Help to solve this exercise

The circles are easy, right?
You can calculate the long diagonal with the Pythagorean Theorem.
Other ideas?
I'm tempted to draw all the radii from the four centers to the nearest rhombus side.
I'm also tempted to draw the four segments between the adjacent circle centers. Is there similarity in there?

Let's see your work. What have you tried?
 
[MATH]bh - 4\pi r^2[/MATH]
where [MATH]b[/MATH] and [MATH]h[/MATH] are the base and the height of the Parallelogram respectively. And [MATH]r[/MATH] is the radius of any circle
 
Finding the base may be the hardest part. A useful first step will be to draw in the short diagonal and realize what kind of triangles you have formed. After that, there are several ways to go. (Eventually you might realize that they gave you too much information, and the 7.5 and 12.85 are only approximations - neither as accurate as claimed.)
 
three_circles.png

with some approximations and adjusting, i found that the angle in the picture above [MATH]\approx 30°[/MATH]
And with some smart geometry, you can prove that it is exactly [MATH]30°[/MATH]
then, the base of the parallelogram is

[MATH]12.85 - 7.5 \tan 30° = 8.52 \ mm[/MATH]
then

the required area [MATH]= bh - 4\pi r^2 = (8.52)(7.5) - 4\pi (2)^2 = 13.63 \ mm^2[/MATH]
 
That's all true, but we'd like the OP to have a chance to do some thinking ...

Did you observe that you can get that 30 degree angle directly from the given lengths, with almost no work? Have you worked out what those lengths should really be, taking only the radius as given?
 
proof that the angle is 30°

three_circles02.png

it becomes obvious that it is 30° when numbers are in front of your eyes

what do you mean by OP?
 
joshuaa said:
proof that the angle is 30°

View attachment 21336

it becomes obvious that it is 30° when numbers are in front of your eyes
Click to expand...
I was referring to getting the same angle (approximately) as arctan(7.5/12.85); but then also that the 7.5 and the 12.85 can be obtained without being given either of them. I imagine the former is probably what the author of the problem intended to be done, but that angle is not the one you need to find the base, so I think you need to see the equilateral triangles anyway. The problem is much more interesting with only the diameter given.
 
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