msu08

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Nov 14, 2006
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a rectangular lid has a length of 5 inches, a width of 3, height of 6 inches. what is the greatest possible distance between any 2 corner points of this solid?

answer is sqrt 70

it would be a diag across this box, but i'm just having difficulty doing the math part. thanks!!
 

MaskItOrCasket

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Aug 20, 2007
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ok try these steps bc ur logic is right so far:

1) draw the box!

2) since you know the height that is half the work! 6 inches

3) now we need to know the distance from the top corner opposite to the far bottom corner. lets get the hypotenuse of the base of the rectangle because that will be the other side of the triangle to figure out the answer. sides 5 and 3 gives sqrt34 from pythag theorem.

4) now we do pythag theorem again! with sides 6 and sqrt 34... and thus the sqrt 34 squared is 34 and 6 squared is 36. 36+34 = 70 and now take the sqrt of all that and u get sqrt 70 as the final answer!
:thumbup:
hope that helps
 

drillbit

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u could also use the equation: sqrt (l^2 + w^2 + h^2) - although you should understand why this equation works - as stated above by Panther85

sqrt (5^2 + 3^2 + 6^2) = sqrt (25 + 9 + 36) = sqrt 70
 
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