CBSE Class 8 Mathematics Worksheet - Square and Square Roots | PDF - Free Printable

Maths Worksheet: Chapter 6 - Square and Square Roots

#### Class VIII

Let's solve each problem step by step.

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Problem 1: Which of the following numbers are perfect squares?

- (a) 3364
- (b) 4489
- (c) 4358

Solution:
A number is a perfect square if it can be expressed as \( n^2 \) for some integer \( n \).

1. Check 3364:
- Find the square root of 3364:
\[
\sqrt{3364} = 58
\]
- Since 58 is an integer, 3364 is a perfect square.

2. Check 4489:
- Find the square root of 4489:
\[
\sqrt{4489} = 67
\]
- Since 67 is an integer, 4489 is a perfect square.

3. Check 4358:
- Find the square root of 4358:
\[
\sqrt{4358} \approx 65.99
\]
- Since 65.99 is not an integer, 4358 is not a perfect square.

Answer:
\[
\boxed{(a) 3364, (b) 4489}
\]

---

Problem 2: Using the property of squares, find the value of the following.

- (a) \( 24^2 - 23^2 \)
- (b) \( 50^2 - 49^2 \)
- (c) \( 105^2 - 104^2 \)

Solution:
We use the difference of squares formula:
\[
a^2 - b^2 = (a - b)(a + b)
\]

1. For \( 24^2 - 23^2 \):
- Here, \( a = 24 \) and \( b = 23 \):
\[
24^2 - 23^2 = (24 - 23)(24 + 23) = 1 \times 47 = 47
\]

2. For \( 50^2 - 49^2 \):
- Here, \( a = 50 \) and \( b = 49 \):
\[
50^2 - 49^2 = (50 - 49)(50 + 49) = 1 \times 99 = 99
\]

3. For \( 105^2 - 104^2 \):
- Here, \( a = 105 \) and \( b = 104 \):
\[
105^2 - 104^2 = (105 - 104)(105 + 104) = 1 \times 209 = 209
\]

Answer:
\[
\boxed{(a) 47, (b) 99, (c) 209}
\]

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Problem 3: Without adding, find the sum.

- (a) \( 1 + 3 + 5 + 7 + 9 \)
- (b) \( 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \)

Solution:
The sum of the first \( n \) odd numbers is given by the formula:
\[
\text{Sum} = n^2
\]

1. For \( 1 + 3 + 5 + 7 + 9 \):
- There are 5 terms (i.e., \( n = 5 \)):
\[
\text{Sum} = 5^2 = 25
\]

2. For \( 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \):
- There are 12 terms (i.e., \( n = 12 \)):
\[
\text{Sum} = 12^2 = 144
\]

Answer:
\[
\boxed{(a) 25, (b) 144}
\]

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Problem 4: Find the square root of the following by prime factorisation method.

- (a) 1024
- (b) 8836
- (c) 8464

Solution:
We find the prime factorisation of each number and then take the square root.

1. For 1024:
- Prime factorisation:
\[
1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}
\]
- Taking the square root:
\[
\sqrt{1024} = \sqrt{2^{10}} = 2^{5} = 32
\]

2. For 8836:
- Prime factorisation:
\[
8836 = 2 \times 2 \times 47 \times 47 = 2^2 \times 47^2
\]
- Taking the square root:
\[
\sqrt{8836} = \sqrt{2^2 \times 47^2} = 2 \times 47 = 94
\]

3. For 8464:
- Prime factorisation:
\[
8464 = 2 \times 2 \times 2 \times 2 \times 23 \times 23 = 2^4 \times 23^2
\]
- Taking the square root:
\[
\sqrt{8464} = \sqrt{2^4 \times 23^2} = 2^2 \times 23 = 4 \times 23 = 92
\]

Answer:
\[
\boxed{(a) 32, (b) 94, (c) 92}
\]

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Problem 5: Find the smallest number by which 2475 must be multiplied to get a perfect square.

Solution:
1. Prime factorisation of 2475:
\[
2475 = 3 \times 3 \times 5 \times 5 \times 11 = 3^2 \times 5^2 \times 11
\]
- For a number to be a perfect square, all prime factors must have even powers.
- Here, the prime factor \( 11 \) has an odd power (1).
- To make it a perfect square, we need to multiply by another \( 11 \).

2. Smallest number to multiply:
\[
11
\]

Answer:
\[
\boxed{11}
\]

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Problem 6: 4096 soldiers are arranged in an auditorium in such a manner that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium?

Solution:
1. Let the number of rows be \( x \). Since the number of soldiers in each row is also \( x \), the total number of soldiers is:
\[
x \times x = x^2
\]
2. We are given that the total number of soldiers is 4096:
\[
x^2 = 4096
\]
3. Taking the square root of both sides:
\[
x = \sqrt{4096}
\]
4. Prime factorisation of 4096:
\[
4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{12}
\]
- Taking the square root:
\[
\sqrt{4096} = \sqrt{2^{12}} = 2^6 = 64
\]

Answer:
\[
\boxed{64}
\]

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Problem 7: Find the least perfect square exactly divisible by each one of the numbers 4, 5, 10.

Solution:
1. Find the LCM of 4, 5, and 10:
- Prime factorisation:
\[
4 = 2^2, \quad 5 = 5, \quad 10 = 2 \times 5
\]
- LCM is the product of the highest powers of all prime factors:
\[
\text{LCM} = 2^2 \times 5 = 20
\]

2. Make the LCM a perfect square:
- The prime factorisation of 20 is:
\[
20 = 2^2 \times 5
\]
- For it to be a perfect square, the power of 5 must be even. Multiply by 5:
\[
20 \times 5 = 100
\]
- Check if 100 is a perfect square:
\[
\sqrt{100} = 10
\]
- Yes, 100 is a perfect square.

Answer:
\[
\boxed{100}
\]

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Problem 8: Find the least square number exactly divisible by each one of the numbers 6, 9, 10, 15, and 20.

Solution:
1. Find the LCM of 6, 9, 10, 15, and 20:
- Prime factorisation:
\[
6 = 2 \times 3, \quad 9 = 3^2, \quad 10 = 2 \times 5, \quad 15 = 3 \times 5, \quad 20 = 2^2 \times 5
\]
- LCM is the product of the highest powers of all prime factors:
\[
\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180
\]

2. Make the LCM a perfect square:
- The prime factorisation of 180 is:
\[
180 = 2^2 \times 3^2 \times 5
\]
- For it to be a perfect square, the power of 5 must be even. Multiply by 5:
\[
180 \times 5 = 900
\]
- Check if 900 is a perfect square:
\[
\sqrt{900} = 30
\]
- Yes, 900 is a perfect square.

Answer:
\[
\boxed{900}
\]

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Final Answers:

1. \(\boxed{(a) 3364, (b) 4489}\)
2. \(\boxed{(a) 47, (b) 99, (c) 209}\)
3. \(\boxed{(a) 25, (b) 144}\)
4. \(\boxed{(a) 32, (b) 94, (c) 92}\)
5. \(\boxed{11}\)
6. \(\boxed{64}\)
7. \(\boxed{100}\)
8. \(\boxed{900}\)