Estimating Square Roots worksheet with a canvas area problem and number line exercises.

Let's solve the problem step by step and explain each part clearly.

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Problem: Estimating Square Roots

#### Jennifer’s Canvas Problem

Given:
- Jennifer wants a square canvas with side length between 8 and 9 inches.
- The canvas has an area of 90 in².

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a. Estimate the dimensions of the canvas. Explain your thinking.

We know:
- Area of a square = side × side = $ s^2 $
- So, $ s = \sqrt{\text{Area}} = \sqrt{90} $

We need to estimate $ \sqrt{90} $.

Step 1: Find perfect squares near 90
- $ 8^2 = 64 $
- $ 9^2 = 81 $
- $ 10^2 = 100 $

So, $ \sqrt{90} $ is between $ \sqrt{81} = 9 $ and $ \sqrt{100} = 10 $.
But wait — we want it between 8 and 9, but $ \sqrt{90} $ is greater than 9?

Wait! Let's double-check:

- $ 9^2 = 81 $
- $ 10^2 = 100 $

So $ \sqrt{90} $ is between 9 and 10, since $ 81 < 90 < 100 $

But Jennifer wants a side length between 8 and 9 inches.

So $ \sqrt{90} \approx ? $

Let’s try:
- $ 9.5^2 = 90.25 $ → very close!
- $ 9.4^2 = (9 + 0.4)^2 = 81 + 2×9×0.4 + 0.16 = 81 + 7.2 + 0.16 = 88.36 $
- $ 9.4^2 = 88.36 $
- $ 9.5^2 = 90.25 $

So $ \sqrt{90} $ is between 9.4 and 9.5, closer to 9.5.

Thus, $ \sqrt{90} \approx 9.49 $ inches

✔ So the side length is about 9.49 inches.

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b. Does the canvas meet Jennifer’s requirements?

She wanted side length between 8 and 9 inches.

But $ \sqrt{90} \approx 9.49 $ inches → which is greater than 9.

✘ So no, the canvas does not meet her requirements.

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Fill in the blanks:

> The area of the canvas above represents a non-perfect square. We can use our knowledge of perfect squares to estimate the square roots of non-perfect squares.

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Perfect Squares vs Non-Perfect Squares

| PERFECT SQUARES | NON-PERFECT SQUARES |
|----------------------|--------------------------|
| Numbers whose square roots are whole numbers | Numbers whose square roots are not whole numbers |
| Examples: 4, 9, 16 | Examples: 2, 5, 10 |

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Now, estimate each square root using number lines and perfect squares

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1. Estimate $ \sqrt{50} $

#### a. Use the number line to locate the whole numbers that $ \sqrt{50} $ should be between.

Find perfect squares around 50:
- $ 7^2 = 49 $
- $ 8^2 = 64 $

So $ \sqrt{50} $ is between 7 and 8

On the number line:
```
<---|-------|-------|-------|-------|--->
6 7 8 9 10
```

Place a point between 7 and 8.

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#### b. Which whole number will $ \sqrt{50} $ be closer to? Explain.

- $ 7^2 = 49 $
- $ 8^2 = 64 $
- $ 50 - 49 = 1 $
- $ 64 - 50 = 14 $

So 50 is only 1 unit away from 49, but 14 units from 64.

✔ Therefore, $ \sqrt{50} $ is much closer to 7.

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#### c. Estimate $ \sqrt{50} $ to the nearest tenth.

Try:
- $ 7.1^2 = 50.41 $
- $ 7.0^2 = 49 $
- $ 7.05^2 = (7 + 0.05)^2 = 49 + 2×7×0.05 + 0.0025 = 49 + 0.7 + 0.0025 = 49.7025 $
- $ 7.07^2 = ? $
- $ 7.07^2 = (7 + 0.07)^2 = 49 + 2×7×0.07 + 0.07^2 = 49 + 0.98 + 0.0049 = 49.9849 $
- $ 7.08^2 = 49 + 2×7×0.08 + 0.08^2 = 49 + 1.12 + 0.0064 = 50.1264 $

So:
- $ 7.07^2 ≈ 49.98 $
- $ 7.08^2 ≈ 50.13 $

Since 50 is between them, and closer to 49.98, so $ \sqrt{50} \approx 7.07 $

To the nearest tenth: look at hundredths place → 7.07 → tenths digit is 0, hundredths is 7 → rounds up?

Wait: 7.07 → tenths place is 0, hundredths is 7 → round up the tenths?

No: 7.07 → the tenths digit is 0, and the hundredths is 7 → so rounding to nearest tenth means look at hundredths: 7 ≥ 5 → round up the tenths digit.

But tenths digit is 0 → becomes 1 → so 7.1

✔ So $ \sqrt{50} \approx \boxed{7.1} $

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2. Estimate $ \sqrt{22} $

Perfect squares:
- $ 4^2 = 16 $
- $ 5^2 = 25 $

So $ \sqrt{22} $ is between 4 and 5

Now, how close?
- $ 22 - 16 = 6 $
- $ 25 - 22 = 3 $

So closer to 5 (since 3 < 6)

Try:
- $ 4.6^2 = (4 + 0.6)^2 = 16 + 2×4×0.6 + 0.36 = 16 + 4.8 + 0.36 = 21.16 $
- $ 4.7^2 = 16 + 2×4×0.7 + 0.49 = 16 + 5.6 + 0.49 = 22.09 $

$ 4.6^2 = 21.16 $, $ 4.7^2 = 22.09 $

22 is between them, and closer to 22.09 → so $ \sqrt{22} \approx 4.69 $

To nearest tenth: $ 4.69 \rightarrow $ tenths digit is 6, hundredths is 9 → round up → 4.7

✔ So:
- $ \sqrt{22} $ is between 4 and 5, but closer to 5
- Decimal estimate: $ \sqrt{22} \approx \boxed{4.7} $

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3. Estimate $ \sqrt{103} $

Perfect squares:
- $ 10^2 = 100 $
- $ 11^2 = 121 $

So $ \sqrt{103} $ is between 10 and 11

Now:
- $ 103 - 100 = 3 $
- $ 121 - 103 = 18 $

So much closer to 10

Try:
- $ 10.1^2 = 100 + 2×10×0.1 + 0.01 = 100 + 2 + 0.01 = 102.01 $
- $ 10.2^2 = 100 + 2×10×0.2 + 0.04 = 100 + 4 + 0.04 = 104.04 $

So:
- $ 10.1^2 = 102.01 $
- $ 10.2^2 = 104.04 $

103 is between them → closer to 102.01 → so $ \sqrt{103} \approx 10.15 $

To nearest tenth: $ 10.15 \rightarrow $ tenths digit is 1, hundredths is 5 → round up → 10.2

✔ So:
- $ \sqrt{103} $ is between 10 and 11, but closer to 10
- Decimal estimate: $ \sqrt{103} \approx \boxed{10.2} $

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✔ Final Answers Summary

#### Jennifer’s Canvas
- a. Side length ≈ $ \sqrt{90} \approx 9.49 $ inches → between 9 and 10
- b. ✘ No, because 9.49 > 9 → not between 8 and 9

#### Definitions
- Perfect squares: whole numbers (e.g., 4, 9, 16)
- Non-perfect squares: irrational or decimal roots (e.g., 2, 5, 10)

#### Estimations
1. $ \sqrt{50} $:
- Between 7 and 8
- Closer to 7
- Estimate: 7.1

2. $ \sqrt{22} $:
- Between 4 and 5, closer to 5
- Estimate: 4.7

3. $ \sqrt{103} $:
- Between 10 and 11, closer to 10
- Estimate: 10.2

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