1. **State the problem:** We have a solid formed by removing a hemisphere (radius $2x$ cm) from a cone (base radius $5x$ cm, height $6x$ cm). The volume of this solid is $6948\pi$ cm³. We need to find the total surface area of the removed hemisphere, rounded to the nearest integer. 2. **Write down the formulas:** - Volume of cone: $$V_{cone} = \frac{1}{3} \pi r^2 h$$ - Volume of hemisphere: $$V_{hemisphere} = \frac{2}{3} \pi r^3$$ - Total volume of solid shape = Volume of cone - Volume of hemisphere = $6948\pi$ - Surface area of a hemisphere (including curved surface and flat circular base): $$A = 3\pi r^2$$ 3. **Express volumes in terms of $x$:** - Cone volume: $$V_{cone} = \frac{1}{3} \pi (5x)^2 (6x) = \frac{1}{3} \pi 25x^2 \cdot 6x = 50\pi x^3$$ - Hemisphere volume: $$V_{hemisphere} = \frac{2}{3} \pi (2x)^3 = \frac{2}{3} \pi 8x^3 = \frac{16}{3} \pi x^3$$ 4. **Set up equation for total volume:** $$V_{solid} = V_{cone} - V_{hemisphere} = 50\pi x^3 - \frac{16}{3} \pi x^3 = 6948\pi$$ Remove $\pi$ from all terms: $$50 x^3 - \frac{16}{3} x^3 = 6948$$ Multiply both sides by 3 to clear denominator: $$150 x^3 - 16 x^3 = 20844$$ Simplify: $$134 x^3 = 20844$$ 5. **Solve for $x^3$:** $$x^3 = \frac{20844}{134} = 155.5820896...$$ 6. **Calculate $x$:** $$x = \sqrt[3]{155.5820896} \approx 5.39$$ 7. **Calculate surface area of removed hemisphere:** Radius of hemisphere = $2x = 2 \times 5.39 = 10.78$ cm Total surface area of hemisphere: $$A = 3 \pi r^2 = 3 \pi (10.78)^2 = 3 \pi (116.27) = 348.81 \pi$$ Numerical value: $$348.81 \times 3.1416 \approx 1095.45$$ 8. **Final answer:** Rounded to nearest integer: $$\boxed{1095}\text{ cm}^2$$