insufficient info. slide AD down along lines AB and CD. shaded area stays the same but area of ABCD increases.

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...

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.
_________________

Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

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.

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

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!

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.

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%)

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%


_________________

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

Find out what's new at GMAT Club - latest features and updates

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)