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If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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09 Jun 2014, 13:29
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If the area of Rectangle ABCD is \(4\sqrt{3}\), then what is the area of the square DEFG ? A) \(\sqrt{3}\) B) \(2\sqrt{3}\) C) 4 D) \(4\sqrt{3}\) E) 12 Attachment:
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Re: If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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09 Jun 2014, 20:44
Great question. This problem illustrates the need to be comfortable with common ratios in geometry (particularly with triangles), and also to be very attentive to details in the wording of problems. Triangle ACD is a 306090 triangle, meaning that the lengths of the sides opposite the angles are in the proportions x, x*sqrt3, 2x. The area of rectangle ABCD is 4*sqrt3, which is equal to (x)*(x*sqrt3), because of the proportions just discussed. This means that X is equal to 2. Triangle CDE is a similar and equal triangle to triangle ACD, meaning that the distance DE is equal to 2. Because the problem says that DEFG is a square, we know that all sides of DEFG must equal 2. Therefore we just multiply 2*2 and come to the conclusion that the area of square DEFG is 4. With geometry problems, it is often helpful to start making any deductions that you can (even when you do not know why you are doing it!). A lot of times, the problem will end up unraveling nicely this way. Veritas does a great job of teaching methodologies that work great with problems like these. I hope this helps!!!
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Re: If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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10 Jun 2014, 15:17
I actually think that the answer is 12.
As Brandon said above, x=2, but the side that is shared with the 454590 triangle is actually 2sqrt(3) because of the 1:1: sqrt2 ratios.
Thus 2 sqrt(3) * 2 sqrt(3) is actually 12, not 4.



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Re: If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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10 Jun 2014, 17:31
Great catch vaj. That is absolutely correct. The triangle on the left is a 306090 triangle, and the solution about how to get to the proportions is in my response above. The triangle on the right, however, is a 454590 triangle (the other common ratio that you need to know). 454590 triangles have sides that correspond to xxx*sqrt 2 (although for this problem you really only need to know that both sides opposite the 45 angles will be equal). In this example, the side opposite the 60 degree angle in the left triangle corresponds to one of the sides opposite a 45 degree angle in the right triangle. As stated above, the side opposite the 60 degree triangle will be equal to x*sqrt3. This means that both sides opposite the 45 degree angles in the right triangle will also be equal to x*sqrt 3, meaning that square DEFG has sides of x*sqrt 3. We know that X = 2, so squaring this we get X^2 * (sqrt3)^2, = 2^2 * (sqrt3)^2 = 4*3 = 12. Hope that helps.
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Re: If the area of Rectangle ABCD is 4√3, then area of the square DEFG is? [#permalink]
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04 Oct 2015, 21:36
Hi amithyarli, While this question is a multishape Geometry question, it's based on a few standard rules (that you probably already know), so you can solve it with some notetaking and math. I'm going to give you a few 'hints' so that you can attempt this question again: 1) What do you notice about Triangle ACD? What would you label its 3 sides? 2) The area formula for a triangle is Area = (1/2)(Base)(height). Using the sides you've just labeled, and the given area, you should be able to figure out the sides of that triangle. 3) What do you know about triangle CDE? From your prior work, you now have 1 of its sides. What do you now know about side DE? 4) Since DEFG is a square and you have DE, how do you figure out the area of that square? GMAT assassins aren't born, they're made, Rich
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Re: If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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05 Oct 2015, 19:10
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amithyarli wrote: If the area of Rectangle ABCD is 4√3, then what is the area of the square DEFG ?
A) √3 B) 2√3 C) 4 D) 4√3 E) 12 This can be solved easily by following a systematic approach: 1. The area of the rectangle ABCD = 4 \(\sqrt{3}\) Hence the area of triangle ACD would be half = \(2 \sqrt{3}\) Now, we need to find the sides of the triangle. We know that the angle ACD = 30. Assuming CD as the base (b), we can say that CD (h) = \(\sqrt{3}\) AD Area of the triangle = (1/2)*b*h = \(2 \sqrt{3}\) Hence we can find the value of b = \(2 \sqrt{3}\) CD is a part of the triangle that is isosceles (two angles are same  this means two sides will be same) Hence CD = DE (the side of the square) Area of the square = DE*DE = \(2 \sqrt{3}\)*\(2 \sqrt{3}\) =12 Option E



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If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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22 Oct 2016, 10:22
OptimusPrepJanielle wrote: amithyarli wrote: If the area of Rectangle ABCD is 4√3, then what is the area of the square DEFG ?
A) √3 B) 2√3 C) 4 D) 4√3 E) 12 This can be solved easily by following a systematic approach: 1. The area of the rectangle ABCD = 4 \(\sqrt{3}\) Hence the area of triangle ACD would be half = \(2 \sqrt{3}\) Now, we need to find the sides of the triangle. We know that the angle ACD = 30. Assuming CD as the base (b), we can say that CD (h) = \(\sqrt{3}\) AD Area of the triangle = (1/2)*b*h = \(2 \sqrt{3}\) Hence we can find the value of b = \(2 \sqrt{3}\) CD is a part of the triangle that is isosceles (two angles are same  this means two sides will be same) Hence CD = DE (the side of the square) Area of the square = DE*DE = \(2 \sqrt{3}\)*\(2 \sqrt{3}\) =12 Option E A 306090 right angled triangle has ratio of sides as 1:sq rt 3 : 2 How did you reach to value of CD (h) = √3 AD ? [ Please explain why AD is being multiplied in √3 AD?]



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Re: If the area of Rectangle ABCD is 4√3, then what is the area of the [#permalink]
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22 Oct 2016, 14:09
Manonamission wrote: OptimusPrepJanielle wrote: amithyarli wrote: If the area of Rectangle ABCD is 4√3, then what is the area of the square DEFG ?
A) √3 B) 2√3 C) 4 D) 4√3 E) 12 This can be solved easily by following a systematic approach: 1. The area of the rectangle ABCD = 4 \(\sqrt{3}\) Hence the area of triangle ACD would be half = \(2 \sqrt{3}\) Now, we need to find the sides of the triangle. We know that the angle ACD = 30. Assuming CD as the base (b), we can say that CD (h) = \(\sqrt{3}\) AD Area of the triangle = (1/2)*b*h = \(2 \sqrt{3}\) Hence we can find the value of b = \(2 \sqrt{3}\) CD is a part of the triangle that is isosceles (two angles are same  this means two sides will be same) Hence CD = DE (the side of the square) Area of the square = DE*DE = \(2 \sqrt{3}\)*\(2 \sqrt{3}\) =12 Option E A 306090 right angled triangle has ratio of sides as 1:sq rt 3 : 2 How did you reach to value of CD (h) = √3 AD ? [ Please explain why AD is being multiplied in √3 AD?]
A right triangle with 306090 can be written as a : a*sqrt(3) : 2a Where 'a' is the side of the smallest length which is opposite to 30 degree Area of rectangle ABCD = a*a*sqrt(3) = 4*sqrt(3) Therefore a=2 CD=2*sqrt(3)= DE Area of square = (2*sqrt(3))^2 = 12 Sent from my iPhone using GMAT Club Forum mobile app



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