![]() ![]() 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. "helpContent": " Full Schedule\n\n Register to participate\n - Watch elite experts fill out applications live\n \n - Best ideas for short answers & red flags to avoid\n \n - Live Q&A\n \n- Free Resources & GMAT Club Tests\n\n Learn more about MBA Walkthroughs\n ",Īs 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. "default_bg": "/forum/images/mba_dashboard/Video.png", ![]() "Hiding it while no sessions are taking place", "name": "Ross Application Walkthrough by PMC", "name": "Rotman Application Walkthrough by GoalisB", "name": "Fuqua Application Walkthrough with Consultant", "name": "Fuqua MBA Application Overview by AdCom", "name": "Wharton Application Walkthrough by PMC", "regUrl": "/mba-application-walkthroughs/", This means that the angle at the top will be 42° (84° ÷ 2)."prizeUrl": "/forum/mba-application-walkthrough-series-free-resources-416112.html", This means that the angle at the centre (the angle that is 84°) will be twice the size of the angle at the circumference, which is the angle at the top (we can also view this as the angle at the top being half the size of the angle at the centre). The first circle theorem stated that the angle that is subtended by an arc at the centre of the circle is twice the angle at the circumference of the circle. However, we do not have any angles for the cyclic quadrilateral, which means that we are unable to use this circle theorem at the moment.Īlso, from looking at the diagram we can see that we have a spaceship like shape, and we saw this shape when we looked at the first circle theorem in this whole section. ![]() We can see from looking at the diagram that we have a cyclic quadrilateral, so we will probably be using the circle theorem whereby opposite angles in a cyclic quadrilateral add up to 180°. We will need to use two different circle theorems in order to be able to answer this question. ![]()
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