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

## Squares and Cube Roots Class 8 RS Aggarwal Exe-3C Goyal Brothers ICSE Maths Solutions Ch-3

Board | ICSE |

Publications | Goyal Brothers Prakshan |

Subject | Maths |

Class | 8th |

Writer | RS Aggarwal |

Book Name | Foundation |

Ch-3 | Squire and Squire Roots Cube and Cube Roots |

Exe-3C | Squire Roots by Division Methods |

Edition | 2024-2025 |

### Squire Roots by Division Methods

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

**Page- 47**

**Exercise- 3C**

Squire Roots by Division Methods of Perfect Squire

**Que-1: Find the square root of each of the following numbers by division method :**

**(i) 576 (ii) 961 (iii) 1444 (iv) 4489 ****(v) 6241 (vi) 5476 (vii) 9025 (viii) 7569**

**Solution- **(i) √576 = 24

(ii) √961 = 31

(iii) √1444 = 38

(iv) √4489 = 67

(v) √6241 = 79

(vi) √5476 = 74

(vii) √9025 = 95

(viii) √7569 = 87

**Que-2: The area of a square field is 7744 sq. metres. Find the perimeter.**

**Solution- **We know that the area of square is a²

⇒ 7744 = a²

⇒ a = √7744

⇒ a = 88m

We know that perimeter of square is 4a

Therefore perimeter = 4×88 = 352m **Ans**.

**Que-3: Find the least number which must be subtracted from 1104 to obtain a perfect square. Find this perfect square and its square root.**

**Solution-**The nearest perfect square below 1104 is (33)^2 = 1089.

So, the difference between 1104 and 1089 is 1104−1089 = 15.

To obtain a perfect square, we need to subtract 15 from 1104.

1104 – 15 = 1089

So, the perfect square is 1089, and its square root is √1089 = 33 **Ans**.

**Que-4: Find the least number which must be subtracted from 7956 to obtain a perfect square. Find this perfect square and its square root.**

**Solution- **The nearest perfect square below 7956 is (89)^2 = 7921.

So, the difference between 7956 and 7921 is 7956−7921 = 35.

To obtain a perfect square, we need to subtract 35 from 7956.

7956 – 35 = 7921

So, the perfect square is 7921, and its square root is √7921 = 89 **Ans**.

**Que-5: ****Find the least number which must be added to 6203 to obtain a perfect square. Find the perfect square and its square root.**

**Solution- **The nearest perfect square above 6203 is (79)^2 = 6241.

So, the difference between 6241 and 6203 is 6241−6203 = 38.

To obtain a perfect square, we need to add 38 to 6203.

6203 + 38 = 6241

So, the perfect square is 6241, and its square root is √6241 = 79 **Ans**.

**Que-6: Find the least number which must be added to 7348 to obtain a perfect square. Find this perfect square and its square root.**

**Solution- **The nearest perfect square above 7348 is (86)^2 = 7396.

So, the difference between 7396 and 7348 is 7396−7348 = 48.

To obtain a perfect square, we need to add 48 to 7348.

7348 + 48 = 7396

So, the perfect square is 7396, and its square root is √7396 = 86 **Ans**.

**Que-7: Find the greatest number of six digits, which is a perfect square. Find the square root of this number.**

**Solution- **The largest perfect square less than 1,000,000 is (999)^2 = 998001.

So, the greatest six-digit number that is a perfect square is 9,98,001.

The square root of 998,001 is √998001 = 999 **Ans**.

**Que-8: Find the least number of four digits which is a perfect square.**

**Solution- **The smallest perfect square greater than 999 is (32)^2 = 1024.

So, the least four-digit number that is a perfect square is 1024 **Ans**.

**Que-9: Find the least number by which 69192 must be (i) decreased (ii) increased (iii) multiplied (iv) divided, to make it a perfect square.**

**Solution- **√69192 = 263.04

i.e., 263² < 69192 < 264²

(i) For decreased number,

Subtract 263² from 69192

i.e., 69192 – 263² = 69192 – 69169 = 23.

(ii) For increased number,

Subtract 69192 from 264²

i.e., 264² – 69192 = 69696 – 69192 = 504.

(iii) The factor of the number 69192 is

69192 = 2 x 2 x 2 x 3 x 3 x 31 x 31

If we see the pairs only one 2 is left alone.

So, If we multiply it with 2 it will make a perfect square.

Therefore, The least multiplied number to make it a perfect square is 2.

(iv) If we see the pairs only one 2 is left alone.

So, If we divide it with 2 it will make a perfect square.

Therefore, The least divide number to make it a perfect square is 2.

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

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

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