1. **State the problem:** We need to find the lateral surface area of a regular quadrilateral pyramid with a base perimeter of 40 cm and a height of 12 cm. 2. **Identify the base shape and dimensions:** A regular quadrilateral pyramid has a square base. The perimeter $P$ of the base is given as 40 cm. 3. **Calculate the side length of the base:** Since the base is a square with 4 equal sides, $$\text{side length} = \frac{P}{4} = \frac{40}{4} = 10 \text{ cm}.$$ 4. **Calculate the slant height $l$:** The height $h$ of the pyramid is 12 cm. The slant height is the height of each triangular lateral face. To find $l$, use the Pythagorean theorem in the right triangle formed by the height, half the base side, and the slant height: $$l = \sqrt{h^2 + \left(\frac{\text{side length}}{2}\right)^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}.$$ 5. **Formula for lateral surface area $A_{lat}$:** $$A_{lat} = \frac{1}{2} \times \text{perimeter} \times \text{slant height} = \frac{1}{2} \times 40 \times 13.$$ 6. **Calculate lateral surface area:** $$A_{lat} = 20 \times 13 = 260 \text{ cm}^2.$$ **Final answer:** The lateral surface area of the pyramid is $260$ cm$^2$.