2/19/2023 0 Comments Parallelogram abcd(2) Line AH, the altitude of parallelogram ABCD, is 5. What is the area of parallelogram ABCD (figure not drawn to scale)? Untitled.pngIn the picture, quadrilateral ABCD is a parallelogram and quadrilateral DEFG is a rectangle. I know that Magoosh has a few hundred math questions, all appropriate difficulty for the GMAT, and each followed but its own video solution. I am not as familiar with Kaplan questions overall, I am not qualified to make a statement about them. You don't have to have be Isaac Newton to answer the hardest questions. The GMAT doesn't expect that, even on 800 level questions. If you've never seen this theorem, there's virtually no way that you will derive the full geometry proof in under a minute, unless you operate at Isaac Newton level. Any GMAT math question, no matter how challenging, is something that someone facile with math would be able to solve in under a minute. I would say a question like this - a question that hinges on a relatively obscure geometry theorem that one probably would have to prove from scratch to answer the question - is something far harder than what they would put on the GMAT. Whereas the questions on GMATPREP seem to be much simpler than this, No? The more I see Kaplan questions, the more I feel the questions can be extremely hard. What is the likelihood of such a question on the GMAT. You just need to substitute side for side.ĭear Mike. I must admit I couldn't get this right, but after reading the explanation of mikemcgarry, I think this way is simpler as you don't have to think and prove similars. So (3) can be rewritten as EQ * DC = EF * DE. But in rectangle DEFG, CP = EF (since DEFG is rectangle, CP perpendicular with DE, so CP must = EF) 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. 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. If you have any queries regarding this article, please ping us through the comment section below and we will get back to you as soon as possible.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. Now you are provided with all the necessary information on the diagonal of a parallelogram formula and we hope this detailed article is helpful to you. Parallelogram Law: The sum of the squares of the sides is equal to the sum of the squares of the diagonals. Each diagonal divides the parallelogram into two congruent triangles.ģ. Diagonals of a parallelogram bisect each other. The diagonals of the parallelogram \(ABCD\) are \(AC\) and \(BD.\) The properties of a parallelogram’s diagonals are as follows:ġ.
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