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ABCD is a square, and EFGH is a square, each vertex of which is on a s
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27 May 2016, 17:05
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ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD. What is the ratio of the area of square EFGH to the area of square ABCD?
Statement #1: AE:AB = 4:7
Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24Geometry is a truly beautiful subject! This question is one of a set of ten practice DS questions about geometry. To see the others, as well as the OE for this particular question, see: GMAT Data Sufficiency Geometry Practice QuestionsMike
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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s
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27 May 2016, 19:53
mikemcgarry wrote: Attachment: Square inside square.png ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD. What is the ratio of the area of square EFGH to the area of square ABCD?
Statement #1: AE:AB = 4:7
Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24Geometry is a truly beautiful subject! This question is one of a set of ten practice DS questions about geometry. To see the others, as well as the OE for this particular question, see: GMAT Data Sufficiency Geometry Practice QuestionsMike we have two squares... the one inside, EFGH, will have least area when its vertex is in center of the sides..... and will become max as it moves closer to the vertex of the bigger square ABCD.. Second point is that all the triangles formed on the vertex of ABCD will be similar... lets see the statement Statement #1: AE:AB = 4:7 let the common ratio be x.... so AE = 4x and AB = 7x, so BE = 3x... side of ABCD = 7x and side of EFGH = HYP of triangle whose sides are 3x and 4x = \(\sqrt{(3x)^2+(4x)^2}\).. ratio =\(\frac{7x*7x}{\sqrt{(3x)^2+(4x)^2}^2}\).. variable x will get cancelled out and we will have a numeric value... Suff Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24 all triangles are similar so their Area = 0.24*Area of EFGH*4..... Let area of square EFGH = x Area of ABCD = Area of 4 triangles + area of EFGH = 0.24*x*4 + x... ratio =\(\frac{0.24*x*4 + x}{x}\) = 1.96/1 Suff D
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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s
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05 Jul 2016, 03:13
chetan2u wrote: mikemcgarry wrote: Attachment: Square inside square.png ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD. What is the ratio of the area of square EFGH to the area of square ABCD?
Statement #1: AE:AB = 4:7
Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24Geometry is a truly beautiful subject! This question is one of a set of ten practice DS questions about geometry. To see the others, as well as the OE for this particular question, see: GMAT Data Sufficiency Geometry Practice QuestionsMike we have two squares... the one inside, EFGH, will have least area when its vertex is in center of the sides..... and will become max as it moves closer to the vertex of the bigger square ABCD.. Second point is that all the triangles formed on the vertex of ABCD will be similar... lets see the statement Statement #1: AE:AB = 4:7 let the common ratio be x.... so AE = 4x and AB = 7x, so BE = 3x... side of ABCD = 7x and side of EFGH = HYP of triangle whose sides are 3x and 4x = \(\sqrt{(3x)^2+(4x)^2}\).. ratio =\(\frac{7x*7x}{\sqrt{(3x)^2+(4x)^2}^2}\).. variable x will get cancelled out and we will have a numeric value... Suff Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24 all triangles are similar so their Area = 0.24*Area of EFGH*4..... Let area of square EFGH = x Area of ABCD = Area of 4 triangles + area of EFGH = 0.24*x*4 + x... ratio =\(\frac{0.24*x*4 + x}{x}\) = 1.96/1 Suff D Dear sir, I don't get the first part... AE/AB=4/7 AE/AE+EB=4/7 AE/EB=4/3 So,AE=4x and EB=3x But then how EBF or any other triangle is a 345 triangle??



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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s
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05 Jul 2016, 10:31
Ashishsteag wrote: Dear sir, I don't get the first part... AE/AB=4/7 AE/AE+EB=4/7 AE/EB=4/3 So,AE=4x and EB=3x But then how EBF or any other triangle is a 345 triangle?? Dear Ashishsteag This is Mike McGarry, the author of this question. I'm happy to respond. So I believe you get that, for some unknown x, AE = 4x and EB = 3x. You see, the entire figure has fourfold symmetry. We could rotate the entire shape by 90 degrees and it would be the same. The four triangles, {AEH, BFE, CGF, and DHG} have to be entirely congruent. Thus, AE = BF = CG = DH = 4x AH = BE = CF = GD = 3x Thus, all of them are 345 triangles. Does this make sense? Mike
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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s
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05 Jul 2016, 22:30
mikemcgarry wrote: Ashishsteag wrote: Dear sir, I don't get the first part... AE/AB=4/7 AE/AE+EB=4/7 AE/EB=4/3 So,AE=4x and EB=3x But then how EBF or any other triangle is a 345 triangle?? Dear Ashishsteag This is Mike McGarry, the author of this question. I'm happy to respond. So I believe you get that, for some unknown x, AE = 4x and EB = 3x. You see, the entire figure has fourfold symmetry. We could rotate the entire shape by 90 degrees and it would be the same. The four triangles, {AEH, BFE, CGF, and DHG} have to be entirely congruent. Thus, AE = BF = CG = DH = 4x AH = BE = CF = GD = 3x Thus, all of them are 345 triangles. Does this make sense? Mike Thanx a lot mike....:D



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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s
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