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05 Dec 2014, 07:35
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D
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Difficulty:
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Question Stats:
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Tough and Tricky questions: Geometry.
A professional painter is painting the window frames of an old Victorian House. The worker has a ladder that is exactly 25 feet in length which he will use to paint two sets of window frames. To reach the first window frame, he places the ladder so that it rests against the side of the house at a point exactly 15 feet above the ground. When he finishes, he proceeds to reposition the ladder to reach the second window so that now the ladder rests against the side of the house at a point exactly 24 feet above the ground. How much closer to the base of the house has the bottom of the ladder now been moved?
A) 7
B) 9
C) 10
D) 13
E) 27
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Source: Chili Hot GMAT
intheend14
GMAT 1: 740 Q49 V41
Posts: 125
05 Dec 2014, 11:10
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Yay for Pythagorean Theorem. This is a double problem involving that concept. In both cases, we have a ladder that is 25ft and is leaned against a wall. So, this is always the hypotenuse for our right triangle. The right triangle, by the way, involves the ladder (hypotenuse), the wall and the ground (the two legs).
In the first case, we're leaning the ladder against a window that is 15 feet up. This means we should have a 3-4-5 right triangle, with the ladder 20 ft from the house wall. In the second case, we're leaning the ladder against a window that is 24 feet up, meaning we have a 7-24-25 triangle. So, the ladder moved from 20 ft out to 7 ft out--a total of 13 ft.
Choice D
In the first case, we're leaning the ladder against a window that is 15 feet up. This means we should have a 3-4-5 right triangle, with the ladder 20 ft from the house wall. In the second case, we're leaning the ladder against a window that is 24 feet up, meaning we have a 7-24-25 triangle. So, the ladder moved from 20 ft out to 7 ft out--a total of 13 ft.
Choice D
05 Dec 2014, 13:28
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A professional painter is painting the window frames of an old Victorian House. The worker has a ladder that is exactly 25 feet in length which he will use to paint two sets of window frames. To reach the first window frame, he places the ladder so that it rests against the side of the house at a point exactly 15 feet above the ground. When he finishes, he proceeds to reposition the ladder to reach the second window so that now the ladder rests against the side of the house at a point exactly 24 feet above the ground. How much closer to the base of the house has the bottom of the ladder now been moved?
A) 7
B) 9
C) 10
D) 13
E) 27
In order to reach the first and second window frame, the painter forms right angle triangle with hypotenuse as the length of the ladder that is 25 feet in length.
The height of the right angle triangle to reach first window = 15 feet
Therefore the distance from base of the house to the bottom of the ladder is \(\sqrt{25^2-15^2}\) \(= 20\)feet
The height of the right angle triangle to reach the second window= 24 feet
Therefore the distance from the base of the house to the bottom of the ladder =\(\sqrt{25^2 - 24^2}=7\) feet
Hence, the ladder was moved 20 feet - 7 feet = 13 feet closer to the base of the house to reach second window.
Answer: D
A) 7
B) 9
C) 10
D) 13
E) 27
In order to reach the first and second window frame, the painter forms right angle triangle with hypotenuse as the length of the ladder that is 25 feet in length.
The height of the right angle triangle to reach first window = 15 feet
Therefore the distance from base of the house to the bottom of the ladder is \(\sqrt{25^2-15^2}\) \(= 20\)feet
The height of the right angle triangle to reach the second window= 24 feet
Therefore the distance from the base of the house to the bottom of the ladder =\(\sqrt{25^2 - 24^2}=7\) feet
Hence, the ladder was moved 20 feet - 7 feet = 13 feet closer to the base of the house to reach second window.
Answer: D
