$\square XYZW$ is a rectangle. If $XY + YZ = 17$ and $XZ + YW = 26$,find $XY$ and $YZ$ (given $XY > YZ$).

(A) In a rectangle,the diagonals are equal in length,so $XZ = YW$. Given $XZ + YW = 26$,we have $2XZ = 26$,which implies $XZ = 13$.
In the right-angled triangle $\triangle XYZ$,by the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2 = 13^2 = 169$.
We are given $XY + YZ = 17$. Squaring both sides,we get $(XY + YZ)^2 = 17^2 = 289$.
Expanding this,$XY^2 + YZ^2 + 2(XY \cdot YZ) = 289$.
Substituting $XY^2 + YZ^2 = 169$,we get $169 + 2(XY \cdot YZ) = 289$,so $2(XY \cdot YZ) = 120$,which means $XY \cdot YZ = 60$.
We need two numbers that add to $17$ and multiply to $60$. These are $12$ and $5$.
Since $XY > YZ$,we have $XY = 12$ and $YZ = 5$.