ML Aggarwal Squares and Squares Roots Exe-3.4 Class 8 ICSE Ch-3 Maths Solutions

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

ml class 8 ch 3 img 37

(ii) √4489 = 67

By division method

ml class 8 ch 3 img 37

(iii) √106929 = 327

By division method

ml class 8 ch 3 img 39

(iv) √167281 = 409

By division method

ml class 8 ch 3 img 39

(v) √53824 = 232

By division method

ml class 8 ch 3 img 41

(vi) √213444 = 462

By division method

ml class 8 ch 3 img 41


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

ml class 8 ch 3 img 43

(ii) √42.25 = 6.5

By division method

ml class 8 ch 3 img 43

(iii) √18.4041 = 4.29

By division method

ml class 8 ch 3 img 45

(iv) √5.774409 = 2.403

By division method

ml class 8 ch 3 img 45

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

ml class 8 ch 3 img 47

(ii) √107.45 = 10.36

It can be written as

ml class 8 ch 3 img 47

(iii) √5.462 = 2.337 = 2.34

It can be written as

ml class 8 ch 3 img 49

(iv) √2 = 1.41

It can be written as

ml class 8 ch 3 img 49

(v) √3 = 1.73

It can be written as

ml class 8 ch 3 img 49

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

ml class 8 ch 3 img 53

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

ml class 8 ch 3 img 53

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

ml class 8 ch 3 img 55

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

ml class 8 ch 3 img 55

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

ml class 8 ch 3 img 57

By taking the square root

412 is less than 1750

So by taking 422

164 – 150 = 14 less

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

(ii) 6412

We know that

ml class 8 ch 3 img 57

By taking the square root

802 is less than 6412

So by taking 812

161 – 12 = 14 less

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

(iii) 6598

We know that

ml class 8 ch 3 img 59

By taking the square root

812 is less than 6598

So by taking 822

324 – 198 = 126 less

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

(iv) 8000

We know that

ml class 8 ch 3 img 59

By taking the square root

892 is less than 8000

So by taking 902

8100 – 8000 = 100 less

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

ml class 8 ch 3 img 61

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

ml class 8 ch 3 img 62

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

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

Take 322 = 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

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

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 = 900.

(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

(i) In a right-angled triangle ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

Substituting the values

= 142 + 482

By further calculation

= 196 + 2304

= 2500

So we get

AC = √2500 = 50 cm

ml class 8 ch 3 img 65

(ii) In a right triangle, ABC

B = 900, AC = 37 cm, BC = 35 cm

ml class 8 ch 3 img 66

Using Pythagoras Theorem

AC2 = AB2 + BC2

Substituting the values

372 = AB2 + 352

By further calculation

1369 = AB2 + 1225

AB2 = 1369 – 1225 = 144

So we get

AB = √144 = 12 cm

ml class 8 ch 3 img 66

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

ml class 8 ch 3 img 68

Here

Number of columns = Number of rows

By taking the square root of 1400

372 < 1400

So take 382 = 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

ml class 8 ch 3 img 69

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

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?

Using Pythagoras theorem

OB2 = OA2 + AB2

Substituting the values

= 162 + 632

By further calculation

= 256 + 3969

= 4225

So we get

OB = √4225 = 65

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?

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

ml class 8 ch 3 img 73

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

Using the Pythagoras theorem,

AB2 = AC2 + BC2

Substituting the values

62 = 4.82 + BC2

By further calculation

BC2 = 62 – 4.82

BC2 = 36 – 23.04 = 12.96

So we get

BC = √12.96 = 3.6 m

ml class 8 ch 3 img 74

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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