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# The shaded area is what percentage of the area of the

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The shaded area is what percentage of the area of the [#permalink]  03 Jul 2008, 18:23
The shaded area is what percentage of the area of the quadrilateral ABCD?

AB || DC and BC || AD

Use 3.14 for $$\pi$$
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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$$. GMAT Club Premium Membership - big benefits and savings Last edited by jallenmorris on 03 Jul 2008, 18:32, edited 1 time in total.  Kaplan Promo Code Knewton GMAT Discount Codes Manhattan GMAT Discount Codes Director Joined: 01 Jan 2008 Posts: 629 Followers: 3 Kudos [?]: 140 [0], given: 1 Re: Geometry Proportions [#permalink] 03 Jul 2008, 18:31 insufficient info. slide AD down along lines AB and CD. shaded area stays the same but area of ABCD increases. SVP Joined: 30 Apr 2008 Posts: 1889 Location: Oklahoma City Schools: Hard Knocks Followers: 35 Kudos [?]: 473 [0], given: 32 Re: Geometry Proportions [#permalink] 03 Jul 2008, 18:33 I guarantee you it's not insufficient info. I don't understand what you are saying as supporting evidence that there is insufficient evidence. maratikus wrote: insufficient info. slide AD down along lines AB and CD. shaded area stays the same but area of ABCD increases. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: Geometry Proportions [#permalink]  03 Jul 2008, 18:59
jallenmorris wrote:
The shaded area is what percentage of the area of the quadrilateral ABCD?

AB || DC and BC || AD

Use 3.14 for $$\pi$$
Attachment:
GMATQuestion.gif

I have solved it somewhere halfway, unable to express the base in terms of radius.

let r be the radius of he circle, then shaded area is = (2*r)^2 - Pi * r^2
= 4 * r^2 - 3 * r^2 ~= r^2

Considering BA to be the base of the parallelogram,
Height equals 4*r

So area of parallelogram = BA * (4 *r)

Ratio of shaded area/ area of parallelogram = r^2/ (BA) * (4*r) .................

I am trying to use the info about angeles now...
SVP
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Re: Geometry Proportions [#permalink]  03 Jul 2008, 19:15
I wrote the question and I'll post my explanation tomorrow.
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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$$. GMAT Club Premium Membership - big benefits and savings SVP Joined: 29 Aug 2007 Posts: 2493 Followers: 59 Kudos [?]: 579 [0], given: 19 Re: Geometry Proportions [#permalink] 03 Jul 2008, 21:32 jallenmorris wrote: I guarantee you it's not insufficient info. I don't understand what you are saying as supporting evidence that there is insufficient evidence. maratikus wrote: insufficient info. slide AD down along lines AB and CD. shaded area stays the same but area of ABCD increases. with out having length and bredth of the quadilateral, you cannot find the ratios of the areas asked in the question. if you slide down AD to futher down, the area of the quadilateral changes without changing the area of shaded region. _________________ GMAT Tutor Joined: 24 Jun 2008 Posts: 1173 Followers: 303 Kudos [?]: 967 [0], given: 4 Re: Geometry Proportions [#permalink] 04 Jul 2008, 05:26 Expert's post jallenmorris wrote: I guarantee you it's not insufficient info. I don't understand what you are saying as supporting evidence that there is insufficient evidence. maratikus wrote: insufficient info. slide AD down along lines AB and CD. shaded area stays the same but area of ABCD increases. It's certainly insufficient info as presented. If the diameter of each circle is 1, we can find the length of BC. Still, CD could be 10 or could be 10,000,000,000. Changing CD will clearly affect the area of the quadrilateral without affecting the area of the shaded region, and therefore clearly changes the ratio in question. _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Current Student Joined: 12 Jun 2008 Posts: 287 Schools: INSEAD Class of July '10 Followers: 6 Kudos [?]: 40 [0], given: 0 Re: Geometry Proportions [#permalink] 04 Jul 2008, 05:50 Information is insufficient. If we call r the radius of the circles, we can find the value of the shaded are. It is (4-Pi)r^2 (area of the square formed by the 4 circle centers - 4* one fourth of the circle areas) Then we can write BC in function of r : BC = 2r/cos(37°) But we have no information about AB or the height of the parallelogram we need. As it was said above, you could slide AD further down along the lines (AB) and (CD) and it would change the area of the parallelogram without changing the shaded area. SVP Joined: 30 Apr 2008 Posts: 1889 Location: Oklahoma City Schools: Hard Knocks Followers: 35 Kudos [?]: 473 [0], given: 32 Re: Geometry Proportions [#permalink] 04 Jul 2008, 06:04 I changed the drawing as I realized I needed to show that point B was directly over point D. Here is my explanation: Now that we know B is directly above D, we know that Triangle ABD and Triangle CDB are identical. (Still same info that AB || DC and BC || AD.) This also shows that Angle BDC is 37 degrees and BCD is 53 degrees, and DBC is 90. [37 + 53 + 90 = 180] These angles are the interior angles of a 3:4:5 triangle. If we make AD = 30, then we can figure out the height of the circles. If you imagine a line running from a point on AB to point C that is perpendicular to AB, that will create another 3:4:5 triangle. Call this point on AB point E. Now we can determine the height of the circles and the other formulas posted are correct from there. The answer should be 2.58% Attachment: GMATQuestion3.gif [ 11.18 KiB | Viewed 1264 times ] Attachments GMATQuestion2.gif [ 9.79 KiB | Viewed 1274 times ] _________________ ------------------------------------ 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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Last edited by jallenmorris on 04 Jul 2008, 06:17, edited 1 time in total.
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Re: Geometry Proportions [#permalink]  04 Jul 2008, 06:13
Clearly there is one thing missing here... the 5 answer choices which are available in real GMAT
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Re: Geometry Proportions [#permalink]  04 Jul 2008, 06:18
Wouldn't that be 5 things missing then?;-)

hi0parag wrote:
Clearly there is one thing missing here... the 5 answer choices which are available in real GMAT

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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$$. GMAT Club Premium Membership - big benefits and savings GMAT Instructor Joined: 04 Jul 2006 Posts: 1266 Location: Madrid Followers: 24 Kudos [?]: 167 [0], given: 0 Re: Geometry Proportions [#permalink] 04 Jul 2008, 06:33 This would make a great GMAT question if you gave the dimensions of triangle ADE or the ratio of AD to DE. Good work, Mr. Morris! Manager Joined: 27 Apr 2008 Posts: 110 Followers: 1 Kudos [?]: 8 [0], given: 0 Re: Geometry Proportions [#permalink] 04 Jul 2008, 06:43 the answer is 2.4%, but i used trigonometry and excel. I dont think it is a GMAT question Intern Joined: 01 Jul 2008 Posts: 41 Followers: 1 Kudos [?]: 9 [0], given: 20 Re: Geometry Proportions [#permalink] 04 Jul 2008, 06:56 I think the assumption over here is that everyone knows that a triangle with 37deg:53deg:90deg angles is approximately a 3-4-5 triangle. And that the line forming the height of the triangle will always form another 3-4-5 triangle. Intern Joined: 01 Jul 2008 Posts: 41 Followers: 1 Kudos [?]: 9 [0], given: 20 Re: Geometry Proportions [#permalink] 04 Jul 2008, 07:09 Someone has done some work on http://www.physicsforums.com/archive/in ... 60382.html and mentions that the mesure of the smallest angle in a 3-4-5 triangle is 36.87 deg Current Student Joined: 12 Jun 2008 Posts: 287 Schools: INSEAD Class of July '10 Followers: 6 Kudos [?]: 40 [0], given: 0 Re: Geometry Proportions [#permalink] 04 Jul 2008, 07:25 jallenmorris wrote: I changed the drawing as I realized I needed to show that point B was directly over point D. In this case, to pursue the reasoning I had above and without any approximation, we have BD = 2r/sin(37°) We had BC = 2r/cos(37°) and shaded area = (4-Pi)*r^2 Then the area of the parallelogram is BD*BC = 4*r^2/(cos(37°)*sin(37°)) The ratio we want is then $$\frac{(4-\Pi)cos(37^o)sin(37^o)}{4}$$ or $$(1-\frac{\Pi}{4})cos(37^o)sin(37^o)$$ which numerically is about 10.3% (sorry I don't find 2.4% or 2.58%) Intern Joined: 01 Jul 2008 Posts: 41 Followers: 1 Kudos [?]: 9 [0], given: 20 Re: Geometry Proportions [#permalink] 04 Jul 2008, 08:02 using the assumption of 3-4-5 triangles and that AD=30 the answer turns out to be 10.32% here are my calculations: Darkened Area= (2r)^2 - pi r^2 = r^2 (4-pi) = 0.86 r^2 = 0.86 (h/4)^2 = 0.86 (6)^2 = 0.86*36 = A1 (say) (as pi-3.14) AD = 30 => AB = 50 (as ABD is a 3-4-5 triangle) AD = 30 => ED = 24 (as EAD is a 3-4-5 triangle) hence, Area of parallelogram ABCD = hl = ED*AB = 24*50 =A2 (say) Therefore, A1/A2*100 = 10.32% SVP Joined: 30 Apr 2008 Posts: 1889 Location: Oklahoma City Schools: Hard Knocks Followers: 35 Kudos [?]: 473 [0], given: 32 Re: Geometry Proportions [#permalink] 04 Jul 2008, 10:03 Here is what I came up with. I'll figure it up again and show all of my work. If AD = 30, and these are 3:4:5 traingles (which 37 & 53 are the nearest integers as degrees rather than to the nearest hundreth of a degree). If AD = 30, then ED will be 24 and AE will be 18. So we know the height of 2 circles = 24, height of 1 = 12, radius of 1 = 6 If you were to draw a line from the midpoint of each circle to the midpoint of the other circles to create a square around the shaded region, one side of that square would be 12, so the area of that square = 144 The area of that square ecloses 1/4 of each circle, or what is equivalent to the area of 1 circle ($$\pi r^2$$) 6^2$$\pi$$ = 36$$\pi$$ so the area of the shaded region should be 144 - 36$$\pi$$. As for the area of the parallelogram, if Angle BDC = 37 degrees and DBC = 90 then BCD = 53. And BCD = DAB, so we know that these larger trianlges are also 3:4:5 traingles. if AD = 30 and it's the "3" side, then BD (height of the parallelogram) is 40. The area of the parallelogram then is 1200. For to answer the question "What percentage is the shaded region of the parallelgram?" We have: $$\frac{144-36\pi}{1200}$$ which reduces to $$\frac{3(4-\pi)}{100}$$ but because the question asks for the percentage, we can do this $$\frac{3(4-\pi)}{100}*\frac{100}{1}$$ and we are left with $$3(4-\pi)$$ 3*.86 = 2.58% Attachment: GMATQuestion3.gif [ 11.18 KiB | Viewed 1167 times ] Attachments GMATQuestion4.gif [ 14.07 KiB | Viewed 1159 times ] _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: Geometry Proportions [#permalink]  04 Jul 2008, 10:21
I have edited the picture yet again. I think this is the best way to ask the question. What do you guys think?

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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: Geometry Proportions [#permalink]  04 Jul 2008, 14:45
jallenmorris wrote:
I have edited the picture yet again. I think this is the best way to ask the question. What do you guys think?

I think it doesn't match with my "no approximation, no assuming of 3:4:5, no picking number" solution

So either I did something wrong, or you did (we have a factor 4 difference)
Re: Geometry Proportions   [#permalink] 04 Jul 2008, 14:45

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