Should there be an elegant solution or is it a big mess of radicals and inverse trig functions?

Ok, for real this time. Let \ell be the length of the rectangle.

We know r\ell = 0.5\pi r^2 and so \ell = 0.5\pi r

Moreover, it'll be useful to write: \ell - r = (0.5\pi - 1)r

The angle of the "pizza slice" that the green region covers is:

\theta = \cos^{-1} \left( \frac{ \ell - r }{r} \right) = \cos^{-1} (0.5\pi -1) 

The area of the pizza slice is 0.5\theta r^2 and the "height" of the green region is r \sin \theta .

Thus:

0.5\theta r^2 - 0.5(\ell - r)r\sin \theta = 10

Or:

\left[ 0.5 \theta - 0.5(0.5\pi - 1)\sin \theta \right] r^2 = 10

Which gives r = 6.3587

The green area is half a circle segment with height h.

The area of the rectangle is r*(r + r - h).

The area of the semicircle is (pi * r^2) / 2.

Since they are equal, you get h = 2*r - (pi * r)/2

Insert that h into the circular segment area based on height h formula and you get

This is another level. I'm not even going to try!

i have a bet with an older friend that i will pass the differential calculus exam when i reach the age at which he failed it (about 40yo), after having the same amount of study time. i shall also pass the integral calculus exam, because it's not about the age, but the amount of cobwebs in the brain.

We should have a button for MathJax syntax like the markdown instructions button in the upper right of the text area input. I am familiar with markdown, but not MathJax.