Squares and Cube Roots Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3. We provide step by step Solutions of council prescribe publication / textbook. Visit official Website **CISCE** for detail information about ICSE Board Class-8 Mathematics.

## Squares and Squares Roots Cube and Cube Roots

Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3

Board | ICSE |

Publications | Goyal Brothers Prakshan |

Subject | Maths |

Class | 8th |

writer | RS Aggarwal |

Book Name | Foundation |

Chapter-3 | Squares and Squares Roots Cube and Cube Roots |

Exe-3A | Method of Finding Squire Roots |

Edition | 2024-2025 |

### Method of Finding Squire Roots

Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3

**Page- 42**

**Exercise- 3A**

Squares and Squares Roots Cube and Cube Roots Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3

**Que-1: Find the square of each of the following numbers :**

**(i) 14 (ii) 137 (iii) 4/17 (iv) 2*(3/4) ****(v) 0.01 (vi) 1.2 (vii) 0.17 (viii) 4.6**

**Solution- **(i) 14² = 196

(ii) 137² = 18769

(iii) (4/17)² = 16/289

(iv) [2*(3/4)]² = (11/4)² = 121/16 = 7*(9/16)

(v) 0.01² = 0.0001

(vi) 1.2² = 1.44

(vii) 0.17² = 0.0289

(viii) 4.6² = 21.16

**Que-2: Using prime factorization method, find which of the following are perfect squares numbers :**

**(i) 252 (ii) 324 (iii) 676 (iv) 1225 ****(iv) 2916 (vi) 3528 (vii) 5625 (viii) 9075**

**Solution- **(i) Prime factorization of 252 = 2^2×3^2×7

the exponent of 7 is not even, so 252 is not a perfect square.

(ii) Prime factorization of 324 = 2^2×3^4

In this case, both exponents are even, so 324 is a perfect square.

(iii) Prime factorization of 676 = 2^2×13^2

Both exponents are even, so 676 is a perfect square.

(iv) Prime factorization of 1225 = 5^2×7^2

Both exponents are even, so 1225 is a perfect square.

(v) Prime factorization of 2916 = 2^2×3^4×17

The exponents of 2 and 3 are even, but the exponent of 17 is not even, so 2916 is not a perfect square.

(vi) Prime factorization of 3528 : 2^3×3^2×7^2

Each prime factor occurs an even number of times (3, 2, 2), so 3528 is a perfect square.

(vii) Prime factorization of 5625 : 3^4×5^2

Each prime factor occurs an even number of times (4, 2), so 5625 is a perfect square.

(viii) Prime factorization of 9075 : 3^2×5^2×11

The power of 11 is odd, so 9075 is not a perfect square

**Que-3: Show that each of the following numbers is a perfect square. Also, find the square root of each :**

**(i) 441 (ii) 784 (iii) 1225 (iv) 3969 ****(v) 2601 (vi) 5929 (vii) 7056 (viii) 8281**

**Solution- **(i) √441 = 21

(ii) √784 = 28

(iii) √1225 = 35

(iv) √3969 = 63

(v) √2601 = 51

(vi) √5929 = 77

(vii) √7056 = 84

(viii) √8281 = 81

**Que-4: Find the smallest number by which each of the following numbers must be multiplied to get a perfect square number. Also, find the square root of the resulting number :**

**(i) 588 (ii) 2592 (iii) 3332 (iv) 3380**

**Solution- **(i) 588 = 2 × 2 × 7 × 7 × 3

Common factors = 2 × 7

Uncommon factor = 3

Now only 3 which is an uncommon factor that is not in pair. So the least number by which 588 be multiplied to get a perfect number is 3.

The square root of resulting number is

2 x 7 x 3 = 42.

(ii) let’s factorise it

factors of 2592: 2×2×2×2×2×9×9

=(2×2)(2×2)(9×9)(2)

so the smallest no by which 2592 should be multiplied to get a perfect square 2.

so the square no is 5184

let’s find its square root :

√5184 = 72.

(iii) 3332 = 2×2×7×7×17

= (2)^2×(7)^2×17

In order to get a perfect square, the given number should be multiplied by 17

The square root of the new number =(2)×(7)×17=238.

(iv) let’s factorise it

factors : 2×2×5×13×13

= (2×2)(13×13)(5)

so the smallest no by which 3380 should be multiplied to get a perfect square is 5

resulted no : 3380×5 = 16900

square root of 16900 = 130

**Que-5: By which least number should be given number be divided to obtain a perfect square number? Also, find the resulting perfect square number and its square root :**

**(i) 1728 (ii) 4500 (iii) 7776 (iv) 8820**

**Solution- **(i) The least number is 3.

1728 ÷ 3 = 576

Square root of 576 is 24.

(ii) Resolve 4500 into prime factors, we get

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

Now to get perfect square we have to divide the above equation by 5 **Ans**.

Then we get, 4500/5 = 2×2×3×3×5×5

New number =(4×9×25)

=(2^2×3^2×5^2) = 900 **Ans**.

Taking squares as common from the above equation we get

∴ New number = (2×3×5)^2

=(30)^2

Hence, the new number is square of 30 **Ans.**

(iii) The least number is 6.

7776 ÷ 6 = 1296

Square root of 1296 is 36.

(iv) The least number is 5.

8820 ÷ 5 = 1764

Square root of 1764 is 42.

**Que-6: The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is Rs 1156, find the strength of a class.**

**Solution-**Let no. of students in the class = x

Then contribution of each student = x

∴x × x = 1156

⇒ x^2 = 1156

∴ x = √1156 = √(2×2)(17×17)

= 2 × 17 = 34 Ans.

**Que-7: In an auditorium, the number of rows is equal to the number of chairs in each row. If the capacity of the auditorium is 2025, find the number of chairs in each row.**

**Solution- ** Let the number of chairs in each row be x.

Then, the number of rows = x.

Total number of chairs in the auditorium

= (x × x) = x²

But, the capacity of the auditorium = 2025 (given).

Therefore, x^2 = 2025

2025 = 5×5×3×3×3×3

= 5^2×3^4

⇒ √(2025)=√(5^2×3^4)

= 5×3×3 = 45

∴x = 5×3×3 = 45 Ans.

**Que-8: Find the least square number which is exactly divisible by each of the numbers 8,9,10 and 15.**

**Solution- **Now, LCM of 8, 9 and 10 = 2 × 2 × 2 × 3 × 3 × 5 = 360

We observe that 2 and 5 do not occur in pairs.

So, 360 is not a perfect square.

Now, 360 must be multiplied by 2 × 5 to get a perfect square

= 360 × 5 × 2

= 360 × 10

= 3600

**— : End of Squares and Cube Roots Class 8 RS Aggarwal Exe-3A Goyal Brothers Brothers Prakashan ICSE Foundation Maths Solutions :–**

**Return to :- ICSE Class -8 RS Aggarwal Goyal Brothers Math Solutions**

Thanks

Please share with yours friends if you find it helpful