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A 10-foot ladder is leaning against a vertical wall. The top
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21 Sep 2018, 09:01
21 Sep 2018, 09:01
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#GREpracticequestion A 10-foot ladder is leaning against a vertical wall.jpg [ 6.55 KiB | Viewed 14538 times ]
A 10-foot ladder is leaning against a vertical wall. The top of the ladder touches the wall at a point 8 feet above the ground. The base of the ladder slips 1 foot away from the wall.
Quantity A |
Quantity B |
1 |
The distance the top of the ladder slides down the wall |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: A 10-foot ladder is leaning against a vertical wall. The top
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22 Sep 2018, 04:33
22 Sep 2018, 04:33
2
Carcass wrote:
Attachment:
wall.jpg
A 10-foot ladder is leaning against a vertical wall. The top of the ladder touches the wall at a point 8 feet above the ground. The base of the ladder slips 1 foot away from the wall.
Quantity A |
Quantity B |
1 |
The distance the top of the ladder slides down the wall |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
1st case:
Length of the ladder = 10 foot
Distance from the ground to point the ladder touches = 8 feet
So the distance of the base of the ladder from the wall =\(\sqrt{10^2 - 8^2} = 6\)
2nd case:
Now the distance of the base slips 1 foot away from the wall i.e = 6 + 1 =7
and the length of the ladder = 10 foot
SO the distance from the base to the point the ladder touches can be calculated by Pythagoras theorem = \(\sqrt{10^2 - 7^2} = \sqrt{51}\)
Now \(\sqrt{49} < \sqrt{51} < \sqrt {64}\)
or \(7 < \sqrt{51} < 8\)
so the new value of the base lies between 7 and 8
so the distance the ladder slides down = 8 - (number greater than 7 but less than 8) = less than 1
SO \(QTYA > QTY B\)
Re: A 10-foot ladder is leaning against a vertical wall. The top
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18 Jan 2019, 16:10
18 Jan 2019, 16:10
By Pythagoras theorem
Ans B
Ans B
