Problem:
In rectangle ADEH, points B and C trisect AD, and points G and F trisect HE. In addition, AH=AC=2. What is the area of quadrilateral WXYZ shown in the figure?
Answer Choices:
A. 21​
B. 22​​
C. 23​​
D. 322​​
E. 323​​
Solution:
First note that since points B and C trisect AD, and points G and F trisect HE, we have HG=GF=FE=AB=BC=CD=1. Also, HG is parallel to CD and HG=CD, so CDGH is a parallelogram. Similarly, AB is parallel to FE and AB=FE, so ABEF is a parallelogram. As a consequence, WXYZ is a parallelogram, and since HG=CD=AB=FE, it is a rhombus.
Since AH=AC=2, the rectangle ACFH is a square of side length 2. Its diagonals AF and CH have length 22​ and form a right angle at X. As a consequence, WXYZ is a square. In isosceles △HXF we have HX=XF=2​. In addition, HG=21​HF. So XW=21​XF=21​2​, and the square WXYZ has area XW2=(A)21​​.
