1. **State the problem:** We need to find the percentage error in the area of a rectangle with length $11.3$ cm and width $9.23$ cm. 2. **Formula for area of a rectangle:** $$\text{Area} = \text{length} \times \text{width}$$ 3. **Calculate the area:** $$\text{Area} = 11.3 \times 9.23 = 104.399$$ cm$^2$ 4. **Percentage error in area:** When multiplying two measurements, the percentage error in the area is approximately the sum of the percentage errors in length and width. 5. **Calculate percentage errors in length and width:** - Length $= 11.3$ cm, correct to 3 significant figures, so the absolute error is about $\pm 0.05$ cm (half of the last digit). - Percentage error in length: $$\frac{0.05}{11.3} \times 100 = 0.4425\%$$ - Width $= 9.23$ cm, correct to 3 significant figures, so the absolute error is about $\pm 0.005$ cm. - Percentage error in width: $$\frac{0.005}{9.23} \times 100 = 0.0542\%$$ 6. **Sum the percentage errors to get the percentage error in area:** $$0.4425\% + 0.0542\% = 0.4967\%$$ 7. **Round to 3 significant figures:** $$0.497\%$$ **Final answer:** The percentage error in the area is approximately $0.497\%$ correct to 3 significant figures.