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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 07 Sep 2008, 07:15
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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. 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

As per my callcualtions value of y should be < Sqrt(2)?????????
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Re: PS - Triangle [#permalink] New post 07 Sep 2008, 08:09
vivektripathi 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

As per my callcualtions value of y should be < Sqrt(2)?????????


Z= hypotenus
area 1=1/2 xy
xy=2
as x<y then it is clearly y>1

options C,D,E are out.. because they contain values y<1..

between A and B.
I can easily ruled out A. because it says y>sqrt(2)... which means it include .. large value... near 1 billion or greater also...

Will go for B.

What is OA.
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Re: PS - Triangle [#permalink] New post 07 Sep 2008, 08:10
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vivektripathi 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

As per my callcualtions value of y should be < Sqrt(2)?????????


A. y > sqrt2.

area = 1/2(xy) cuz z is hypotenous. therefore x and y must be p and b.
1= 1/2(xy)
2 = xy
since xy = 2, and x < y, x and y cannot be equal but they are equal when each is sqrt2.
since y > x, y should be > sqrt2.
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Re: PS - Triangle [#permalink] New post 07 Sep 2008, 08:37
vivektripathi 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

As per my callcualtions value of y should be < Sqrt(2)?????????


A

(1/2)*x*y = 1
x*y = 2

If x < y, then y must be greater than the value at which x = y and x*y = 2.
This occurs at sqrt(2), so y must be greater than sqrt(2).

There is no upper bound for y. As y --> infininty, x --> 0 such that x*y = 2
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Re: PS - Triangle [#permalink] New post 07 Sep 2008, 11:27
zoinnk wrote:
vivektripathi 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

As per my callcualtions value of y should be < Sqrt(2)?????????


A

(1/2)*x*y = 1
x*y = 2

If x < y, then y must be greater than the value at which x = y and x*y = 2.
This occurs at sqrt(2), so y must be greater than sqrt(2).

There is no upper bound for y. As y --> infininty, x --> 0 such that x*y = 2


YOU ARE RIGHT!! SILLY MISTAKE AGAIN.
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Re: PS - Triangle [#permalink] New post 08 Sep 2008, 19:42
Thanks GMAT TIGER, very good explanation
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Re: PS - Triangle [#permalink] New post 08 Sep 2008, 21:32
IMO A

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

now, y>x=> y > 2/y => y^2> 2 => y > sqrt(2)
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Re: PS - Triangle [#permalink] New post 09 Sep 2008, 00:43
ssandeepan wrote:
IMO A

area =1/2*xy=1 => xy=2 = > x=2/ynow, y>x=> y > 2/y => y^2> 2 => y > sqrt(2)


hell man!!!
i did the same similar step, but from there on did not know how to proceed. thanks sandeep.
Re: PS - Triangle   [#permalink] 09 Sep 2008, 00:43
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