# RS Aggarwal Class 8 Math Third Chapter Squares and Square Roots Exercise 3A Solution

## EXERCISE 3A

**(1) Using the prime factorization method, find which of the following numbers are perfect squares:**

(i) 441 = 3 × 3 × 7 × 7 = 3^{2} × 7^{2}

(ii) 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 2^{6} × 3^{2}

(iii) 11025 = 5 × 5 × 3 × 3 × 7 × 7 = 5^{2} × 3^{2} × 7^{2}

(iv) 1176 = 2 × 2 × 2 × 3 × 7 × 7

(v) 5625 = 3 × 3 × 5 × 5 × 5 × 5 = 3^{2} × 5^{4}

(vi) 9075 = 3 × 5 × 5 × 11 × 11

(vii) 4225 = 5 × 5 × 13 × 13 = 5^{2} × 13^{2}

(viii) 1089 = 3 × 3 × 11 × 11 = 3^{2} × 11^{2}

**(2) Show that each of the following numbers is a perfect square. In each case, find the number whose square is the given number:**

**(i) 1225** = 5 × 5 × 7 × 7 = 5^{2} × 7^{2}

Thus, 1225 is the product of pairs of equal factors.

∴ 1225 is a perfect square.

Also = (5 × 7)^{2} = (35)^{2}

Hence, 35 is the number whose square is 1225.

**(ii) 2601** = 3 × 3 × 17 × 17 = 3^{2} × 17^{2}

Thus, 2601 is the product of pairs of equal factors.

∴ 2601 is a perfect square.

Also = (3 × 17)^{2} = (51)^{2}

Hence, 51 is the number whose square is 2601.

**(iii) 5929** = 7 × 7 × 11 × 11 = 7^{2} × 11^{2}

Thus, 5929 is the product of pairs of equal factors.

∴ 5929 is a perfect square.

Also = (7 × 11)^{2} = (77)^{2}

Hence, 77 is the number whose square is 5929.

**(iv) 7056** = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 = 2^{2 }× 2^{2} × 3^{2} × 7^{2}

Thus, 7056 is the product of pairs of equal factors.

∴ 7056 is a perfect square.

Also = (2 × 2 × 3 × 7)^{2} = (84)^{2}

Hence, 84 is the number whose square is 7056.

**(v) 8281** = 7 × 7 × 13 × 13 = 7^{2} × 13^{2}

Thus, 8281 is the product of pairs of equal factors.

∴ 8281 is a perfect square.

Also = (7 × 13)^{2} = (91)^{2}

Hence, 91 is the number whose square is 8281.

**(3) By what least number should the given number be multiplied to get a perfect square number? In each case, find the number whose square is the new number.**

**(i) 3675**

Solution: Resolving 3675 into prime factors, we get

3675 = 3 × 5 × 5 × 7 × 7 = (3 × 5^{2} × 7^{2})

Thus, to get a perfect square number, the given number should be multiplied by 3.

New number = (3^{2} × 5^{2} × 7^{2}) = (3 × 5 × 7)^{2} = (105)^{2}

Hence, the number whose square is the new number = 105.

**(ii) 2156**

Solution: Resolving 2156 into prime factors, we get

2156 = 2 × 2 × 7 × 7 × 11

Thus, to get a perfect square number, the given number should be multiplied by 11.

New number = (2^{2} × 7^{2} × 11^{2}) = (2 × 7 × 11)^{2} = (154)^{2}

Hence, the number whose square is the new number = 154.

**(iii) 3332**

Solution: Resolving 3332 into prime factors, we get

3332 = 2 × 2 × 7 × 7 × 17

Thus, to get a perfect square number, the given number should be multiplied by 17.

New number = (2^{2} × 7^{2} × 17^{2}) = (2 × 7 × 17)^{2} = (238)^{2}

Hence, the number whose square is the new number = 238.

**(iv) 2925**

Solution: Resolving 2925 into prime factors, we get

2925 = 3 × 3 × 5 × 5 × 13

Thus, to get a perfect square number, the given number should be multiplied by 13.

New number = (3^{2} × 5^{2} × 13^{2}) = (3 × 5 × 13)^{2} = (195)^{2}

Hence, the number whose square is the new number = 195.

**(v) 9075**

Solution: Resolving 9075 into prime factors, we get

9075 = 3 × 5 × 5 × 11 × 11

Thus, to get a perfect square number, the given number should be multiplied by 3.

New number = (3^{2} × 5^{2} × 11^{2}) = (3 × 5 × 11)^{2} = (165)^{2}

Hence, the number whose square is the new number = 165.

**(vi) 7623**

Solution: Resolving 7623 into prime factors, we get

7623 = 3 × 3 × 7 × 11 × 11

Thus, to get a perfect square number, the given number should be multiplied by 7.

New number = (3^{2} × 7^{2} × 11^{2}) = (3 × 7 × 11)^{2} = (231)^{2}

Hence, the number whose square is the new number = 231.

**(vii) 3380**

Solution: Resolving 3380 into prime factors, we get

3380 = 2 × 2 × 5 × 13 × 13

Thus, to get a perfect square number, the given number should be multiplied by 5.

New number = (2^{2} × 5^{2} × 13^{2}) = (2 × 5 × 13)^{2} = (130)^{2}

Hence, the number whose square is the new number = 130.

**(viii) 2475**

Solution: Resolving 2475 into prime factors, we get

2475 = 3 × 3 × 5 × 5 × 11

Thus, to get a perfect square number, the given number should be multiplied by 11.

New number = (3^{2} × 5^{2} × 11^{2}) = (3 × 5 × 11)^{2} = (165)^{2}

Hence, the number whose square is the new number = 165.

**(4) By what least number should the given number be divided to get a perfect square number? In each case, find the number whose square is the new number.**

**(i) 1575**

Solution: Resolving 1575 into prime factors, we get

1575 = 3 × 3 × 5 × 5 × 7 = (3^{2} × 5^{2} × 7)

Thus, to get a perfect square number, the given number should be divided by 7.

New number obtained = (3^{2} × 5^{2}) = (3 × 5)^{2} = (15)^{2}

Hence, the number whose square is the new number = 15.

**(ii) 9075**

Solution: Resolving 9075 into prime factors, we get

9075 = 3 × 5 × 5 × 11 × 11 = (3 × 5^{2} × 11^{2})

Thus, to get a perfect square number, the given number should be divided by 3.

New number obtained = (5^{2} × 11^{2}) = (5 × 11)^{2} = (55)^{2}

Hence, the number whose square is the new number = 55.

**(iii) 4851**

Solution: Resolving 4851 into prime factors, we get

4851 = 3 × 3 × 7 × 7 × 11 = (3^{2} × 7^{2} × 11)

Thus, to get a perfect square number, the given number should be divided by 11.

New number obtained = (3^{2} × 7^{2}) = (3 × 7)^{2} = (21)^{2}

Hence, the number whose square is the new number = 21.

**(iv) 3380**

Solution: Resolving 3380 into prime factors, we get

3380 = 2 × 2 × 5 × 13 × 13 = (2^{2} × 5 × 13^{2})

Thus, to get a perfect square number, the given number should be divided by 5.

New number obtained = (2^{2} × 13^{2}) = (2 × 13)^{2} = (26)^{2}

Hence, the number whose square is the new number = 26.

**(v) 4500**

Solution: Resolving 4500 into prime factors, we get

4500 = 2 × 2 × 3 × 3 × 5 × 5 × 5 = (2^{2} × 3^{2 }× 5 × 5^{2})

Thus, to get a perfect square number, the given number should be divided by 5.

New number obtained = (2^{2} × 3^{2} × 5^{2}) = (2 × 3 × 5)^{2} = (30)^{2}

Hence, the number whose square is the new number = 30.

**(vi) 7776**

Solution: Resolving 7776 into prime factors, we get

7776 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 = (2^{2} × 2^{2 }× 2 × 3 × 3^{2} × 3^{2})

Thus, to get a perfect square number, the given number should be divided by 2 × 3.

New number obtained = (2^{2 }× 2^{2} × 3^{2} × 3^{2}) = (2 × 2 × 3 × 3)^{2} = (36)^{2}

Hence, the number whose square is the new number = 36.

**(vii) 8820**

Solution: Resolving 8820 into prime factors, we get

8820 = 2 × 2 × 3 × 3 × 5 × 7 × 7 = (2^{2} × 3^{2 }× 5 × 7^{2})

Thus, to get a perfect square number, the given number should be divided by 5.

New number obtained = (2^{2} × 3^{2} × 7^{2}) = (2 × 3 × 7)^{2} = (42)^{2}

Hence, the number whose square is the new number = 42.

**(viii) 4056**

Solution: Resolving 4056 into prime factors, we get

4500 = 2 × 2 × 2 × 3 × 13 × 13 = (2^{2} × 2 × 3 × 13^{2})

Thus, to get a perfect square number, the given number should be divided by 2 × 3.

New number obtained = (2^{2} × 13^{2}) = (2 × 13)^{2} = (26)^{2}

Hence, the number whose square is the new number = 26.

**(5) Find the largest number of 2 digits which is a perfect square.**

Ans: The largest 2 digits number is 99.

Square of 10 = 100 > 99, thus the number would be less than 10.

And the largest whole number less than 10 is 9.

Therefore, 9 × 9 = 81

**(6) Find the largest number of 3 digits which is a perfect square.**

Ans: The largest three digits number is 999. But 961 is a largest three digits number, is a perfect square.

961 = 31 × 31

Here, for easy to understand we take the before and after number of 31. Those are 30 and 32 respectively.

Now, 30 × 30 = 900 and 32 × 32 = 1024.

Hence, we can write 961 is largest three numbers has a perfect square.

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