Isn't this a pentagon?
I guess you can set the angle E or C to be 180 degrees, and you will have a quadrilateral with the same surface as the figure.
But I how do we prove that the surface of rectangle EQE1 = the Surface of the Rectangle QDC
The way question is worded does not prevent using an isoscelesCode:E E F D A D A C B C B
triangle on top of a square, then made into a rectangle.
Square ABCD sides = 6, AE = DE = 5 ; so area ABCDE = 36 + 12 = 48.
Rectangle becomes a 8 by 6 (ADEF being 6 by 2), so area 48.