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

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Manager
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A certain right triangle has sides of length x, y, and z [#permalink]  17 Jan 2010, 01:39
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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. $$y<\frac{\sqrt{3}}{4}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-right-triangle-has-sides-of-length-x-y-and-z-107872.html
[Reveal] Spoiler: OA
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Re: Possible Values of side of Right Angle Triangle [#permalink]  17 Jan 2010, 01:55
vibhaj 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 > sqr root (2)
B. sqr root (3)/2 < y < sqr root (2)
C. sqr root (2)/3 < y < sqr root (3)/2
D. sqr root (3)/4 < y < sqr root (2)/3
E. y < sqr root (3)/4

OA after discussion.

--Edited to denote the square roots more clearly

IMO it should be A,

area of the triangle=1 = 1/2 X*Y

or X*Y = 2

Now , also X<Y , now we by using the POE, we can see that apart from 1 option rest all doesn't satisfy this equation.

Cheers,
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Re: Possible Values of side of Right Angle Triangle [#permalink]  17 Jan 2010, 06:51
Thanks Nitish.is there a better way to solve this , without POE ..?
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Re: Possible Values of side of Right Angle Triangle [#permalink]  17 Jan 2010, 07:06
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vibhaj wrote:
Thanks Nitish.is there a better way to solve this , without POE ..?

I am not sure about that but I guess for the purpose of GMAT I found this one quickest.

Anyway, going back to the options if you see, the options are like sqr root of 3 or 2

and from the equation XY =2, we know Y can have a value of 2 ( X=1, Y=2)

So thats why eliminated the rest of the options.

Cheers
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Re: Possible Values of side of Right Angle Triangle [#permalink]  17 Jan 2010, 11:30
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vibhaj wrote:
Thanks Nitish.is there a better way to solve this , without POE ..?

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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Geometry..?? [#permalink]  29 Apr 2010, 15:28
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. $$(sqrt 3)/2$$ < y <$$sqrt 2$$
C. $$(sqrt 2)/3$$ < y < $$(sqrt 3)/2$$
D. $$(sqrt 3)/4$$ < y < $$(sqrt 2)/3$$
E. y < $$(sqrt 3)/4$$

I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!
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Re: Geometry..?? [#permalink]  29 Apr 2010, 16:07
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nverma 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. $$(sqrt 3)/2$$ < y <$$sqrt 2$$
C. $$(sqrt 2)/3$$ < y < $$(sqrt 3)/2$$
D. $$(sqrt 3)/4$$ < y < $$(sqrt 2)/3$$
E. y < $$(sqrt 3)/4$$

I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!

area = 1/2 * x* y = 1 => xy=1

since y>x => $$y^2 > xy$$ => $$y^2 > 2$$

=>$$y > \sqrt{2}$$

Hence A...whats OA?
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Manager
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1.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. sqrt.3/2 < y < sqrt. 2
C. sqrt.2/3 < y < sqrt.3/2
D. sqrt.3/4 < y < sqrt.2/3
E. y < sqrt.3/4

Thanks
Manager
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ok question 1 doesnt seem GMATesque but i have come up with the solution ...let me know what the source is and what the OA is

1) we know that area of triangle is 2 so xy/2 = 1 so xy = 2 moreover x < y .....so think about the two graphs ...y =2/x and x=y ....there will be a point on x-axis beyond which y<x ...to find this point lets equate the two eqns

2/x = x --> x = sqrt(2)

beyond this point y<x ...so y > sqrt(2)

so answer should be A ...in all other choices the value of y is less than sqrt(2)
Manager
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can you please post each question as a separate thread ...it would help you and others to post answers to your questions easily
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A certain right triangle [#permalink]  21 May 2010, 02:17
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 ?
A. y > sqrt.2
B. sqrt.3/2 < y < sqrt. 2
C. sqrt.2/3 < y < sqrt.3/2
D. sqrt.3/4 < y < sqrt.2/3
E. y < sqrt.3/4
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Re: A certain right triangle [#permalink]  21 May 2010, 02:24
dimitri92 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. sqrt.3/2 < y < sqrt. 2
C. sqrt.2/3 < y < sqrt.3/2
D. sqrt.3/4 < y < sqrt.2/3
E. y < sqrt.3/4

possible-values-of-side-of-right-angle-triangle-89294.html
Manager
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Re: Possible Values of side of Right Angle Triangle [#permalink]  22 May 2010, 07:27
After all this information and discussions.... I think this one is one of most simple questions.... it seems the answer is obvious....
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Re: Possible Values of side of Right Angle Triangle [#permalink]  10 Mar 2011, 12:20
as x<y<z so z is the hypotenuse
so, 1/2*xy=1
=> xy=2
y>x
=> y^2>xy
=>y^2>2

so y>√2
Ans. A
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Re: Possible Values of side of Right Angle Triangle [#permalink]  10 Mar 2011, 18:39
Solved this intuitively - pls verify this reasoning.
The above stem means that one angle > 45 and another angle is < 45

At x = y (hypothetically) i.e. Both angles equal.
x = z sin (45) = z/ sqrt(2)
y = z cos (45) = z/ sqrt(2)

1/2 x y = 1
Hence z = 2

x = sqrt(2)
y = sqrt(2)

If one angle > 45 and another angle is < 45, And knowing x < y < z
So 0 < x < sqrt(2)
and sqrt(2) < y < Infinity .i.e. when x -> 0 then y -> Infinity
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Re: Possible Values of side of Right Angle Triangle [#permalink]  10 Mar 2011, 20:03
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gmat1220 wrote:
Solved this intuitively - pls verify this reasoning.
The above stem means that one angle > 45 and another angle is < 45

At x = y (hypothetically) i.e. Both angles equal.
x = z sin (45) = z/ sqrt(2)
y = z cos (45) = z/ sqrt(2)

1/2 x y = 1
Hence z = 2

x = sqrt(2)
y = sqrt(2)

If one angle > 45 and another angle is < 45, And knowing x < y < z
So 0 < x < sqrt(2)
and sqrt(2) < y < Infinity

Yes, it can be solved intuitively. It is a very interesting way of thinking. There is one small thing: z may not be 2. e.g. When x = 1 and y = 2, z = sqrt(5)

Think this way:
We know x < y and (1/2)xy = 1 (z is the hypotenuse)
So xy = 2
If x = y, then x = y = sqrt(2)
But since x < y, x is less than sqrt(2) and y is greater than sqrt(2). Makes sense. Good thinking.
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Re: Possible Values of side of Right Angle Triangle   [#permalink] 10 Mar 2011, 20:03
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