1. **State the problem:** We have a rectangle with one side length 20 cm and the other side length $x$ cm. The area is given as 400 cm$^2$, both area and length are accurate to 2 significant figures. 2. **Formula used:** The area $A$ of a rectangle is given by $$A = \text{length} \times \text{width}$$ Here, $A = 400$ cm$^2$, one side is 20 cm, and the other side is $x$ cm. 3. **Calculate the bounds for the area:** Since the area is accurate to 2 significant figures, the bounds are $$\text{Lower bound of area} = 395$$ $$\text{Upper bound of area} = 405$$ 4. **Calculate the bounds for the length 20 cm:** Since 20 cm is accurate to 2 significant figures, $$\text{Lower bound of length} = 19.5$$ $$\text{Upper bound of length} = 20.5$$ 5. **Calculate the bounds for $x$:** Using the formula $A = 20 \times x$, rearranged to $$x = \frac{A}{20}$$ Calculate the lower bound for $x$ by dividing the lower bound of area by the upper bound of length: $$x_{\text{lower}} = \frac{395}{20.5}$$ Calculate the upper bound for $x$ by dividing the upper bound of area by the lower bound of length: $$x_{\text{upper}} = \frac{405}{19.5}$$ 6. **Simplify the bounds:** $$x_{\text{lower}} = \frac{395}{20.5} = \frac{\cancel{395}}{\cancel{20.5}} = 19.26829268... \approx 19.3$$ $$x_{\text{upper}} = \frac{405}{19.5} = \frac{\cancel{405}}{\cancel{19.5}} = 20.76923077... \approx 20.8$$ 7. **Final answer:** The length $x$ is between approximately 19.3 cm and 20.8 cm, considering the bounds from the given accuracies.