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

Problem: Solve the given problems related to squares and square roots.

#### Problem 1: Find the square root of 6400.

To find the square root of 6400, we can use the prime factorization method or direct calculation.

- Prime factorization of 6400:
\[
6400 = 2^7 \times 5^2
\]
- Group the factors into pairs:
\[
6400 = (2^3 \times 5)^2 = 80^2
\]
- Therefore, the square root of 6400 is:
\[
\sqrt{6400} = 80
\]

Answer:
\[
\boxed{80}
\]

---

#### Problem 2: Is 90 a perfect square?

A perfect square is a number that can be expressed as the square of an integer. To check if 90 is a perfect square, we can try to find its square root.

- The square root of 90 is approximately:
\[
\sqrt{90} \approx 9.4868
\]
- Since 9.4868 is not an integer, 90 is not a perfect square.

Answer:
\[
\text{No, 90 is not a perfect square.}
\]

---

#### Problem 3: Is 2352 a perfect square? If not, find the smallest multiple of 2352 which is a perfect square. Find the square root of the new number.

- First, check if 2352 is a perfect square by finding its prime factorization:
\[
2352 = 2^4 \times 3 \times 7^2
\]
- For a number to be a perfect square, all the exponents in its prime factorization must be even. Here, the exponent of 3 is 1, which is odd.
- To make 2352 a perfect square, we need to multiply it by 3 (to make the exponent of 3 even):
\[
2352 \times 3 = 2^4 \times 3^2 \times 7^2 = (2^2 \times 3 \times 7)^2 = 84^2
\]
- The smallest multiple of 2352 that is a perfect square is 7056.
- The square root of 7056 is:
\[
\sqrt{7056} = 84
\]

Answer:
\[
\text{No, 2352 is not a perfect square. The smallest multiple of 2352 that is a perfect square is 7056, and its square root is 84.}
\]

---

#### Problem 4: Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient.

- First, find the prime factorization of 9408:
\[
9408 = 2^6 \times 3 \times 7^2
\]
- For the quotient to be a perfect square, all the exponents in the prime factorization must be even. Here, the exponent of 3 is 1, which is odd.
- To make the quotient a perfect square, we need to divide 9408 by 3:
\[
\frac{9408}{3} = 2^6 \times 7^2 = (2^3 \times 7)^2 = 56^2
\]
- The smallest number by which 9408 must be divided is 3.
- The square root of the quotient (56) is:
\[
\sqrt{56^2} = 56
\]

Answer:
\[
\text{The smallest number by which 9408 must be divided is 3, and the square root of the quotient is 56.}
\]

---

#### Problem 5: Without doing any calculation, find the numbers which are surely not perfect squares.

- A perfect square cannot end with certain digits. Specifically, a perfect square cannot end with 2, 3, 7, or 8.
- Check each option:
- i) 153: Ends with 3 → Not a perfect square.
- ii) 257: Ends with 7 → Not a perfect square.
- iii) 408: Ends with 8 → Not a perfect square.
- iv) 441: Ends with 1 → Could be a perfect square (but we are not calculating).

Answer:
\[
\text{The numbers which are surely not perfect squares are: i) 153, ii) 257, iii) 408.}
\]

---

#### Problem 6: Find the square root of the following numbers by the prime factorisation method.

- i) 400:
\[
400 = 2^4 \times 5^2 = (2^2 \times 5)^2 = 20^2 \quad \Rightarrow \quad \sqrt{400} = 20
\]
- ii) 9604:
\[
9604 = 2^2 \times 7^4 = (2 \times 7^2)^2 = 98^2 \quad \Rightarrow \quad \sqrt{9604} = 98
\]
- iii) 8100:
\[
8100 = 2^2 \times 3^4 \times 5^2 = (2 \times 3^2 \times 5)^2 = 90^2 \quad \Rightarrow \quad \sqrt{8100} = 90
\]
- iv) 1764:
\[
1764 = 2^2 \times 3^2 \times 7^2 = (2 \times 3 \times 7)^2 = 42^2 \quad \Rightarrow \quad \sqrt{1764} = 42
\]
- v) 5929:
\[
5929 = 7^2 \times 11^2 = (7 \times 11)^2 = 77^2 \quad \Rightarrow \quad \sqrt{5929} = 77
\]
- vi) 9216:
\[
9216 = 2^{10} \times 3^2 = (2^5 \times 3)^2 = 96^2 \quad \Rightarrow \quad \sqrt{9216} = 96
\]

Answers:
\[
\text{i) } 20, \text{ ii) } 98, \text{ iii) } 90, \text{ iv) } 42, \text{ v) } 77, \text{ vi) } 96
\]

---

#### Problem 7: For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square. Also find the square root of the square number so obtained.

- i) 252:
\[
252 = 2^2 \times 3^2 \times 7 \quad \Rightarrow \quad \text{Multiply by 7: } 252 \times 7 = 1764 = 42^2 \quad \Rightarrow \quad \sqrt{1764} = 42
\]
- ii) 2925:
\[
2925 = 3^2 \times 5^2 \times 13 \quad \Rightarrow \quad \text{Multiply by 13: } 2925 \times 13 = 38025 = 195^2 \quad \Rightarrow \quad \sqrt{38025} = 195
\]
- iii) 396:
\[
396 = 2^2 \times 3^2 \times 11 \quad \Rightarrow \quad \text{Multiply by 11: } 396 \times 11 = 4356 = 66^2 \quad \Rightarrow \quad \sqrt{4356} = 66
\]
- iv) 2028:
\[
2028 = 2^2 \times 3 \times 13^2 \quad \Rightarrow \quad \text{Multiply by 3: } 2028 \times 3 = 6084 = 78^2 \quad \Rightarrow \quad \sqrt{6084} = 78
\]
- v) 1458:
\[
1458 = 2 \times 3^6 \quad \Rightarrow \quad \text{Multiply by 2: } 1458 \times 2 = 2916 = 54^2 \quad \Rightarrow \quad \sqrt{2916} = 54
\]
- vi) 768:
\[
768 = 2^8 \times 3 \quad \Rightarrow \quad \text{Multiply by 3: } 768 \times 3 = 2304 = 48^2 \quad \Rightarrow \quad \sqrt{2304} = 48
\]

Answers:
\[
\text{i) } 7, 42; \text{ ii) } 13, 195; \text{ iii) } 11, 66; \text{ iv) } 3, 78; \text{ v) } 2, 54; \text{ vi) } 3, 48
\]

---

#### Problem 8: For each of the following number, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

- i) 252:
\[
252 = 2^2 \times 3^2 \times 7 \quad \Rightarrow \quad \text{Divide by 7: } \frac{252}{7} = 36 = 6^2 \quad \Rightarrow \quad \sqrt{36} = 6
\]
- ii) 180:
\[
180 = 2^2 \times 3^2 \times 5 \quad \Rightarrow \quad \text{Divide by 5: } \frac{180}{5} = 36 = 6^2 \quad \Rightarrow \quad \sqrt{36} = 6
\]
- iii) 1008:
\[
1008 = 2^4 \times 3^2 \times 7 \quad \Rightarrow \quad \text{Divide by 7: } \frac{1008}{7} = 144 = 12^2 \quad \Rightarrow \quad \sqrt{144} = 12
\]
- iv) 2028:
\[
2028 = 2^2 \times 3 \times 13^2 \quad \Rightarrow \quad \text{Divide by 3: } \frac{2028}{3} = 676 = 26^2 \quad \Rightarrow \quad \sqrt{676} = 26
\]
- v) 1458:
\[
1458 = 2 \times 3^6 \quad \Rightarrow \quad \text{Divide by 2: } \frac{1458}{2} = 729 = 27^2 \quad \Rightarrow \quad \sqrt{729} = 27
\]
- vi) 768:
\[
768 = 2^8 \times 3 \quad \Rightarrow \quad \text{Divide by 3: } \frac{768}{3} = 256 = 16^2 \quad \Rightarrow \quad \sqrt{256} = 16
\]

Answers:
\[
\text{i) } 7, 6; \text{ ii) } 5, 6; \text{ iii) } 7, 12; \text{ iv) } 3, 26; \text{ v) } 2, 27; \text{ vi) } 3, 16
\]

---

#### Problem 9: The students of class VIII of a school donated Rs 2401 in all, for prime minister's national relief fund. Each student donated as many rupees as the no. of students in the class. Find the no. of students in the class.

- Let the number of students in the class be \( n \).
- Each student donated \( n \) rupees.
- Total donation = \( n \times n = n^2 \).
- Given that the total donation is Rs 2401:
\[
n^2 = 2401
\]
- Find the square root of 2401:
\[
\sqrt{2401} = 49
\]
- Therefore, the number of students in the class is 49.

Answer:
\[
\boxed{49}
\]

---

#### Problem 10: 2118 plants are to be planted in a garden in such a way that each row contains as many plants as the no. of rows. Find the no. of rows and the no. of plants in each row.

- Let the number of rows be \( x \).
- Each row contains \( x \) plants.
- Total number of plants = \( x \times x = x^2 \).
- Given that the total number of plants is 2118:
\[
x^2 = 2118
\]
- However, 2118 is not a perfect square. This means it is impossible to arrange 2118 plants in such a way that each row contains as many plants as the number of rows.

Answer:
\[
\text{It is not possible to arrange 2118 plants in the required manner.}
\]

---

Final Answers:

\[
\boxed{80, \text{No}, 7056, 84, 3, 56, \text{i) 153, ii) 257, iii) 408}, 20, 98, 90, 42, 77, 96, 7, 42; 13, 195; 11, 66; 3, 78; 2, 54; 3, 48, 7, 6; 5, 6; 7, 12; 3, 26; 2, 27; 3, 16, 49, \text{Not possible}}
\]