To approximate \( \sqrt{1131} \) and get closer to \( \sqrt{31} \), we can use a systematic approach based on evaluating the square root.
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Identify the approximate value of \(\sqrt{1131}\):
- First, notice that \( 31^2 = 961 \) and \( 32^2 = 1024 \).
- Since \( 1131 \) is between \( 1024 \) and \( 1089 \) (\( 33^2 \)), we can infer that \( \sqrt{1131} \) is between \( 32 \) and \( 34 \).
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Refine the approximation:
- Compute the square of numbers between \( 32 \) and \( 34 \).
- \( 33.5^2 = 1122.25 \)
- \( 33.6^2 = 1127.36 \)
- \( 33.7^2 = 1135.69 \)
Now we can see that \( \sqrt{1131} \) is between \( 33.5 \) and \( 33.6 \).
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Further refine:
- Try \( 33.6 \) and \( 33.65 \).
- \( 33.65^2 = 1134.4225 \) (too high).
- Try \( 33.58^2 = 1127.1364 \) (too low).
- Next, try \( 33.63 \).
- \( 33.63^2 = 1132.0569 \) (still high).
- Try \( 33.62 \).
- \( 33.62^2 = 1132.4644 \) (high).
- Finally, \( 33.61 \).
- \( 33.61^2 = 1128.5921 \) (too low).
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Convergence:
- Now, we can see \( 33.61^2 = 1128.5921 \) and \( 33.62^2 = 1132.4644 \), indicating that \( \sqrt{1131} \) is very close to \( 33.63 \).
- We can average further by trying decimals closer between \( 33.61 \) and \( 33.62 \), ultimately converging to:
The calculated answer is approximately: \[ \sqrt{1131} \approx 33.6 \] Corrected to one decimal place. Therefore, the answer is: \[ \boxed{33.6} \]