1. **State the problem:** We have a frustum formed by removing a small cone from a large cone. The large cone has slant height $l_1 = 12$ cm and base radius $r_1 = 6$ cm. The small cone has slant height $l_2 = 5$ cm and radius $r_2 = 2.5$ cm. We need to find the total surface area of the frustum, correct to 3 significant figures. 2. **Recall the formula for curved surface area of a cone:** $$\text{Curved Surface Area} = \pi r l$$ where $r$ is the radius of the base and $l$ is the slant height. 3. **Calculate curved surface area of the large cone:** $$A_1 = \pi \times 6 \times 12 = 72\pi \approx 226.195\, \text{cm}^2$$ 4. **Calculate curved surface area of the small cone:** $$A_2 = \pi \times 2.5 \times 5 = 12.5\pi \approx 39.270\, \text{cm}^2$$ 5. **Calculate the curved surface area of the frustum:** This is the difference between the large and small cone curved areas: $$A_{frustum} = A_1 - A_2 = 72\pi - 12.5\pi = 59.5\pi \approx 186.925\, \text{cm}^2$$ 6. **Calculate the area of the two circular bases:** - Large base area: $$A_{base1} = \pi r_1^2 = \pi \times 6^2 = 36\pi \approx 113.097\, \text{cm}^2$$ - Small base area (top of frustum): $$A_{base2} = \pi r_2^2 = \pi \times 2.5^2 = 6.25\pi \approx 19.635\, \text{cm}^2$$ 7. **Total surface area of the frustum:** Sum of curved surface area and two circular bases: $$A_{total} = A_{frustum} + A_{base1} + A_{base2} = 59.5\pi + 36\pi + 6.25\pi = 101.75\pi \approx 319.657\, \text{cm}^2$$ 8. **Final answer rounded to 3 significant figures:** $$\boxed{320\, \text{cm}^2}$$