To begin, the total of the perimeters of rectangles X Y and Z is 46 cm. This is equal to 2(A+B+C+2D+E). So A+B+C+2D+E is 23 centimetres. If we can find the value of D, we can solve the problem. Knowing that rectangles X and Y have equal perimeters and X and Z have equal areas, we can form a system of equations below. There are only 2 possible values of D, 3 cm and 13 cm. 13 cm obviously doesn’t work. D=3, and so A+B+C+D+E = 20 cm.
A+B=C+D\\AB=DE\\A+B+C+2D+E=23\\AB+CD+DE=39\\[8pt] \\D(C+2E)=39\\C,D,E\in\Z\\\Rightarrow D=3Month: January 2020
Radical Rectangles
This challenges covers geometry and algebra.
We have 3 rectangles, X Y and Z. Rectangle X has sides with lengths A and B. Rectangle Y, C and D, and rectangle Z, D and E. Rectangle X has the same perimeter as rectangle Y and the same area as rectangle Z. The total perimeter of the three rectangles is 46 centimetres, and the total area, 39 square centimetres. All sides have integral centimetre lengths. Find the sum A+B+C+D+E.