In triangle HAC, using Pythagoras's Theorem, the hypotenuse HC =Sqrt(8)
In triangle AHX, using Pythagoras's Theorem, HX=Sqrt(2)
In triangle BCY, using Pythagoras's Theorem, CY=1/sqrt(2)
Therefore, XY =HC - HX - CY =XY =sqrt(8) - sqrt(2) - 1/sqrt(2) =1/sqrt(2) =WZ
By similar reasoning, XW=YZ = 1/sqrt(2)
Therefore, the area of the quadrilateral WXYZ =[1/sqrt(2)]^2 =1/2