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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink]
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pranab01 wrote:

SO the distance from the base to the point the ladder touches can be calculated by prognathous theorem = \(\sqrt{10^2 - 7^2} = \sqrt{51}\)


Please fix the typo
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink]
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considering the original condition, a 10 ft. ladder resting on a wall with its top 8 ft. above the ground. A right-angle triangle with a hypotenuse of 10 ft. (length of ladder) and height of 8 ft. can be drawn, by applying Pythagoras theorem we can derive the base of the triangle as 6 ft.
After the ladder has shifted by 1 ft. away from the wall (as shown), we can conclude that the top of the ladder has also slid down the wall (lets assume that distance as "d"). In the new triangle, the hypotenuse remains 10 ft., the base becomes (6+1) = 7 ft. and the height will be now (8-d) ft. Again on applying the Pythagoras theorem, we can calculate d=~0.86 ft., which is less than 1.

Hence, column (A) is greater than (B). Right answer is (A)
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink]
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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A 10-foot ladder is leaning against a vertical wall. The top [#permalink]
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