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Tommieboi918 Just gettin' started!
Joined: 23 Jan 2008 Posts: 4
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Posted: Tue Jul 01, 2008 8:23 am Post subject: Can any one work this out and give a explaination? |
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A rectangilar wooden crate has inside dimensions 3 meters by 4 meters by 12 meters. What is the length, in meters, of the longest, straight, inflexible rod of negligible diameter that can be placed completely within the crate?
a) 12
b) 12.6
c) 13
d) 19
e) 24 |
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somail Just gettin' started!
Joined: 01 Jul 2008 Posts: 22
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Location: San Diego
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Posted: Tue Jul 01, 2008 1:08 pm Post subject: |
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This is my first post, so I would rely on the answer's of others before mine.
The question is basically asking for the longest line located within the box. In a three dimentional box, this would be an upper corner and an opposite side's lower corner. Verbally its hard for me to explain, so if you can draw 3d pictures I suggest doing it and then try to follow my explanation.
You need to solve for two triangles. The first (1) being located on the cubes base, and the second (2) being made up of the cube's height, hypotenuse of the base (hypotenuse) from (1), and the hypotenuse from the upper corner to opposite lower corner.
To make the sides, you have 3 x 4 x 12.
Make the base triangle have sides 3 and 4. This will give the cube a base of 4 x 3 and height of 12.
Using a^2 + b^2 + c^2 will give you a (1) triangle of 4x3x5. You now have two sides of the second triangle (height 12 and base 5).
Now to find the longest point, which is the hypotenuse of (2). Once again, a^2 + b^2 + c^2, and you will have a triaangle of 5x 12 x 13. 13 is your answer.
Again this is my first post, so I would wait till other confirm of correct my answer. |
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asigheartau Just gettin' started!
Joined: 22 Jun 2008 Posts: 18
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Location: NY Target GMAT Score: 700+
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Posted: Tue Jul 01, 2008 1:36 pm Post subject: |
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somail, you are correct.
the answer is 13.
Tommieboi918,
Consider drawing a 3d rectangular with length 3, width 4, and height 12.
Now try to imagine a box with the same dimensions. it is obvious that if you were to put a rod inside of it it could be around 12 meters high. One's first instinct is to say it`s 12 automatically, but pay attention.
It is not mentioned how you would place the rod inside the box so you have to account for any possibility. If you were to put the rod diagonally it could be longer than 12 m because diagonals in a rectangle are longer than the sides.
somail did a good job explaining |
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simpdimp Just gettin' started!
Joined: 19 Jan 2008 Posts: 7
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Posted: Wed Jul 02, 2008 11:24 pm Post subject: |
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| There is a simple formula to solve this problem - sqrt(3sq+4sq+12sq) and the answer is 13. |
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sibbineni Really wants to Beat The GMAT!
Joined: 04 Jan 2008 Posts: 223
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Posted: Sun Jul 06, 2008 8:21 pm Post subject: |
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This is simple formula
for rectangular box
The longest length of the rod is
sqrt(l^2+w^2+h^2)
for rectangle:
the longest length is
sqrt(l^2+w^2) |
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