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

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A certain right triangle has sides of length x, y, and z [#permalink]  16 Jan 2011, 15:01
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57% (02:25) correct 42% (01:56) wrong based on 139 sessions
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}
[Reveal] Spoiler: OA
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]  16 Jan 2011, 15:12
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Expert's post
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.

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

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]  09 Feb 2012, 03:42
Expert's post
rohansharma wrote:
While solving the question ,I assumed it to be the special 90,60,30 triangle.
Am I wrong in following that approach ?

We are just told that the triangle is right, not that it's a special kind like 30-60-90 or 45-45-90.
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Re: A certain right triangle has sides of length x, y, and z [#permalink]  09 Feb 2012, 05:54
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: A certain right triangle has sides of length x, y, and z [#permalink]  13 Mar 2012, 14:11
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: Tricky Triangle Inequalities. QR #157 PS [#permalink]  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: Tricky Triangle Inequalities. QR #157 PS [#permalink]  28 Feb 2013, 09:01
Expert's post
2flY wrote:
Bunuel wrote:
x=\frac{2}{y}. As x<y, then \frac{2}{y}<y

Could you explain why that is?

Just substitute x with \frac{2}{y} in x<y to get \frac{2}{y}<y.
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]  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.

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.

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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink]  01 Mar 2013, 01:50
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.

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.

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, ...

Hope it's clear.
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Re: A certain right triangle has sides of length x, y, and z [#permalink]  04 Mar 2013, 02:20
it is simple but hard enough to kill us

this is NOT og questions.
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Re: A certain right triangle has sides of length x, y, and z [#permalink]  05 Nov 2013, 10:04
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: A certain right triangle has sides of length x, y, and z [#permalink]  10 Nov 2013, 18:20
Buneul has quite literally owned this problem. Great solution!
Re: A certain right triangle has sides of length x, y, and z   [#permalink] 10 Nov 2013, 18:20
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