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A certain right triangle has sides of length x, y, and z, wh [#permalink]
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The Official Guide For GMAT® Quantitative Review, 2ND EditionA certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? (A) \(y >\sqrt{2}\) (B) \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) (C) \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) (D) \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) (E) \(y<\frac{\sqrt{3}}{4}\) Problem Solving Question: 157 Category: Geometry; Algebra Triangles; Area; Inequalities Page: 83 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: A certain right triangle has sides of length x, y, and z, wh [#permalink]
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SOLUTIONA certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?(A) \(y >\sqrt{2}\) (B) \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) (C) \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) (D) \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) (E) \(y<\frac{\sqrt{3}}{4}\) The area of the triangle is \(\frac{xy}{2}=1\) (\(x<y<z\) means that hypotenuse is \(z\)) > \(x=\frac{2}{y}\). As \(x<y\), then \(\frac{2}{y}<y\) > \(2<y^2\) > \(\sqrt{2}<y\). Answer: A. Also note that max value of \(y\) is not limited at all. For example \(y\) can be \(1,000,000\) and in this case \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\).
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A certain right triangle has sides of length x, y, and z, wh [#permalink]
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20 Nov 2015, 13:11
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionA certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? (A) \(y >\sqrt{2}\) (B) \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) (C) \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) (D) \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) (E) \(y<\frac{\sqrt{3}}{4}\) This was a tough problem for me, to be honest I couldn't find a solution like Bunuel did, but I have tested aswer choices to derive at the correct answer. We know that \(x*y=2\) actually let's square this expression \(x^2*y^2=4\) so let's test the values given in the answer choices with Min/Max approach (we ca also square values in the answer choices, as sides of a triangle can not be ve) (B) so \(y^2\) can be max 2 and \(x^2\)= max \(\frac{3}{4}\)> multiply \(x^2*y^2\)=1,5. So B is out because we need a 4 when \(x^2 and y^2\) are multiplied (C) \(y^2\) max = \(\frac{3}{4}\), \(x^2\) max=2/9 > \(\frac{2}{9}*\frac{3}{4} < 4\) , C is out (D) \(\frac{2}{9}*\frac{3}{16} < 4\) (E) \(y^2\) < \(\frac{3}{16}\), so we know that y > x, thus when y < \(\frac{3}{16}\) \(x^2\) is also < \(\frac{3}{16}\) and their product is < 4, E is also out and we are left with Answer Choice A (A)
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Re: A certain right triangle has sides of length x, y, and z, wh [#permalink]
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27 Apr 2016, 16:59
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionA certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? (A) \(y >\sqrt{2}\) (B) \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) (C) \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) (D) \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) (E) \(y<\frac{\sqrt{3}}{4}\) Thank you![/textarea] Hi, from the area of triangle we know x*y = 2. So there is one possibility that x could be 1 and y could be 2, but none of the answer choices ,except A confirms to y =2 possibility. So A is our answer.



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A certain right triangle has sides of length x, y, and z, wh [#permalink]
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Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionA certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? (A) \(y >\sqrt{2}\) (B) \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) (C) \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) (D) \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) (E) \(y<\frac{\sqrt{3}}{4}\) You can think about it in terms of transition points too. x < y< z so this means that x and y are the legs of the right triangle and z is the hypotenuse. The area of the triangle will be (1/2)*xy = 1 xy = 2 Now, if x = y, then both x and y would be equal to \(\sqrt{2}\). But y is greater than x, so y would be at least a slight bit greater than \(\sqrt{2}\) and x would be a slight bit less than \(\sqrt{2}\). In all options other than (A), y takes values less than 1.414. Answer (A)
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A certain right triangle has sides of length x, y, and z, wh [#permalink]
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Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionA certain right triangle has sides of length x, y, and z, where x < y < z, If the area of this triangular region is 1, which of the following indicates all of the possible values of y ? (A) \(y >\sqrt{2}\) (B) \(\frac{\sqrt{3}}{2}<y<\sqrt{2}\) (C) \(\frac{\sqrt{2}}{3}<y<\frac{\sqrt{3}}{2}\) (D) \(\frac{\sqrt{3}}{4} < y <\frac{\sqrt{2}}{3}\) (E) \(y<\frac{\sqrt{3}}{4}\) First step will be to breakdown the options into recognizable decimal representations (assuming \(\sqrt{2} \approx 1.4\), \(\sqrt{3} \approx 1.7\)) A) y>1.4 B) 0.8<y<1.4 C) 0.5<y<0.8 D) 0.4<y<0.5 E) y<0.4 We are given that x<y<z and that 0.5*x*y=1 > x*y=2 Now from the relation xy=2 > go back to the options and test for y=1. You get x=2 but we are given that x<y > y MUST be > \(\approx\)1.4 such that x < y For any value of y < 1.4 , you will end up getting x>y (try with y=0.5 or 0.75 etc). Only A satisfies this condition and is hence the correct answer. Hope this helps.



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