Class 8 Maths Square and Square Roots Worksheet - Free Printable

Let's solve each problem step by step with clear explanations.

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1. Fill in the blanks

(i) The digit at unit’s place of square of 239 = _______

- We need to find the unit digit of $ 239^2 $.
- The unit digit of a square depends only on the unit digit of the number.
- Unit digit of 239 is 9.
- $ 9^2 = 81 $ → unit digit is 1.
- ✔ Answer: 1

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(ii) $ 29 \times 31 = \boxed{?} - 1 $

- Use identity: $ (a - b)(a + b) = a^2 - b^2 $
- $ 29 \times 31 = (30 - 1)(30 + 1) = 30^2 - 1^2 = 900 - 1 = 899 $
- So, $ 29 \times 31 = 900 - 1 $
- ✔ Answer: 900

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(iii) $ 17 \times 23 = \boxed{?} - 3^2 $

- First compute $ 17 \times 23 $
- $ 17 \times 23 = (20 - 3)(20 + 3) = 20^2 - 3^2 = 400 - 9 = 391 $
- So, $ 17 \times 23 = 391 $
- Now, $ 391 = ? - 9 $ → $ ? = 391 + 9 = 400 $
- ✔ Answer: 400

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(iv) $ (151)^2 - (150)^2 = ? $

- Use identity: $ a^2 - b^2 = (a - b)(a + b) $
- $ a = 151, b = 150 $
- $ = (151 - 150)(151 + 150) = (1)(301) = 301 $
- ✔ Answer: 301

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(v) The sum of first five odd numbers = ________

- First five odd numbers: 1, 3, 5, 7, 9
- Sum = $ 1 + 3 + 5 + 7 + 9 = 25 $
- Alternatively: Sum of first $ n $ odd numbers = $ n^2 $
- $ 5^2 = 25 $
- ✔ Answer: 25

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(vi) If $ 6^x = 1296 $, then $ x = ? $

- $ 1296 \div 6 = 216 $
- $ 216 \div 6 = 36 $
- $ 36 \div 6 = 6 $
- $ 6 \div 6 = 1 $
- So, $ 6^4 = 1296 $
- ✔ Answer: 4

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2. Write the perfect square numbers between 100 and 150

- Perfect squares:
- $ 10^2 = 100 $ → not included (since "between" means >100 and <150)
- $ 11^2 = 121 $
- $ 12^2 = 144 $
- $ 13^2 = 169 $ → too big
- So, perfect squares between 100 and 150 are: 121, 144
- ✔ Answer: 121, 144

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3. Write 17² as sum of two consecutive integers

- $ 17^2 = 289 $
- Let the two consecutive integers be $ x $ and $ x+1 $
- Then: $ x + (x+1) = 289 $
- $ 2x + 1 = 289 $
- $ 2x = 288 $ → $ x = 144 $
- So, numbers are 144 and 145
- Check: $ 144 + 145 = 289 $
- ✔ Answer: 144 and 145

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4. Find the Pythagorean triplet whose smallest number is 10

- A Pythagorean triplet satisfies $ a^2 + b^2 = c^2 $
- Given smallest number is 10 → assume $ a = 10 $
- Try to find $ b $ and $ c $ such that $ 10^2 + b^2 = c^2 $
- $ 100 + b^2 = c^2 $
- Try small values:
- $ b = 24 $: $ 100 + 576 = 676 = 26^2 $
- So, $ 10, 24, 26 $
- Check: $ 10^2 + 24^2 = 100 + 576 = 676 = 26^2 $ → Yes!
- ✔ Answer: (10, 24, 26)

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5. Find the smallest number by which 192 must be multiplied to make the product a perfect square

- Prime factorization of 192:
- $ 192 = 64 \times 3 = 2^6 \times 3 $
- For a perfect square, all exponents must be even.
- Here, exponent of 2 is 6 (even), exponent of 3 is 1 (odd)
- So, multiply by 3 to make it $ 2^6 \times 3^2 $
- $ 192 \times 3 = 576 = (24)^2 $
- ✔ Answer: 3

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6. Find the square root of the following

(i) $ \sqrt{10609} $

- Try estimating:
- $ 100^2 = 10000 $
- $ 103^2 = (100 + 3)^2 = 10000 + 600 + 9 = 10609 $
- ✔ Answer: 103

(ii) $ \sqrt{33.64} $

- Note: $ \sqrt{3364} = 58 $ (since $ 58^2 = 3364 $)
- So $ \sqrt{33.64} = \frac{\sqrt{3364}}{10} = \frac{58}{10} = 5.8 $
- ✔ Answer: 5.8

(iii) $ \sqrt{0.4489} $

- $ \sqrt{4489} = 67 $ (since $ 67^2 = 4489 $)
- So $ \sqrt{0.4489} = \frac{67}{100} = 0.67 $
- ✔ Answer: 0.67

(iv) $ \sqrt{\frac{289}{361}} $

- $ \sqrt{289} = 17 $, $ \sqrt{361} = 19 $
- So $ \frac{17}{19} $
- ✔ Answer: $ \frac{17}{19} $

(v) $ \sqrt{1\frac{7}{9}} = \sqrt{\frac{16}{9}} = \frac{4}{3} $
- $ 1\frac{7}{9} = \frac{9+7}{9} = \frac{16}{9} $
- $ \sqrt{\frac{16}{9}} = \frac{4}{3} $
- ✔ Answer: $ \frac{4}{3} $

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7. Find:

(i) $ \sqrt{55} \times \sqrt{220} $

- $ = \sqrt{55 \times 220} $
- $ 55 \times 220 = 55 \times 22 \times 10 = (55 \times 22) \times 10 $
- But better: $ 55 \times 220 = 12100 $
- $ \sqrt{12100} = 110 $
- ✔ Answer: 110

(ii) $ \sqrt{0.25} \times \sqrt{0.09} $

- $ \sqrt{0.25} = 0.5 $, $ \sqrt{0.09} = 0.3 $
- $ 0.5 \times 0.3 = 0.15 $
- ✔ Answer: 0.15

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8. Find the smallest number of four digits which is a perfect square

- Smallest 4-digit number = 1000
- Find smallest perfect square ≥ 1000
- $ \sqrt{1000} \approx 31.62 $
- So next integer = 32
- $ 32^2 = 1024 $
- ✔ Answer: 1024

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9. Find the smallest square number divisible by 4, 12, and 16

- Find LCM of 4, 12, 16
- $ 4 = 2^2 $
- $ 12 = 2^2 \times 3 $
- $ 16 = 2^4 $
- LCM = $ 2^4 \times 3 = 16 \times 3 = 48 $
- Now find smallest perfect square divisible by 48
- Factor: $ 48 = 2^4 \times 3 $
- To make it a perfect square, all exponents must be even → 3 has exponent 1 → need one more 3
- Multiply by 3 → $ 48 \times 3 = 144 $
- $ 144 = 12^2 $ → perfect square
- ✔ Answer: 144

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10. Find $ x $ if $ \frac{x}{16} = \sqrt{\frac{9}{16}} $

- $ \sqrt{\frac{9}{16}} = \frac{3}{4} $
- So: $ \frac{x}{16} = \frac{3}{4} $
- Multiply both sides by 16: $ x = 16 \times \frac{3}{4} = 12 $
- ✔ Answer: 12

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11. Area of square field = 4225 m². Girl cycles along boundary twice. How much distance?

- Area = $ s^2 = 4225 $
- $ s = \sqrt{4225} $
- $ \sqrt{4225} = 65 $ (since $ 65^2 = 4225 $)
- Perimeter = $ 4 \times 65 = 260 $ m
- She covers boundary twice, so distance = $ 2 \times 260 = 520 $ m
- ✔ Answer: 520 meters

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12. Arrange 1024 students in rows such that each row has as many students as number of rows

- Let number of rows = $ x $, students per row = $ x $
- Total students = $ x \times x = x^2 $
- $ x^2 = 1024 $
- $ x = \sqrt{1024} $
- $ \sqrt{1024} = 32 $ (since $ 32^2 = 1024 $)
- ✔ Answer: 32 students in each row

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✔ Final Answers Summary:

1.
(i) 1
(ii) 900
(iii) 400
(iv) 301
(v) 25
(vi) 4

2. 121, 144

3. 144 and 145

4. (10, 24, 26)

5. 3

6.
(i) 103
(ii) 5.8
(iii) 0.67
(iv) $ \frac{17}{19} $
(v) $ \frac{4}{3} $

7.
(i) 110
(ii) 0.15

8. 1024

9. 144

10. 12

11. 520 meters

12. 32 students per row

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