1.9.1 Worksheet · Homework - Free Printable

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? Let's check:

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

So $ \sqrt{90} $ is between 9 and 10.

Wait! That means the side length is greater than 9 inches, which is not between 8 and 9.

Let’s double-check:

- $ 9^2 = 81 $
- $ 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 $
- $ 9.45^2 = ? $
- $ (9.4 + 0.05)^2 = 9.4^2 + 2(9.4)(0.05) + 0.05^2 = 88.36 + 0.94 + 0.0025 = 89.3025 $
- $ 9.48^2 = ? $
- $ (9.4 + 0.08)^2 = 88.36 + 2(9.4)(0.08) + 0.0064 = 88.36 + 1.504 + 0.0064 = 89.8704 $
- $ 9.49^2 = (9.48 + 0.01)^2 ≈ 89.8704 + 2(9.48)(0.01) + 0.0001 ≈ 89.8704 + 0.1896 + 0.0001 = 90.06 $

So $ \sqrt{90} \approx 9.48 $ inches

✔ So the side length is approximately 9.48 inches.

Answer to (a):
The dimensions of the canvas are approximately 9.48 inches by 9.48 inches, since $ \sqrt{90} \approx 9.48 $. I estimated this by finding perfect squares near 90: $ 9^2 = 81 $, $ 10^2 = 100 $, so $ \sqrt{90} $ is between 9 and 10. Then I used trial values like $ 9.4^2 = 88.36 $, $ 9.5^2 = 90.25 $, and narrowed it down to about 9.48.

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b. Does the canvas meet Jennifer’s requirements?

Jennifer wants side length between 8 and 9 inches.

But we found $ \sqrt{90} \approx 9.48 $ inches → greater than 9 inches.

✘ No, the canvas does not meet her requirements.

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Fill-in-the-blank:

> 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, 3, 5 |

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Now, estimate the square roots 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 near 50:
- $ 7^2 = 49 $
- $ 8^2 = 64 $

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

On the number line: mark 7 and 8.

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

- $ 50 - 49 = 1 $
- $ 64 - 50 = 14 $

So 50 is much closer to 49 than to 64.

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

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

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

So $ \sqrt{50} $ is between 7.07 and 7.08.

Since $ 7.07^2 = 49.9849 $ (very close to 50), and $ 7.08^2 = 50.1264 $, we round to the nearest tenth.

→ $ \sqrt{50} \approx 7.1 $ (since 7.07 is closer to 7.1 than 7.0)

✔ Answer: $ \sqrt{50} \approx 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

Which is it closer to?
- $ 22 - 16 = 6 $
- $ 25 - 22 = 3 $

Closer to 25 → closer to 5

Try:
- $ 4.6^2 = 21.16 $
- $ 4.7^2 = 22.09 $

So $ \sqrt{22} $ is between 4.6 and 4.7

$ 4.6^2 = 21.16 $, $ 4.7^2 = 22.09 $, and 22 is very close to 22.09

So $ \sqrt{22} \approx 4.7 $

Answer:
- $ \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

Which is it closer to?
- $ 103 - 100 = 3 $
- $ 121 - 103 = 18 $

Closer to 100 → closer to 10

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

So $ \sqrt{103} $ is between 10.1 and 10.2

Since $ 10.1^2 = 102.01 $, and $ 103 - 102.01 = 0.99 $

Try $ 10.15^2 = (10 + 0.15)^2 = 100 + 2(10)(0.15) + 0.0225 = 100 + 3 + 0.0225 = 103.0225 $

Very close!

So $ \sqrt{103} \approx 10.15 $ → to the nearest tenth: 10.2

But let’s see:
- $ 10.1^2 = 102.01 $
- $ 10.15^2 = 103.0225 $
- Since 103 is just below 103.0225, $ \sqrt{103} \approx 10.14 $ or so

To the nearest tenth: look at hundredths place.

$ 10.14 $ → tenths digit is 1, hundredths is 4 → rounds to 10.1

Wait: 10.14 is closer to 10.1 than 10.2?

No: 10.14 is less than 10.15, so rounds to 10.1

But let’s confirm:

- $ 10.1^2 = 102.01 $
- $ 10.2^2 = 104.04 $
- $ \sqrt{103} $ is between them, closer to 10.1

But earlier we saw $ 10.15^2 = 103.0225 $, so $ \sqrt{103} \approx 10.149 $

So to the nearest tenth, we look at the hundredths digit: 4 → less than 5 → round down.

So $ \sqrt{103} \approx \boxed{10.1} $

But wait — is it really closer to 10.1 or 10.2?

Let’s calculate midpoint between 10.1 and 10.2: 10.15

Since $ \sqrt{103} \approx 10.149 < 10.15 $, it's closer to 10.1

✔ So $ \sqrt{103} \approx 10.1 $

But some might argue it's closer to 10.2? No — 10.149 is only 0.049 away from 10.1, but 0.051 away from 10.2 → actually, no:

Wait: 10.149 – 10.1 = 0.049
10.2 – 10.149 = 0.051 → so yes, closer to 10.1

✔ Final answer: $ \sqrt{103} \approx \boxed{10.1} $

But let’s recheck: is $ 10.1^2 = 102.01 $, and $ 103 - 102.01 = 0.99 $, so we’re missing 0.99

Try $ 10.14^2 = (10 + 0.14)^2 = 100 + 2(10)(0.14) + 0.0196 = 100 + 2.8 + 0.0196 = 102.8196 $

Still low

$ 10.15^2 = 103.0225 $ as before

So $ \sqrt{103} \approx 10.149 $ → to the nearest tenth → 10.1

Yes.

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✔ Final Answers Summary:

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#### Jennifer’s Canvas:
- a. Side length ≈ $ \sqrt{90} \approx 9.48 $ inches → 9.48 in × 9.48 in
- b. No, because 9.48 > 9 → does not meet requirement.

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#### Fill-ins:

- The area represents a non-perfect square.
- Perfect Squares: Numbers whose square roots are whole numbers; examples: 4, 9, 16
- Non-Perfect Squares: Numbers whose square roots are not whole numbers; examples: 2, 3, 5

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#### Estimations:

1. $ \sqrt{50} $
- a. Between 7 and 8
- b. Closer to 7 (since 50 is closer to 49 than 64)
- c. $ \sqrt{50} \approx \boxed{7.1} $

2. $ \sqrt{22} $
- Between 4 and 5, closer to 5
- Decimal estimate: $ \sqrt{22} \approx \boxed{4.7} $

3. $ \sqrt{103} $
- Between 10 and 11, closer to 10
- Decimal estimate: $ \sqrt{103} \approx \boxed{10.1} $

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✔ All problems solved and explained!