I think it makes a pretty fine problem

http://farm4.static.flickr.com/3167/...722c9d.jpg?v=0

Let's say the length of rectangle ABCD is L, and the width is W. If r is the radius of circle E, then the length of rectangle EFDG is (L-r) and the width is (W-r).

We can split triangle ABC into three smaller triangles: AEC, AEB and BEC. Each has an altitude of r.

http://farm4.static.flickr.com/3192/...70c17f.jpg?v=0

area(ABC) = area(AEC) + area(AEB) + area(BEC)

$\displaystyle

\frac {LW} {2} = \frac {r \sqrt {L^2 + W^2} }{2} + \frac {Lr} {2} + \frac {Wr} {2}

$

$\displaystyle

r = \frac {LW} {L + W + \sqrt {L^2 + W^2} }

$

So the area of rectangle EFDG is

$\displaystyle

A_{EFDG} = (L - \frac {LW} {L + W + \sqrt {L^2 + W^2} } ) (W - \frac {LW} {L + W + \sqrt {L^2 + W^2} } )

$

Whoa. So I suppose unless we are provided with the L/W ratio, we will not have a pretty answer for this problem.