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# PS: Geometry

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VP
Joined: 21 Jul 2006
Posts: 1485

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19 Nov 2008, 07:40
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A certain right triangle has sides of length x, y, amd z where x sqrt(2)[/m]

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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Intern
Joined: 19 Nov 2008
Posts: 43

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19 Nov 2008, 08:21
1
KUDOS
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.
Manager
Joined: 14 Oct 2008
Posts: 160

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19 Nov 2008, 09:11
1
KUDOS
Good explanation GMatEnemy. Tarek can you confirm the QA pls ?
Manager
Joined: 23 Jul 2008
Posts: 189

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19 Nov 2008, 09:35
GmatEnemy wrote:
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
I almost panicked
Thanks
Manager
Joined: 08 Aug 2008
Posts: 228

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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.
VP
Joined: 05 Jul 2008
Posts: 1367

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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.
SVP
Joined: 30 Apr 2008
Posts: 1850
Location: Oklahoma City
Schools: Hard Knocks

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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:
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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J Allen Morris
**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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Intern
Joined: 19 Nov 2008
Posts: 43

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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

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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