1. **Problem statement:** Find the area swept by a door that opens as a sector of a circle with radius $80$ cm and central angle $45^\circ$. 2. **Formula for the area of a sector:** The area $A$ of a sector with radius $r$ and central angle $\theta$ (in degrees) is given by: $$A = \frac{\theta}{360} \times \pi r^2$$ 3. **Substitute the given values:** $$r = 80, \quad \theta = 45$$ $$A = \frac{45}{360} \times \pi \times 80^2$$ 4. **Simplify the fraction:** $$\frac{45}{360} = \frac{\cancel{45}}{\cancel{360}} = \frac{1}{8}$$ 5. **Calculate the area:** $$A = \frac{1}{8} \times \pi \times 80^2 = \frac{1}{8} \times \pi \times 6400$$ 6. **Simplify further:** $$A = \pi \times \frac{6400}{8} = \pi \times 800$$ 7. **Final answer:** The area swept by the door is: $$\boxed{800\pi \text{ cm}^2}$$ This means the door sweeps an area of $800\pi$ square centimeters as it opens through $45^\circ$.