1. **State the problem:** We have a square VWXY with vertices labeled such that side YX has length $a+26$ and side XW has length $3a$. We need to find the length of side $XY$. 2. **Recall properties of a square:** All sides of a square are equal in length. Therefore, the lengths of sides YX and XW must be equal. 3. **Set up the equation:** Since $YX = a+26$ and $XW = 3a$, we have $$a + 26 = 3a$$ 4. **Solve for $a$:** $$a + 26 = 3a$$ $$26 = 3a - a$$ $$26 = 2a$$ $$a = \frac{26}{2}$$ $$a = 13$$ 5. **Find the length of side $XY$:** Since $XY$ is a side of the square, it equals $YX$ or $XW$. Using $YX = a + 26$: $$XY = 13 + 26 = 39$$ **Final answer:** $$\boxed{39}$$