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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. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

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.

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%

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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$$.

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.

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)

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%

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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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