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A certain right triangle has sides of length x, y, and z
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17 Jan 2010, 02:39
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A 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. \(y<\frac{\sqrt{3}}{4}\) OPEN DISCUSSION OF THIS QUESTION IS HERE: acertainrighttrianglehassidesoflengthxyandz107872.html
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Re: Possible Values of side of Right Angle Triangle
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17 Jan 2010, 02:55
vibhaj wrote: A 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 > sqr root (2) B. sqr root (3)/2 < y < sqr root (2) C. sqr root (2)/3 < y < sqr root (3)/2 D. sqr root (3)/4 < y < sqr root (2)/3 E. y < sqr root (3)/4
OA after discussion.
Edited to denote the square roots more clearly IMO it should be A, area of the triangle=1 = 1/2 X*Y or X*Y = 2 Now , also X<Y , now we by using the POE, we can see that apart from 1 option rest all doesn't satisfy this equation. Cheers,



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Re: Possible Values of side of Right Angle Triangle
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17 Jan 2010, 07:51
Thanks Nitish.is there a better way to solve this , without POE ..?



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Re: Possible Values of side of Right Angle Triangle
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17 Jan 2010, 08:06
vibhaj wrote: Thanks Nitish.is there a better way to solve this , without POE ..? I am not sure about that but I guess for the purpose of GMAT I found this one quickest. Anyway, going back to the options if you see, the options are like sqr root of 3 or 2 and from the equation XY =2, we know Y can have a value of 2 ( X=1, Y=2) So thats why eliminated the rest of the options. Cheers



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Re: Possible Values of side of Right Angle Triangle
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17 Jan 2010, 12:30
vibhaj wrote: Thanks Nitish.is there a better way to solve this , without POE ..? 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 \(\frac{2}{y}<y\) > \(2<y^2\) > \(\sqrt{2}<y\). Also note that max value of \(y\) is not limited at all. For example \(y\) can be \(1,000,000\) and in this case \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\). Hope it helps.
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Geometry..??
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29 Apr 2010, 16:28
A 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. \((sqrt 3)/2\) < y <\(sqrt 2\) C. \((sqrt 2)/3\) < y < \((sqrt 3)/2\) D. \((sqrt 3)/4\) < y < \((sqrt 2)/3\) E. y < \((sqrt 3)/4\)
I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!



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Re: Geometry..??
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29 Apr 2010, 17:07
nverma wrote: A 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. \((sqrt 3)/2\) < y <\(sqrt 2\) C. \((sqrt 2)/3\) < y < \((sqrt 3)/2\) D. \((sqrt 3)/4\) < y < \((sqrt 2)/3\) E. y < \((sqrt 3)/4\)
I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!! area = 1/2 * x* y = 1 => xy=1 since y>x => \(y^2 > xy\) => \(y^2 > 2\) =>\(y > \sqrt{2}\) Hence A...whats OA?
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Some questions .... please help
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07 May 2010, 22:05
1.A 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. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4 Thanks



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Re: Some questions .... please help
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07 May 2010, 23:42
ok question 1 doesnt seem GMATesque but i have come up with the solution ...let me know what the source is and what the OA is
1) we know that area of triangle is 2 so xy/2 = 1 so xy = 2 moreover x < y .....so think about the two graphs ...y =2/x and x=y ....there will be a point on xaxis beyond which y<x ...to find this point lets equate the two eqns
2/x = x > x = sqrt(2) beyond this point y<x ...so y > sqrt(2)
so answer should be A ...in all other choices the value of y is less than sqrt(2)



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Re: Some questions .... please help
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07 May 2010, 23:45
can you please post each question as a separate thread ...it would help you and others to post answers to your questions easily



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A certain right triangle
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21 May 2010, 03:17
A 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. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4



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Re: A certain right triangle
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21 May 2010, 03:24
dimitri92 wrote: A 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. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4 This has already been discussed, please follow the link possiblevaluesofsideofrightangletriangle89294.html



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Re: Possible Values of side of Right Angle Triangle
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22 May 2010, 08:27
After all this information and discussions.... I think this one is one of most simple questions.... it seems the answer is obvious....



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Re: Possible Values of side of Right Angle Triangle
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10 Mar 2011, 13:20



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Re: Possible Values of side of Right Angle Triangle
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10 Mar 2011, 19:39
Solved this intuitively  pls verify this reasoning. The above stem means that one angle > 45 and another angle is < 45
At x = y (hypothetically) i.e. Both angles equal. x = z sin (45) = z/ sqrt(2) y = z cos (45) = z/ sqrt(2)
1/2 x y = 1 Hence z = 2
x = sqrt(2) y = sqrt(2)
If one angle > 45 and another angle is < 45, And knowing x < y < z So 0 < x < sqrt(2) and sqrt(2) < y < Infinity .i.e. when x > 0 then y > Infinity



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Re: Possible Values of side of Right Angle Triangle
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10 Mar 2011, 21:03
gmat1220 wrote: Solved this intuitively  pls verify this reasoning. The above stem means that one angle > 45 and another angle is < 45
At x = y (hypothetically) i.e. Both angles equal. x = z sin (45) = z/ sqrt(2) y = z cos (45) = z/ sqrt(2)
1/2 x y = 1 Hence z = 2
x = sqrt(2) y = sqrt(2)
If one angle > 45 and another angle is < 45, And knowing x < y < z So 0 < x < sqrt(2) and sqrt(2) < y < Infinity Yes, it can be solved intuitively. It is a very interesting way of thinking. There is one small thing: z may not be 2. e.g. When x = 1 and y = 2, z = sqrt(5) Think this way: We know x < y and (1/2)xy = 1 (z is the hypotenuse) So xy = 2 If x = y, then x = y = sqrt(2) But since x < y, x is less than sqrt(2) and y is greater than sqrt(2). Makes sense. Good thinking.
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Re: A certain right triangle has sides of length x, y, and z
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