1. The problem is to approximate $\sqrt{11}$ to one decimal place. 2. We know that $3^2 = 9$ and $4^2 = 16$, so $3 < \sqrt{11} < 4$ because $9 < 11 < 16$. 3. The number 11 is 2 units away from 9 and 5 units away from 16 on the number line. 4. To estimate how far $\sqrt{11}$ is from 3 (which corresponds to 9), we use the ratio of the distances: $$\text{Ratio} = \frac{\text{distance from } 9 \text{ to } 11}{\text{distance from } 9 \text{ to } 16} = \frac{2}{2 + 5} = \frac{2}{7} \approx 0.3$$ 5. This means $\sqrt{11}$ is about 0.3 units above 3, so: $$\sqrt{11} \approx 3 + 0.3 = 3.3$$ 6. Checking the approximation: $$3.3^2 = 10.89 \approx 11$$ This confirms the estimate is reasonable. The ratio part works by comparing how far 11 is between 9 and 16, then applying that fraction to the interval between 3 and 4 to find the approximate square root.