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A certain right triangle has sides of length x, y, and z [#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: Tricky Triangle Inequalities. QR #157 PS [#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 [#permalink]
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13 Mar 2012, 15:11
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Is there a reason we can completely ignore z in the inequality while solving for y? That is the only part I don't understand. (im rusty)



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Re: A certain right triangle has sides of length x, y, and z [#permalink]
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05 Nov 2013, 11:04
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tanviet wrote: it is simple but hard enough to kill us
this is NOT og questions. It is the Quant Review 2nd Ed. #157



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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]
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06 Aug 2011, 04:41
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: Tricky Triangle Inequalities. QR #157 PS [#permalink]
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09 Feb 2012, 04:42



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Re: A certain right triangle has sides of length x, y, and z [#permalink]
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09 Feb 2012, 06:54
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: Tricky Triangle Inequalities. QR #157 PS [#permalink]
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28 Feb 2013, 09:55
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: Tricky Triangle Inequalities. QR #157 PS [#permalink]
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]
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01 Mar 2013, 02:21
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: Tricky Triangle Inequalities. QR #157 PS [#permalink]
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01 Mar 2013, 02:50
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 [#permalink]
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10 Nov 2013, 19:20
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 [#permalink]
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15 Dec 2013, 08:04
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 [#permalink]
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16 Dec 2013, 14:17
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 [#permalink]
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Re: A certain right triangle has sides of length x, y, and z [#permalink]
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23 Aug 2015, 07:02
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 [#permalink]
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12 Feb 2016, 19:34
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 [#permalink]
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25 May 2017, 11:23
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




Re: A certain right triangle has sides of length x, y, and z
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