Approximating Square Roots | Worksheet - Free Printable

Problem Analysis:

The task involves approximating square roots of numbers that are not perfect squares. The goal is to identify the two whole numbers between which the square root of a given number falls. This requires understanding how to estimate square roots by comparing them to perfect squares.

Solution Approach:

1. Identify Perfect Squares: Recall or calculate the perfect squares (e.g., \(1^2 = 1\), \(2^2 = 4\), \(3^2 = 9\), etc.).
2. Locate the Given Number: For each number in the problem, determine which two consecutive perfect squares it lies between.
3. Approximate the Square Root: The square root of the given number will lie between the square roots of these two perfect squares.

Step-by-Step Solution:

#### Example: Approximating \(\sqrt{22}\)
- The perfect squares closest to 22 are \(16\) (\(4^2\)) and \(25\) (\(5^2\)).
- Since \(16 < 22 < 25\), it follows that \(\sqrt{16} < \sqrt{22} < \sqrt{25}\).
- Therefore, \(4 < \sqrt{22} < 5\).

#### Generalizing the Process:
For any number \(N\):
1. Find the largest perfect square less than \(N\) and the smallest perfect square greater than \(N\).
2. The square root of \(N\) will lie between the square roots of these two perfect squares.

#### Applying to Each Problem:
We will now solve each part of the worksheet using this method.

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Worksheet Solutions:

#### Part 1: Approximating Square Roots
1. \(\sqrt{8}\)
- Perfect squares: \(4\) (\(2^2\)) and \(9\) (\(3^2\)).
- Since \(4 < 8 < 9\), \(\sqrt{4} < \sqrt{8} < \sqrt{9}\).
- Therefore, \(2 < \sqrt{8} < 3\).

2. \(\sqrt{22}\)
- Perfect squares: \(16\) (\(4^2\)) and \(25\) (\(5^2\)).
- Since \(16 < 22 < 25\), \(\sqrt{16} < \sqrt{22} < \sqrt{25}\).
- Therefore, \(4 < \sqrt{22} < 5\).

3. \(\sqrt{30}\)
- Perfect squares: \(25\) (\(5^2\)) and \(36\) (\(6^2\)).
- Since \(25 < 30 < 36\), \(\sqrt{25} < \sqrt{30} < \sqrt{36}\).
- Therefore, \(5 < \sqrt{30} < 6\).

4. \(\sqrt{50}\)
- Perfect squares: \(49\) (\(7^2\)) and \(64\) (\(8^2\)).
- Since \(49 < 50 < 64\), \(\sqrt{49} < \sqrt{50} < \sqrt{64}\).
- Therefore, \(7 < \sqrt{50} < 8\).

5. \(\sqrt{77}\)
- Perfect squares: \(64\) (\(8^2\)) and \(81\) (\(9^2\)).
- Since \(64 < 77 < 81\), \(\sqrt{64} < \sqrt{77} < \sqrt{81}\).
- Therefore, \(8 < \sqrt{77} < 9\).

6. \(\sqrt{88}\)
- Perfect squares: \(81\) (\(9^2\)) and \(100\) (\(10^2\)).
- Since \(81 < 88 < 100\), \(\sqrt{81} < \sqrt{88} < \sqrt{100}\).
- Therefore, \(9 < \sqrt{88} < 10\).

7. \(\sqrt{98}\)
- Perfect squares: \(81\) (\(9^2\)) and \(100\) (\(10^2\)).
- Since \(81 < 98 < 100\), \(\sqrt{81} < \sqrt{98} < \sqrt{100}\).
- Therefore, \(9 < \sqrt{98} < 10\).

8. \(\sqrt{108}\)
- Perfect squares: \(100\) (\(10^2\)) and \(121\) (\(11^2\)).
- Since \(100 < 108 < 121\), \(\sqrt{100} < \sqrt{108} < \sqrt{121}\).
- Therefore, \(10 < \sqrt{108} < 11\).

9. \(\sqrt{125}\)
- Perfect squares: \(121\) (\(11^2\)) and \(144\) (\(12^2\)).
- Since \(121 < 125 < 144\), \(\sqrt{121} < \sqrt{125} < \sqrt{144}\).
- Therefore, \(11 < \sqrt{125} < 12\).

10. \(\sqrt{130}\)
- Perfect squares: \(121\) (\(11^2\)) and \(144\) (\(12^2\)).
- Since \(121 < 130 < 144\), \(\sqrt{121} < \sqrt{130} < \sqrt{144}\).
- Therefore, \(11 < \sqrt{130} < 12\).

11. \(\sqrt{150}\)
- Perfect squares: \(144\) (\(12^2\)) and \(169\) (\(13^2\)).
- Since \(144 < 150 < 169\), \(\sqrt{144} < \sqrt{150} < \sqrt{169}\).
- Therefore, \(12 < \sqrt{150} < 13\).

12. \(\sqrt{198}\)
- Perfect squares: \(196\) (\(14^2\)) and \(225\) (\(15^2\)).
- Since \(196 < 198 < 225\), \(\sqrt{196} < \sqrt{198} < \sqrt{225}\).
- Therefore, \(14 < \sqrt{198} < 15\).

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & 2 < \sqrt{8} < 3 \\
2. & 4 < \sqrt{22} < 5 \\
3. & 5 < \sqrt{30} < 6 \\
4. & 7 < \sqrt{50} < 8 \\
5. & 8 < \sqrt{77} < 9 \\
6. & 9 < \sqrt{88} < 10 \\
7. & 9 < \sqrt{98} < 10 \\
8. & 10 < \sqrt{108} < 11 \\
9. & 11 < \sqrt{125} < 12 \\
10. & 11 < \sqrt{130} < 12 \\
11. & 12 < \sqrt{150} < 13 \\
12. & 14 < \sqrt{198} < 15 \\
\end{array}
}
\]

Explanation for Challenge Question:

If you were to approximate \(\sqrt{50}\), you would find that it lies between \(7\) and \(8\). Since \(50\) is closer to \(49\) (\(7^2\)) than to \(64\) (\(8^2\)), \(\sqrt{50}\) is closer to \(7\). This reasoning can be applied to any number by assessing its proximity to the nearest perfect squares.