The area of square $A B C D$ is 64 and $A X=B W=C Z=D Y=2$. What is the area of square WXYZ?
(A) 58
(B) 52
(C) 48
(D) 40
(E) 36

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-082.jpg?height=309&width=306&top_left_y=371&top_left_x=1408)

Part C: Each correct answer is worth 8.
Solution 1

Since the area of square $A B C D$ is 64 , then the side length of square $A B C D$ is 8 . Since $A X=B W=C Z=D Y=2$, then $A W=B Z=C Y=D X=6$. Thus, each of triangles $X A W, W B Z, Z C Y$ and $Y D X$ is right-angled with one leg of length 2 and the other of length 6 . Therefore, each of these four triangles has area $\frac{1}{2}(2)(6)=6$. Therefore, the area of square $W X Y Z$ is equal to the area of square $A B C D$ minus the sum of the areas of the four triangles, or $64-4(6)=40$.

## Solution 2

Since the area of square $A B C D$ is 64 , then the side length of square $A B C D$ is 8 . Since $A X=B W=C Z=D Y=2$, then $A W=B Z=C Y=D X=6$.

By the Pythagorean Theorem,

$X W=W Z=Z Y=Y X=\sqrt{2^{2}+6^{2}}=\sqrt{4+36}=\sqrt{40}$.

Therefore, the area of square WXYZ is $(\sqrt{40})^{2}=40$.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-354.jpg?height=732&width=272&top_left_y=747&top_left_x=1400)

Answer: (D)

## Part C