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Length of a Triangle leg

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Length of a Triangle leg

by joconnor » Mon Dec 19, 2011 6:22 am
A certain right triangle has sides of length x, y, z where x<y<z. Area of tri is 1. Which indicates all the possible values of y?

y > sq rt 2

sq rt 3 / 2 < y < sq rt 2

sq rt 2 / 3 < y < sq rt 3 / 2

sqrt3 / 4 < y < sqrt2 / 3

y < sqrt3 / 4

thanks!
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by user123321 » Mon Dec 19, 2011 6:52 am
joconnor wrote:A certain right triangle has sides of length x, y, z where x<y<z. Area of tri is 1. Which indicates all the possible values of y?

y > sq rt 2

sq rt 3 / 2 < y < sq rt 2

sq rt 2 / 3 < y < sq rt 3 / 2

sqrt3 / 4 < y < sqrt2 / 3

y < sqrt3 / 4

thanks!
[spoiler]Is it A?[/spoiler]

in a right angled triangle, the area with above conditions is 1/2 * x * y
=> xy/2 = 1 (from given)
xy = 2
since x<y
y should be atleast greater than sqrt(2) to follow that condition.

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by neelgandham » Mon Dec 19, 2011 6:56 am
A certain right triangle has sides of length x, y, z where x<y<z. Area of tri is 1. Which indicates all the possible values of y?.

Since x<y<z and length of Hypotenuse > length of the side(any side), we know that z is the length of the hypotenuse.
Area of triangle = 1/2 * base * height = 1/2 (Side X) * (Side Y) = (1/2)*x*y = 1 Implies xy = 2 and x = 2/y

From the question x<y.
Implies 2/y<y
Implies y-(2/y) > 0
Implies (y^2-2)/y > 0
Implies (y-√2)(y+√2)/y > 0
Checkpoints here are at y = √2, -√2 and 0
Since y is length of a side of a triangle, y > 0, so we need not check for y < 0
Condition 1
0<y<√2
(y-√2)(y+√2)/y < 0, doesn't satisfy the inequality.
Condition 2
y>√2
(y-√2)(y+√2)/y > 0, satisfies the inequality.

IMO option A
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by joconnor » Mon Dec 19, 2011 7:33 am
answer is a

thanks!
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by Brent@GMATPrepNow » Mon Dec 19, 2011 7:42 am
joconnor wrote:A certain right triangle has sides of length x, y, z where x<y<z. Area of tri is 1. Which indicates all the possible values of y?
y > sq rt 2
sq rt 3 / 2 < y < sq rt 2
sq rt 2 / 3 < y < sq rt 3 / 2
sqrt3 / 4 < y < sqrt2 / 3
y < sqrt3 / 4
thanks!
A quick way to solve this question is to recognize that the area of 1 has no effect on limiting the maximum length of y.

As the other two posts suggest, we can see that z is the length of the hypotenuse, and y is the longer of the two legs of the right triangle.
If the area is 1, then xy/2 = 1 with means xy=2
Given this, we could have x=0.1 and y=20, or we could have x=0.001 and y=2000, and so on.

As you can see, we can make y as long as we wish.
As such, we can eliminate answer choices B, C, D and E since they all limit the value of y.

Cheers,
Brent
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