text: If the four angles at P where PA, PB, PC, and PD meet -- i.e., the four angles APB, BPC, CPD, and DPA -- are all right angles, then using the Pythagorean formula you get\n\nAB = sqrt(1600) = 40\nBC = sqrt(1808) = sqrt(16 * 113) = 4 sqrt(113)\nCD = sqrt(2809) = 53\nDA = sqrt(2601) = 51\n\nArea = (1/2)(PA PB + PB PC + PC PD + PD PA) = 2002\nPerimeter = AB + BC + CD + DA = 144 + 4sqrt(113)\n = 4 ( 36 + sqrt(113) )\n\nActually I iterated a steepest descent search for the best values for those four angles to minimize the sum of the squared differences of the resulting area from 2002 and the sum of angles from 2 pi.  The result converged to all four angles being equal, i.e., being right angles.  Then I just made that assumption and did the math.\n\nWas there a diagram with the original problem that drew APC and BPD as straight lines intersecting at right angles?