I dropped a perpendicular from D to CE (call it DF) to make another 3-4-5 triangle with hypotenuse on the right side of the square. Then I dropped a perpendicular from A to DF (AG) to form a 3-4-5 triangle with hypotenuse on the top of square. This gives you two 3-4-5 triangles inscribed in the quadrilateral (CDF and ADG), and the remainder forms a trapezoid with bases 1 & 4, height of 1 (AEFG). Areas are 6 + 6 + 2.5 = 14.5, and that's the answer. EDIT: This could also be considered using triangle rotation on the solve, since you are just forming two more congruent triangles.
Another solution is: 1. Find area of BEC=(3*4)/2=6 2. Draw perpendicular line from side BC to point E, EF 3. Find value of EF using area of BEC, EF is 2.4 4. Calculate BF using Pythagorean theorem, it is 1.8 5. BF is height in AEB from side AB, can find area of AEB=(5*1.8)/2=4.5 6. Area of AECD=25-4.5-6=14.5
