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PS: Geometry

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PS: Geometry [#permalink]

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New post 19 Nov 2008, 07:40
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A certain right triangle has sides of length x, y, amd 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: PS: Geometry [#permalink]

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New post 19 Nov 2008, 08:21
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The Answer is A.
Just calm down when you encounter such questions. THey will be very simple as long as we
use our common sense. Moment we panic our mind dont work and end up skipping the questions.

The answer explanation goes like this. It is a right angled triangle. Highest side is hypotenuse
SO Z = Hypotenuse. Now area is 1 unit which has to be x*y(Base * height).
and one more condition is y > x . SO .. Y should always be greater than 1 to satisfy
x*y = 1 where y > x. (Just try plugging in numbers)

A is the only choice to give y always greater than 1.

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Re: PS: Geometry [#permalink]

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New post 19 Nov 2008, 09:11
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Good explanation GMatEnemy. Tarek can you confirm the QA pls ?

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Re: PS: Geometry [#permalink]

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New post 19 Nov 2008, 09:35
GmatEnemy wrote:
The Answer is A.
Just calm down when you encounter such questions. THey will be very simple as long as we
use our common sense. Moment we panic our mind dont work and end up skipping the questions.

The answer explanation goes like this. It is a right angled triangle. Highest side is hypotenuse
SO Z = Hypotenuse. Now area is 1 unit which has to be x*y(Base * height).
and one more condition is y > x . SO .. Y should always be greater than 1 to satisfy
x*y = 1 where y > x. (Just try plugging in numbers)

A is the only choice to give y always greater than 1.


more than the solution i liked the tip :-D
I almost panicked
Thanks

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Re: PS: Geometry [#permalink]

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New post 26 Nov 2008, 03:30
A is correct.
infact \(x*y=2\) .assume \(y=x\) from a threshold limit, we get \(y=x=sqrt(2)\)
but we know y>x, so \(y>sqrt(2)\).

also POE results only in A as \(y>1\) as gmatenemy pointed out.

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Re: PS: Geometry [#permalink]

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New post 26 Nov 2008, 15:15
prasun84 wrote:
A is correct.
infact \(x*y=2\) .assume \(y=x\) from a threshold limit, we get \(y=x=sqrt(2)\)
but we know y>x, so \(y>sqrt(2)\).

also POE results only in A as \(y>1\) as gmatenemy pointed out.


Great explanation. I fell short after I reached xy=2. I was trying to come with the opposite angle but badly missed the y>x part. That was easy once I saw your note.

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Re: PS: Geometry [#permalink]

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New post 26 Nov 2008, 17:44
I'm not sure I follow this explanation.

if this is a right triangle, I agree that z is the longest and the other two must be the base and height, but the area of a triangle is

base * height * 0.5

So if the area = 1, then x * y = 2 because: \(\frac{X*Y}{2}=Area\)
Area = 1, then x * y = 2, not 1.

GmatEnemy wrote:
The Answer is A.
Just calm down when you encounter such questions. THey will be very simple as long as we
use our common sense. Moment we panic our mind dont work and end up skipping the questions.

The answer explanation goes like this. It is a right angled triangle. Highest side is hypotenuse
SO Z = Hypotenuse. Now area is 1 unit which has to be x*y(Base * height).
and one more condition is y > x . SO .. Y should always be greater than 1 to satisfy
x*y = 1 where y > x. (Just try plugging in numbers)

A is the only choice to give y always greater than 1.

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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: PS: Geometry [#permalink]

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New post 27 Nov 2008, 00:32
Yep u r rite..

I am sorry for wrong formula. even after it is x*y =2 only A adheres to it

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Re: PS: Geometry   [#permalink] 27 Nov 2008, 00:32
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