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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] New post 16 Jan 2011, 15:01
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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}

e. y < \frac {\sqrt {3}}{4}
[Reveal] Spoiler: OA
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink] New post 16 Jan 2011, 15:12
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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}.

Hope it helps.
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink] New post 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: Tricky Triangle Inequalities. QR #157 PS [#permalink] New post 09 Feb 2012, 03:42
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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] New post 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] New post 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] New post 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] New post 28 Feb 2013, 09:01
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink] New post 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 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
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Re: Tricky Triangle Inequalities. QR #157 PS [#permalink] New post 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.

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, ...

Hope it's clear.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A certain right triangle has sides of length x, y, and z [#permalink] New post 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] New post 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] New post 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 [#permalink] New post 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 [#permalink] New post 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}
Re: A certain right triangle has sides of length x, y, and z   [#permalink] 16 Dec 2013, 13:17
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