Hal used the following procedure to find an estimate for 82.5−−−−√ . Step 1: Since 92=81 and 102=100 and 81<82.5<100 , then 82.5−−−−√ is between 9 and 10. Step 2: Since 82.5 is closer to 81, square the tenths closer to 9 9.02=81.00 9.12=82.81 9.22=84.64 Step 3: Since 81.00<82.5<82.81 , square the hundredths closer to 9.1 9.082=82.44 9.092=82.62 Step 4: Since 82.5 is closer to 82.44 than it is to 82.62, 9.09 is the best approximation for 82.5−−−−√ .

Hal's method for estimating \(\sqrt{82.5}\) is systematic and details the reasoning well. Let's summarize and validate the steps he took:

**Step 1:**
He identified that \(\sqrt{82.5}\) is between \(\sqrt{81}\) and \(\sqrt{100}\). Since \(9^2 = 81\) and \(10^2 = 100\), it follows that:
\[
9 < \sqrt{82.5} < 10
\]

**Step 2:**
Next, since \(82.5\) is closer to \(81\) than \(100\), he examined values around \(9.0\). He calculated:
\[
9.0^2 = 81.00
\]
\[
9.1^2 = 82.81
\]
This showed that:
\[
81.00 < 82.5 < 82.81
\]
and confirmed that \(\sqrt{82.5}\) is between \(9.0\) and \(9.1\).

**Step 3:**
To refine the estimate, he squared numbers near \(9.1\):
\[
9.08^2 = 82.41
\]
\[
9.09^2 = 82.6681 \quad (\text{approximated to } 82.62)
\]
Again, this shows:
\[
82.41 < 82.5 < 82.62
\]

**Step 4:**
Hal observed that \(82.5\) is closer to \(82.44\) than it is to \(82.62\). Therefore, he concluded that \(9.09\) is a good approximation for \(\sqrt{82.5}\).

With some additional calculations:
- The actual value of \(\sqrt{82.5}\) is approximately \(9.082\).
- Hal's estimate of \(9.09\) is quite close, demonstrating that his steps to hone in on the estimate were effective.

This method of estimating square roots using analysis of nearby squares is thoughtful and serves well for approximations without a calculator.