Now, for triangle DEC, consider ED as base and CP as height, we have Area of DEC = 1/2 CP * DE (2)įrom (1) and (2), the 2 area is the same, we have EQ * DC = CP * DE (3). Now, from C, draw a line CP that is perpendicular with DE with P is on DE. The tricky part is how to link it with the rectangle DEFG.

$what is the area of parallelogram abcd$

It also equals 1/2 Area ABCD (area of parallelogram is base * height). I use the same diagram that mikemcgarry provided.įirst, we all agree that by considering DC as base and EQ as height, Area DEC = 1/2 * EQ * DC (1). I think it's doesn't need to be that complicated. Hi, mikemcgarry's is good but it uses similar triangles to prove. Please let me know if you have any questions on what I've said here. Here's another geometry DS, a little closer to the actual level of difficulty of the GMAT itself.

Therefore, Statement #2 is insufficient.ĭoes all this (including everything in the pdf) make sense? If we know the altitude and not the base, that's not enough. Statement #2: Line AH, the altitude of parallelogram ABCD, is 5.Īrea of a parallelogram = (base)*(altitude). Leaving those details aside for the moment, Statement #1 is sufficient. Why do the rectangle and parallelogram have equal area? You will see the complete geometric argument in the pdf attachment to this post. Well, to cut to the chase, this statement is sufficient because the rectangle and the parallelogram must have equal area. Statement #1: The area of rectangle DEFG is 8√5. As a geometry geek myself, I found this a very cool geometry problem, but I will say - it is WAY harder than anything you would be expected to figure out for yourself on the real GMAT.

What is the area of parallelogram abcd