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A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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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. \(y < \frac {\sqrt {3}}{4}\)
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
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 \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\). Hope it helps.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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06 Aug 2011, 03:41
Bunuel wrote: tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
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 \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\). Hope it helps. Dear Bunuel, While solving the question ,I assumed it to be the special 90,60,30 triangle. Am I wrong in following that approach ?



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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09 Feb 2012, 03:42



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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09 Feb 2012, 05:54
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OA is A since this is a rt angled triangle so z is th hypotnuse and given xy = 2 so as x decreased y increases. Now if x is 1 then y is 2, when x is 1/2 y is 4. Only option A supports this result.



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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28 Feb 2013, 08:55
Bunuel wrote: \(x=\frac{2}{y}\). As \(x<y\), then \(\frac{2}{y}<y\) Could you explain why that is?



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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28 Feb 2013, 09:01



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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01 Mar 2013, 01:21
rohansharma wrote: Bunuel wrote: tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
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 \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\). Hope it helps. Dear Bunuel, While solving the question ,I assumed it to be the special 90,60,30 triangle. Am I wrong in following that approach ? Hi Bunuel, The Q stem says that sides are x<y<z and it is a right angle triangle. So we can assume that it will be 306090 triangle. Had it been 454590 triangle then the 2 sides ie base and perpendicular would have been equal and therefore x=y and x,y<z I guess it should be okay to assume that it is 3060 90 triangle. Please confirm
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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01 Mar 2013, 01:50
mridulparashar1 wrote: rohansharma wrote: Bunuel wrote: 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 \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\).
Hope it helps. Dear Bunuel, While solving the question ,I assumed it to be the special 90,60,30 triangle. Am I wrong in following that approach ? Hi Bunuel, The Q stem says that sides are x<y<z and it is a right angle triangle. So we can assume that it will be 306090 triangle. Had it been 454590 triangle then the 2 sides ie base and perpendicular would have been equal and therefore x=y and x,y<z I guess it should be okay to assume that it is 3060 90 triangle. Please confirm Yes, if it were 454590, then we would have that x=y<z. BUT, knowing that it's not a 454590 right triangle does NOT mean that it's necessarily 306090 triangle: there are numerous other right triangles. For example, 108090, 117990, 256590, ... Hope it's clear.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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10 Nov 2013, 18:20
Buneul has quite literally owned this problem. Great solution!



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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15 Dec 2013, 07:04
i did some guesstimates to arrive at this choice
here we go 
if we assume this to be a isosceles right angled triangle then the area would be maximum.
and xy/2=1
y^2=1 (since it is an isosceles triangle) y=Sq Root 2
now we know y>x and area =1; y has to be > sq root 2 and x < sq root 2
fortunately in this case, only one option had this range.



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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16 Dec 2013, 13:17
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?
We are told that this is a right triangle which right off the bat tells me one of two things, either we need to solve with some variation of a^2 + b^2 = c^2 or that we can find the area with base*height.
Because this is a right triangle and x < y < z we know that z is the hypotenuse and that x is the shortest leg. The area = 1 so:
a=1/2 b*h 1=1/2 b*h 2=b*h.
Y is the second longest measurement in this right triangle which means it must be longer than x but shorter than z. If we run through a few possible combinations of a and b we see that there isn't a limit on the length of y so long as y*x = 2 and y<x. For example, x=1 and y = 4 and z can = 5. This means that there is no upward limit on the value of y so answer choice E is out. This also means that D, C and B are out as well because all contain upward limits on the value of y can be any number so long as y*x = 2 and y<x. Therefore, A is the only answer choice.
Answer: a. y > \sqrt {2}



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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23 Aug 2015, 06:02
I think I've made it more comlicated then it is... xy=2, z^2=x^2+y^2 (x+y)^2=z^2+4 and just stucked at this point ....
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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12 Feb 2016, 18:34
One other way that I noticed to solve this problem is to check the length of \(y\) when \(x=y\), i.e. 45,45,90. In that case \(x=y=\sqrt{2}\), however as \(y>x\), it'd always need to be \(>\sqrt{2}\).
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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25 May 2017, 10:23
tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
c. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)
d. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)
e. \(y < \frac {\sqrt {3}}{4}\) Here we can solve this problem as follows 1/2 *xy = 1 xy =2 x = 2/y 2/y < y 2<y^2 sqrt(2) < y



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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11 Sep 2017, 12:54
tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
c. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)
d. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)
e. \(y < \frac {\sqrt {3}}{4}\) There are infinitely many right triangles that have an area of 1. So, one approach is to find a triangle that meets the given conditions, and see what conclusions we can draw. Here's one such right triangle: This meets the conditions that the area is 1 AND x < y < z With this triangle, y = 4 When we check the answer choices, only one (answer choice A) allows for y to equal 4 Answer: Cheers, Brent
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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28 Sep 2017, 07:39
tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
C. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)
D. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)
E. \(y < \frac {\sqrt {3}}{4}\) Since y is the longest of all sides, we can conclude that it's hypotenuse. Hence, x & y are relevant for Area. Since Area= 1, we can 1/2* x*y = 1. Therefore, xy = 1. Notice there could be any combination of x & y. What needs to be taken care of is x<y. Hence it could be 1 & 2, 1/2 & 4 etc. Option A : It says Y> Sq.root 2 If we substitute same in the equation we get x also as sq.root 2. However, we know that x cannot equal to y. Hence, y can be anything greater than sq.root 2. This option satisfies the criteria. This doesn't leave any scope for Y value to differ as Y cannot be equal to less than sq.root of 2.



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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29 Sep 2017, 10:22
tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
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 answer is A Area =1/2 x*y as it is right triangle and according to the condition x < y < z therefore the two legs of the triangle are x and y so x*y=2 Now taking the condition x < y < z Let us take x<y Multiply both sides by y we have xy<y^2 or 0<y^2x*y 0<y^2 2 or 0<(y√2)*(y+√2) As length can not be negative therefore we have y>√2
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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14 Dec 2017, 00:20
Bunuel wrote: tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
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 \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) > \(x=\frac{2}{1,000,000}\). Hope it helps. Thanks for explaining this clearly and concisely, Bunuel! It reminded me again that we should NOT assume that if x < y < z on a right triangle, it's automatically a 306090 triangle. The answer choices also added to the confusion as the sqrt {3} is in there. During the exams, we should really be aware of these mindsets



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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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14 Dec 2017, 04:17
tonebeeze 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. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)
C. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)
D. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)
E. \(y < \frac {\sqrt {3}}{4}\) Since it is a right triangle, z is the greatest side and hence the hypotenuse. So area of the triangle will be (1/2)*xy = 1 xy = 2 Note that x < y. If x were equal to y, \(x = y = \sqrt{2}\) Since x is less than y, \(y > \sqrt{2}\) and \(x < \sqrt{2}\) Answer (A)
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