1. **State the problem:** We have quadrilateral PQRS with vertical sides PQ = 3 and RS = 3, and the total area is 6. We need to find the length of diagonal PR. 2. **Analyze the problem:** Since PQ and RS are vertical and equal, PQRS is a trapezoid with bases PS and QR parallel. The height (distance between PQ and RS) is 3. 3. **Use the area formula for trapezoid:** $$\text{Area} = \frac{(\text{base}_1 + \text{base}_2)}{2} \times \text{height}$$ Here, bases are PS and QR, height is 3, and area is 6. 4. **Set up the equation:** $$6 = \frac{(PS + QR)}{2} \times 3$$ 5. **Solve for sum of bases:** $$6 = \frac{3}{2} (PS + QR)$$ Multiply both sides by \frac{2}{3}: $$\cancel{6} \times \frac{2}{3} = \cancel{\frac{3}{2}} (PS + QR) \times \frac{2}{3}$$ $$4 = PS + QR$$ 6. **Find diagonal PR:** Since PQRS is a trapezoid with vertical sides PQ and RS both 3, and bases PS and QR summing to 4, diagonal PR forms a right triangle with vertical leg 3 and horizontal leg equal to the difference between PS and QR. 7. **Assuming PS and QR are equal (each 2) for simplicity:** Then horizontal distance between P and R is 2. 8. **Calculate diagonal PR using Pythagoras theorem:** $$PR = \sqrt{(\text{vertical side})^2 + (\text{horizontal side})^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$$ **Final answer:** $$\boxed{\sqrt{13}}$$