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

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.”

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

Hello from the GMAT Club BumpBot!

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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 \(>\sqrt{2}\).

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

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13 Feb 2017, 01:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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

gmatclubot

Re: A certain right triangle has sides of length x, y, and z
[#permalink]
25 May 2017, 11:23

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