1. **State the problem:** We have rectangle WXYZ with side XY = 10 units and diagonal XZ = 26 units. We need to find how much longer the perimeter of the rectangle is than the diagonal XZ, rounding to the nearest tenth. 2. **Recall properties and formulas:** In a rectangle, opposite sides are equal, and the diagonal forms a right triangle with the sides. Using the Pythagorean theorem: $$XZ^2 = XY^2 + YZ^2$$ where YZ is the other side length. 3. **Find the length of side YZ:** $$26^2 = 10^2 + YZ^2$$ $$676 = 100 + YZ^2$$ $$YZ^2 = 676 - 100 = 576$$ $$YZ = \sqrt{576} = 24$$ 4. **Calculate the perimeter of the rectangle:** Perimeter $P = 2(XY + YZ) = 2(10 + 24) = 2 \times 34 = 68$ 5. **Find how much longer the perimeter is than the diagonal:** $$68 - 26 = 42$$ 6. **Final answer:** The perimeter is 42 units longer than the diagonal XZ. **Answer:** 42 units