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Triangle Problem

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Master | Next Rank: 500 Posts
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by hariharakarthi » Sun Jul 26, 2009 1:34 pm
A cube has sides measuring 6 inches. What is the greatest
possible (straight-line) distance, in inches, between any
two points on the box?
(A) 2sqrt(6)
(B) 3sqrt(6)
(C) 6sqrt(2)
(D) 6sqrt(3)
(E) 12

OA D
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by shibal » Sun Jul 26, 2009 1:39 pm
6sqrt3

the greatest possible measure in a cube is the diagonal line inside the cube that comes from the top part to the bottom one... try to picture a cube and from the high left corner draw a line to the bottom back corner.....

therefore it'll make a triangle with height 6, base 6sqrt2 (it forms a 90-45-45 triangle). then just solve for a^2=6^2+(6sqrt2)^2

hope it is clear
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by truplayer256 » Sun Jul 26, 2009 4:56 pm
There's a formula for these kinds of problems. Whenever you have a problem asking you for the greatest possible straight line distance between any two points in a 3 dimensional figure, you use the formula:

sqrt(L^2+W^2+H^2)

In this problem in particular, the greatest possible distance between any two points would be:

sqrt( 36*3)=sqrt(36)*sqrt(3)=6sqrt(3)
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