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I have a certain side ratio of small rectangle, and a side ratio for a large rectangle. How can I write an equation that gives me the number of small rectangles that will make the large rectangle? Is there a formula for that?

Apologies for the title -- it was a last-ditch attempt.

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The side ratio is not enough, because it does not tell you how small/large the rectangles are. E.g, a 2-by-1 rectangle has the same side ratio as a 200-by-100 rectangle. What do you mean by "make"? –  Jonas Meyer Mar 21 '11 at 23:08

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up vote 2 down vote accepted

I think that you might mean you have the ratios $x$ = (side 1 of small rectangle)/(side 1 of large rectangle) and $y$ = (side 2 of small rectangle)/(side 2 of large rectangle), in which case your problem is solvable. The area of the small rectangle is (side 1 of small rectangle)(side 2 of small rectangle) and the area of the large rectangle is (side 1 of large rectangle)(side 2 of large rectangle), thus (# of small rectangles it takes to cover large rectangle) = (area of large rectangle)/(area of small rectangle) = ((side 1 of large rectangle)(side 2 of large rectangle))/((side 1 of small rectangle)(side 2 of small rectangle)) = $(1/x)(1/y)$ = $1/(xy)$.

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This is fine as long as $1/x$ and $1/y$ are naturals. –  Ross Millikan Mar 21 '11 at 23:52
    
@Ross: Yes, I am assuming that a solution exists. –  Alex Becker Mar 22 '11 at 0:04

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