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The answer is "D".
Stat(1): AC=EG. Let EG=x. then EH=x/sqrt(2). AB=x/2(sqrt2).With this we can find the side BC(one side of the rectangle and the diagonal are known).Hence we can find the area in terms of 'x'. Hence suffic.
Stat(2): As the diagonal divides both into two equal parts, if area of triangles is not same, then definitely the areas of rectangles are not same.
Hence suffic....

DS: Square Rectangle [#permalink]
14 Dec 2004, 08:34

I agree with Gladiator: D, with the same reasoning.
For 1, I did EG=sqrt(2), so EH=1, and AB=1/2 AC=sqrt(2). The rectangle is 7/8 the area of the square

2 is obvious, if the area of triangles are not equal, twice the area of each is not going to make them equal

one thing: I did this by plugging in numbers. The square is a perfect figure, and the rectangle's side is half the square, so I knew that by plugging in just one value I'd get a very good idea of what's happening in the problem. Once I did, I saw everything clearly.

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