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This one is driving me crazy. I'm sure it's been posted

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This one is driving me crazy. I'm sure it's been posted [#permalink] New post 17 Oct 2005, 05:41
This one is driving me crazy. I'm sure it's been posted before but I can't seem to find it. Here goes:

A rectangular box has length 10 in, width 10 in and heigh 5 in. What is the length of the line drawn from the greatest distance from one edge of the box to the other. (not exact wording but I think you get the idea).

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 [#permalink] New post 17 Oct 2005, 05:47
Imagine a 3=D box and draw the longest hypotenuse from the front bottom left hand corner to the top rear right hand corner. This is definately an ubiquitous GMAT problem, but I didn`t see it on my last test.
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 [#permalink] New post 17 Oct 2005, 05:53
well the diagonal is the longest possible line in a cubic strucutre...

so D=Sqrt(10^2+10^2+5^2)

sqrt(225)=15
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explain further? [#permalink] New post 17 Oct 2005, 07:10
could you explain this further? why is it simply the sqr rt of the each side squared? I was trying to draw 2 triangles and get the hypotenuses by pythagorean thereom and add the 2 hypotenueses together. is that wrong?

Also- general question- when a line is drawn from a square or rectangle angle (so that you know the angle is a 90 degrees angle)- can you assume the line always splits the angle in half making it a 45 degrees angle....or could it be anything- 30 degrees, 70 degrees, etc?

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Re: explain further? [#permalink] New post 17 Oct 2005, 07:28
you could do it your way...or memorize my shortcut...any rectangular box sructure the longest distance btw any 2 points is the diagnol and is given by D=Sqrt(a^2+b^2+c^2)

Jennif102 wrote:
could you explain this further? why is it simply the sqr rt of the each side squared? I was trying to draw 2 triangles and get the hypotenuses by pythagorean thereom and add the 2 hypotenueses together. is that wrong?

Also- general question- when a line is drawn from a square or rectangle angle (so that you know the angle is a 90 degrees angle)- can you assume the line always splits the angle in half making it a 45 degrees angle....or could it be anything- 30 degrees, 70 degrees, etc?

thx
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 [#permalink] New post 17 Oct 2005, 07:59
Jennifer

The triangle you are trying to crack uses the diagonal from one corner to the opposite lower corner as the hypotenus, the diagonal of the bottom face is the base, and height is the height. So first try to figure out the length of the base.

You know you have a 10x10 square so therefore you know the diagonal is 10*sqrt(2).

So now you have a right triangle with base 10*sqrt(2) and height 5.

So...

A^2 + B^2 =C^2
(10*sqrt(2))^2 + (5)^2 = 225
sqrt(225) = 15

Brendan.
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Re: This one is killing me... [#permalink] New post 17 Oct 2005, 08:40
Jennif102 wrote:
This one is driving me crazy. I'm sure it's been posted before but I can't seem to find it. Here goes:

A rectangular box has length 10 in, width 10 in and heigh 5 in. What is the length of the line drawn from the greatest distance from one edge of the box to the other. (not exact wording but I think you get the idea).

Thanks!


Jenn,

this is very easy once you conceptualize a box. the longest point between two points in a box is the upper corner to the lower corner - straight across the diagonal line of the box. Thus first find the diagonal floor of the box. In this case since you have a square it's 10*sqrt(2) - remember your ratios. Then find the diagonal that we talked about (which is the hypotenuse of the diagnol of the floor and the height of the box). Thus, 5^+10sqrt2^2=225, sqrt225=15

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thanks [#permalink] New post 17 Oct 2005, 08:47
Thanks guys...I'm still having trouble conceptualizing the box- I'm not a very artistic/visual person. I get the floor part- but why is it the diagnol of the floor sqaured plus the height squared? I don't get why you don't have to form a triangle with the height and figure out the hypotenuse of that triangle.
Either way, I doubt the test would have anything more difficult than this so as long as I remember this discussion I should be ok.

thanks!
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Re: thanks [#permalink] New post 17 Oct 2005, 08:54
Jennif102 wrote:
Thanks guys...I'm still having trouble conceptualizing the box- I'm not a very artistic/visual person. I get the floor part- but why is it the diagnol of the floor sqaured plus the height squared? I don't get why you don't have to form a triangle with the height and figure out the hypotenuse of that triangle.
Either way, I doubt the test would have anything more difficult than this so as long as I remember this discussion I should be ok.

thanks!


If you have the Official Guide 10th edition, look at Page 181 problem 215. This should be able to help you. THere's a diagram and everything.
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 [#permalink] New post 17 Oct 2005, 09:33
I agree with fresinha... its a formula worth memorizing (its simple too)... helps you solve in less than 30secs.
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perfect [#permalink] New post 17 Oct 2005, 09:48
Thanks guys- I do have the Official Guide and I was so sure I had seen it in there but couldn't find it again while flipping through last night.
Thanks for pointing out where it is- will be sure to check that one out.
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 [#permalink] New post 17 Oct 2005, 10:10
You are calculating this diagonal.

B.

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  [#permalink] 17 Oct 2005, 10:10
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