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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]

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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]

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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 $$\frac{2}{y}<y$$ --> $$2<y^2$$ --> $$\sqrt{2}<y$$.

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]

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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]

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09 Feb 2012, 03:42
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]

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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]

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13 Mar 2012, 14:11
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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: Tricky Triangle Inequalities. QR #157 PS [#permalink]

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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]

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28 Feb 2013, 09:01
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]

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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]

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01 Mar 2013, 01:50
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 $$\frac{2}{y}<y$$ --> $$2<y^2$$ --> $$\sqrt{2}<y$$.

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]

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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]

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05 Nov 2013, 10:04
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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: A certain right triangle has sides of length x, y, and z [#permalink]

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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]

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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]

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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}
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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, 06:02
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, 18:34
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
Re: A certain right triangle has sides of length x, y, and z   [#permalink] 12 Feb 2016, 18:34
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