-foot ladder is placed against a vertical wall of a building [#permalink]

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New post 01 Sep 2017, 08:32
mahu101, No shorter method to solve it!

The ladder forms a right angled triangle with the wall.
The length of the ladder is the hypotenuse(25 feet)
Since the distance from the wall is 7 feet, the distance
from the base of the building can be calculated using the Pythagorean theorem.
Let this distance be x.
\(x^2 = 625 - 49 = 576\)
\(x = 24\)

Since the ladder is going down by 4 feet, the ladder is 20 feet from base of the building.
Let the distance from the wall be y.
Using the same property of the right triangle,
\(y^2 = 625 - 400 =225\)
\(y = 15\)

From 7 feet, the ladder moves a further 8 feet to move a total of 15 feet
Hence, the distance is 8 feet(Option E)
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Re: A 25-foot ladder is placed against a vertical wall of a building [#permalink]

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New post 01 Sep 2017, 10:18
Thank you pushpitkc.

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Re: A 25-foot ladder is placed against a vertical wall of a building   [#permalink] 01 Sep 2017, 10:18
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A 25-foot ladder is placed against a vertical wall of a building

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