Hey bajwa,
Love the question - quick thing, though...I think the designations may be slightly off from the attached diagram. Do you mean to say that right triangle ADE has an area of 7? The way the diagonal is drawn from A to E in the figure seems to suggest that. And then in that case I'd predict that EC (the line from the edge of the triangle at the bottom of the screen across to the right corner of the rectangle) would be 3 times then length of DE (the side of the triangle that connects to line EC).
If that's the case, then you'd first want to start by identifying what you know. The area of the triangle is equal to 1/2 bh, but because it's isosceles we can say that b = h (using the two legs of the triangle as the base and height, since they form a 90-degree angle). So we'd have:
Area = 7 = 1/2 b^2
14 = b^2
sqrt 14 = base (or one side of the triangle, including side DE).
So one side of the rectangle (AD) is sqrt 14, and so is segment DE. Never fear at this point - yes, sqrt 14 is an ugly, ugly number, but remember that we'll be multiplying the length and width of the rectangle, so as long as both sides include sqrt 14 it will multiply itself out! The GMAT loves to set up calculations taht look ugly at first, but that clean up nicely for those who are patient.
Then, if the remainder of the long side, segment EC, is 3(DE), then EC = 3(sqrt 14), and we can add the two segments together to find that the long side has a length of 4(sqrt 14).
So for the area of the rectangle, we'd multiply sqrt 14 * 4(sqrt 14) = 4 * 14 = 56.
Again, this assumes that I've predicted the correct segments for those measurements, but given the look of the diagram I think that should be the case...
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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