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A certain right triangle has sides of length x, y, and z [#permalink]

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16 Jan 2011, 16:01

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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. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)

Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]

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16 Jan 2011, 16:12

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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 \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) --> \(x=\frac{2}{1,000,000}\).

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 ?

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.

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 30-60-90 triangle. Had it been 45-45-90 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 30-60 -90 triangle.

Please confirm _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]

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01 Mar 2013, 02:50

Expert's post

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 30-60-90 triangle. Had it been 45-45-90 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 30-60 -90 triangle.

Please confirm

Yes, if it were 45-45-90, then we would have that x=y<z. BUT, knowing that it's not a 45-45-90 right triangle does NOT mean that it's necessarily 30-60-90 triangle: there are numerous other right triangles. For example, 10-80-90, 11-79-90, 25-65-90, ...

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.

Re: A certain right triangle has sides of length x, y, and z [#permalink]

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31 Jan 2015, 09:25

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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 \(>\sqrt{2}\).

HTH

gmatclubot

Re: A certain right triangle has sides of length x, y, and z
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12 Feb 2016, 19:34

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