I would like to know whether the following stuff is known. Well known: There does not exist a set of rectangles which will tile both a 1 by 2 rectangle and a square of side sqrt(2) (although they both have the same area). We say these sets are not EQUI-TILABLE. GENERAL PROBLEM: Let S be a union of a finite set of rectangles and let S' be another such set. Under what conditions are S and S' equi-tilable. ANSWER: Think of reals as a vector space over Q. For a rectangle of width a and height b we can form their TENSOR PRODUCT a(x)b. The TENSOR AREA of the set S is summation(a_i(x)b_i) summed over the rectangles of S. THEOREM. The sets S and S' are equi-tilable if and only if they have the same tensor area. The necessity of the condition is easy and probably known. It's the sufficiency which is interesting. EXAMPLE; a,b,c,d,e are rationally independent positive numbers S consists of two rectangles with (width, height) (a,b-c) and (a-c,d) S' is three rectangles (e,b),(a-e,b-c),(a-e-d,c) . One verifies the tensor areas are equal. Our proof of sufficiency is constructive showing how to find the tiles that do the job. (I have some pretty pictures if anyone is interested) [ Actually it matters whether we or not we use SYMMETRIC tensor product where we have a(x)b=b(x)a. This corresponds to allowing interchanging height and width in the tiling. (or of rotating tiles through 90 degrees). Refernces?? David G