ML Aggarwal Squares and Squares Roots Exe-3.4 Class 8 ICSE Ch-3 Maths Solutions. We Provide Step by Step Answer of Exe-3.4 Questions for Squares and Squares Roots as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

## ML Aggarwal Squares and Squares Roots Exe-3.4 Class 8 ICSE Maths Solutions

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-3 | Squares and Squares Roots |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-3.4 Questions |

Edition | 2023-2024 |

**Squares and Squares Roots Exe-3.4**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-62

**Question 1. ****Find the square root of each of the following by division method:**

(i) 2401

(ii) 4489

(iii) 106929

(iv) 167281

(v) 53824

(vi) 213444

**Answer:**

**(i) √2401 = 49**

By division method

**(ii) √4489 = 67**

By division method

**(iii) √106929 = 327**

By division method

**(iv) √167281 = 409**

By division method

**(v) √53824 = 232**

By division method

**(vi) √213444 = 462**

By division method

**Squares and Squares Roots Exe-3.4**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-63

**Question 2. ****Find the square root of the following decimal numbers by division method:**

(i) 51.84

(ii) 42.25

(iii) 18.4041

(iv) 5.774409

**Answer:**

**(i) √51.84 = 7.2**

By division method

**(ii) √42.25 = 6.5**

By division method

**(iii) √18.4041 = 4.29**

By division method

**(iv) √5.774409 = 2.403**

By division method

**Question 3. ****Find the square root of the following numbers correct to two decimal places:**

(i) 645.8

(ii) 107.45

(iii) 5.462

(iv) 2

(v) 3

**Answer:**

**(i) √645.8 = 25.41**

It can be written as

**(ii) √107.45 = 10.36**

It can be written as

**(iii) √5.462 = 2.337 = 2.34**

It can be written as

**(iv) √2 = 1.41**

It can be written as

**(v) √3 = 1.73**

It can be written as

**Question 4. Find the square root of the following fractions correct to two decimal places:**

(i) 11(3/8)

(ii) 5(5/11)

(iii) 7(1/3)

**Answer: update soon**

**Question 5. ****Find the least number which must be subtracted from each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:**

(i) 2000

(ii) 984

(iii) 8934

(iv) 11021

**Answer:**

**(i) 2000**

We know that

By taking the square root, 64 is left as the remainder

Subtracting 64 from 2000

We get 1936 which is a perfect square, and its square root is 44.

**(ii) 984**

We know that

By taking the square root, 23 is left as the remainder

Subtracting 23 from 984

We get 961 which is a perfect square, and its square root is 31.

**(iii) 8934**

We know that

By taking the square root, 98 is left as the remainder

Subtracting 98 from 894

We get 8934 – 98 = 8836 which is a perfect square, and its square root is 94.

**(iv) 11021**

We know that

By taking the square root, 205 is left as the remainder

Subtracting 205 from 11021

We get 11021 – 205 = 10816 which is a perfect square, and its square root is 104.

**Question 6. ****Find the least number which must be added to each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:**

(i) 1750

(ii) 6412

(iii) 6598

(iv) 8000

**Answer:**

**(i) 1750**

We know that

By taking the square root

41^{2} is less than 1750

So by taking 42^{2}

164 – 150 = 14 less

Adding 14, we get a square of 42 which is 1764.

**(ii) 6412**

We know that

By taking the square root

80^{2} is less than 6412

So by taking 81^{2}

161 – 12 = 14 less

Adding 149, we get a square of 81 which is 6561.

**(iii) 6598**

We know that

By taking the square root

81^{2} is less than 6598

So by taking 82^{2}

324 – 198 = 126 less

Adding 126, we get a square of 82 which is 6724.

**(iv) 8000**

We know that

By taking the square root

89^{2} is less than 8000

So by taking 90^{2}

8100 – 8000 = 100 less

Adding 100, we get a square of 90 which is 8100.

**Question 7. ****Find the smallest four-digit number, which is a perfect square.**

**Answer:**

**It is given that**

Smallest four–digit number = 1000

We know that

By taking the square root, we find that 39 is left.

If we subtract any number from 1000, we get 3 digit number

Take 32^{2} = 1024

Here 1024 – 1000 = 24 is to be added to get a perfect square of least 4 digit number

Therefore, the required 4-digit smallest number is 1024.

**Question 8. ****Find the greatest number of six digits, which is a perfect square.**

**Answer:**

It is given that

Greatest six-digit number = 999999

We know that

By taking the square root, we find that 1998 is left

If we subtract 1998 from 999999, we get 998001 which is a perfect square.

Therefore, the required six-digit greatest number is 998001.

**Question 9. ****In a right triangle ABC, ∠B = 90**^{0}.

^{0}.

(i) If AB = 14 cm, BC = 48 cm, find AC.

(ii) If AC = 37 cm, BC = 35 cm, find AB.

**Answer:**

**(i) In a right-angled triangle ABC**

It is given that

AB = 14 cm and BC = 48 cm

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

= 14^{2} + 48^{2}

By further calculation

= 196 + 2304

= 2500

So we get

AC = √2500 = 50 cm

**(ii) In a right triangle, ABC**

B = 90^{0}, AC = 37 cm, BC = 35 cm

Using Pythagoras Theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

37^{2} = AB^{2} + 352

By further calculation

1369 = AB^{2} + 1225

AB^{2} = 1369 – 1225 = 144

So we get

AB = √144 = 12 cm

**Question 10. ****A gardener has 1400 plants. He wants to plant these in such a way that the number of rows and columns remains the same. Find the minimum number of plants he needs more for this.**

**Answer:**

It is given that

Total number of plants = 1400

We know that

Here

Number of columns = Number of rows

By taking the square root of 1400

37^{2} < 1400

So take 38^{2} = 1444

We need 1444 – 1400 = 44 plants more

Therefore, the minimum number of plants he needs more for this is 44.

**Question 11. ****There are 1000 children in a school. For a P.T. drill, they have to stand in such a way that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?**

**Answer:**

It is given that

No. of total children in a school = 1000

For a P.T. drill, children have to stand in such a way that

No. of rows = No. of columns

Take the square root of 1000

39 is left as the remainder

Left out children = 39

Hence, 39 children would be left out in this arrangement.

**Question 12. ****Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?**

**Answer:**

It is given that

Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend’s house

Consider O as the house and A and B as the places

OA = 16 m, AO = 63 m

Using Pythagoras theorem

OB^{2} = OA^{2} + AB^{2}

Substituting the values

= 16^{2} + 63^{2}

By further calculation

= 256 + 3969

= 4225

So we get

OB = √4225 = 65

Therefore, Amit has to walk 65 m to reach his house.

**Question 13. ****A ladder 6 m long leaned against a wall. The ladder reaches the wall to a height of 4.8 m. Find the distance between the wall and the foot of the ladder.**

**Answer:**

It is given that

Length of ladder = 6 m

The ladder reaches the wall to a height of 4.8 m

Consider AB as the ladder and AC as the height of the wall

AB = 6 m and AC = 4.8 m

The distance between the foot of the ladder and the wall is BC

Using the Pythagoras theorem,

AB^{2} = AC^{2} + BC^{2}

Substituting the values

6^{2} = 4.8^{2} + BC^{2}

By further calculation

BC^{2} = 6^{2} – 4.8^{2}

BC^{2} = 36 – 23.04 = 12.96

So we get

BC = √12.96 = 3.6 m

Hence, the distance between the wall and the foot of the ladder is 3.6 m.

— : End of ML Aggarwal Squares and Squares Roots Exe-3.4 Class 8 ICSE Maths Solutions :–

Return to **– ML Aggarwal Maths Solutions for ICSE Class -8**

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