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The shaded area is what percentage of the area of the

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Re: Geometry Proportions [#permalink] New post 04 Jul 2008, 15:05
I think it is me who made made a mistake ;)

Actually it is not BC = 2r/cos(37°) but BC = 4r/cos(37°). Same thing BD = 4r/sin(37°)

(hence the factor 4 difference: I wrote radius for diameter...)

Then the area of the parallelogram is BD*BC = 16*r^2/(cos(37°)*sin(37°))

The ratio we want is then \frac{(4-\Pi)cos(37^o)sin(37^o)}{16} (exact answer, no approximation of any kind)

which numerically is about 2.58% :-D
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Re: Geometry Proportions [#permalink] New post 04 Jul 2008, 15:11
Hi, jallenmorris,

The last version of the pic., again, does not give sufficient info:
you can change the angle BAD, and obtain different values of parallelogram’s area.

The fact that in your picture B is strictly above D does not help: it could be merely a coincidence. So in the end the picture doesn’t let us to correctly infer that ABD is right triangle. On real GMAT pictures, all measures that intended to be fixed, are explicitly fixed.

So I think this would be somewhat more proper picture:

Attachment:
GMATQuestionBestPicture_edit.gif
GMATQuestionBestPicture_edit.gif [ 6.48 KiB | Viewed 350 times ]


Also, you need to state that ABCD is parallelogram.
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Re: Geometry Proportions [#permalink] New post 04 Jul 2008, 19:03
If you read the first post, I do state it's a parallelogram by stating AD || BC and AB || DC.

Attachment:
GmatQuestion_v_3.14.gif
GmatQuestion_v_3.14.gif [ 6.8 KiB | Viewed 336 times ]


greenoak wrote:
Hi, jallenmorris,

The last version of the pic., again, does not give sufficient info:
you can change the angle BAD, and obtain different values of parallelogram’s area.

The fact that in your picture B is strictly above D does not help: it could be merely a coincidence. So in the end the picture doesn’t let us to correctly infer that ABD is right triangle. On real GMAT pictures, all measures that intended to be fixed, are explicitly fixed.

So I think this would be somewhat more proper picture:

Attachment:
The attachment GMATQuestionBestPicture_edit.gif is no longer available


Also, you need to state that ABCD is parallelogram.

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Re: Geometry Proportions [#permalink] New post 06 Jul 2008, 01:19
Oski wrote:
I think it is me who made made a mistake ;)

Question for you english speakers: Is this correct?

Do you say "it is me who made made a mistake" or do you say "it is I who made made a mistake"?

Thanks!
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Re: Geometry Proportions [#permalink] New post 06 Jul 2008, 06:35
I believe this is the correct format:

"It is I who made a mistake." - Use I and only 1 "made".
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Re: Geometry Proportions [#permalink] New post 06 Jul 2008, 06:50
jallenmorris wrote:
I believe this is the correct format:

"It is I who made a mistake." - Use I and only 1 "made".

Sorry for the double made ;)

And thank you for your answer for "I" :-D !
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Re: Geometry Proportions [#permalink] New post 06 Jul 2008, 06:54
also, realize this....even though the proper form is "It is I who made a mistake." I doubt many native english speakers actually say this. Most people are going to say "It was me that made the mistake." or they'll use any other form of saying it is their fault.
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Re: Geometry Proportions [#permalink] New post 06 Jul 2008, 06:59
jallenmorris wrote:
also, realize this....even though the proper form is "It is I who made a mistake." I doubt many native english speakers actually say this. Most people are going to say "It was me that made the mistake." or they'll use any other form of saying it is their fault.

"It is I who made a mistake" but "It is me that made a mistake"... No wonders I have difficulties with the sentence correction questions...

But thanks for your patience with this ! :-D
Re: Geometry Proportions   [#permalink] 06 Jul 2008, 06:59
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