1. **State the problem:** We need to find the total surface area of the pentagonal prism (pedestal) to determine how much tile is required to cover it. 2. **Identify the shape and dimensions:** The prism has a height of $20$ cm. The base is a pentagon composed of: - Two adjacent sides of $12$ cm each, - One side of $14$ cm, - Two small triangles on top and bottom with side length $8$ cm. From the net, the central rectangle is $20$ cm by $14$ cm. 3. **Surface area components:** - The lateral faces correspond to the rectangle and attached triangles. - The two pentagonal bases (top and bottom) are congruent. 4. **Calculate the area of the rectangular face:** $$\text{Area}_{rectangle} = 20 \times 14 = 280 \text{ cm}^2$$ 5. **Calculate the area of the two large triangles on the sides:** Each triangle has sides $12$, $12$, and $14$ cm. Use Heron's formula: $$s = \frac{12 + 12 + 14}{2} = 19$$ $$\text{Area} = \sqrt{s(s-12)(s-12)(s-14)} = \sqrt{19 \times 7 \times 7 \times 5} = \sqrt{4655} \approx 68.24 \text{ cm}^2$$ Two such triangles: $$2 \times 68.24 = 136.48 \text{ cm}^2$$ 6. **Calculate the area of the two small triangles on top and bottom:** Each has base $14$ cm and height $8$ cm. $$\text{Area} = \frac{1}{2} \times 14 \times 8 = 56 \text{ cm}^2$$ Two such triangles: $$2 \times 56 = 112 \text{ cm}^2$$ 7. **Calculate the area of the pentagonal bases:** The base is composed of the rectangle $14 \times 20$ plus the two small triangles $56$ each. But since the net shows the rectangle and triangles as lateral faces, the bases are pentagons formed by the sides $12$, $12$, $14$, and the small triangle parts. We approximate the pentagonal base area as the sum of the rectangle and two small triangles: $$280 + 112 = 392 \text{ cm}^2$$ Two bases: $$2 \times 392 = 784 \text{ cm}^2$$ 8. **Total surface area:** Sum of lateral faces and two bases: $$280 + 136.48 + 112 + 784 = 1312.48 \text{ cm}^2$$ 9. **Final answer:** The tile needed to cover the pedestal is approximately $$\boxed{1312.48 \text{ cm}^2}$$