AP Board Class 8 Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Book Answers |

## Andhra Pradesh Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbooks Solutions PDF

Andhra Pradesh State Board STD 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbooks. These Andhra Pradesh State Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.## Andhra Pradesh State Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Books Solutions

Board | AP Board |

Materials | Textbook Solutions/Guide |

Format | DOC/PDF |

Class | 8th |

Subject | Maths |

Chapters | Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 |

Provider | Hsslive |

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## AP Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 8th Maths Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:Question 1.

Find the square roots of the following numbers by Prime factorization method.

(1) 441

(ii) 784

(iii) 4096

(iv) 7056

Solution:

Question 2.

Find the smallest number by which 3645 must be multiplied to get a perfect square.

Solution:

The prime factorization of 3645

= (3 × 3) × (3 × 3) (3 × 3) × 5

∴ Deficiency of one ‘5’ is appeared in the above product.

∴ 3645 is multiplied with 5 then we will get a perfect square.

Question 3.

Find the smallest number by which 2400 is to be multiplied to get a perfect square and also find the square root of the resulting number.

Solution:

The prime factorization of 2400

=(2 × 2) × (2 × 2) × 2 × (5 × 5) × 3

∴ 2,3 are needed to form a pair

∴ 2 × 3 = 6

∴ 6 should be multiplied with 2400 then we will get a perfect square number.

∴ 2400 × 6 = 14400

∴ 14400‾‾‾‾‾‾√ = 120

Question 4.

Find the smallest number by which 7776 is to be divided to get a perfect square.

Solution:

The prime factorization of 7776

=(2 × 2) × (2 × 2) × 2 × (3 × 3) × (3 × 3) × 3

∴ 2, 3 are needed to form a pair

∴ 2 × 3 = 6

∴ 7776 should be divided by 6 then we will get a perfect square number.

Question 5.

1521 trees are planted in a garden in such a way that there are as many trees in each row as there are rows in the garden. Find the number of rows and number of trees in each row.

Solution:

Let the no. of trees planted in a garden for each row = x say.

No. of rows in the garden = x

∴ Total no. of trees in the garden = x × x = x^{2}

According to the sum x^{2} = 1521

x = 1521‾‾‾‾‾√=39×39‾‾‾‾‾‾‾√ = 39

∴ No. of trees for each row = 39

No. of rows in the garden = 39

Question 6.

A school collected ₹ 2601 as fees from its students. If fee paid by each student and number students in the school were equal, how many students were there in the school?

Solution:

Let the no. of students in a school = x say

The (amount) fee paid by each student = ₹ x

Amount collected by all the students

= x × x = x^{2}

According to the sum

∴ x^{2} = 2601

x = 2601‾‾‾‾‾√=51×51‾‾‾‾‾‾‾√ = 51

∴ x = 51

∴ No. of students in the school = 51

Question 7.

The product of two numbers is 1296. If one number is 16 times the other, find the two numbers?

Solution:

Given that the product of two numbers = 1296.

Let the second number = x say

Then first number = 16 × x = 16x

∴ The product of two numbers

= x × 16x= 16x^{2}

According to the sum

16x^{2} = 1296

⇒ x^{2} = 129616 = 81

⇒ x^{2} = 81

⇒ x = 8‾√1⎯⎯⎯=9×9‾‾‾‾‾√ = 9

⇒ x = 9

∴ The first number = 16x

= 16 × 9

=144

The second number = x = 9

Question 8.

7921 soldiers sat in an auditorium in such a way that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium’?

Solution:

Let the number of soldiers sat in an auditorium for each row = x say

∴ No. of rows in an auditorium = x

∴ Total no. of soldiers = x × x = x^{2}

According to the sum,

x^{2} = 7921

x = 7921‾‾‾‾‾√=89×89‾‾‾‾‾‾‾√ = 89

∴ No. of rowS in an auditorium = 89

Question 9.

The area of a square field is 5184 m^{2}. Find the area of a rectangular field, whose perimeter is equal to the perimeter of the square field and whose length is twice of its breadth.

Solution:

Area of a square field = 5184 m^{2}

A = s^{2} = 5184

:. s = 5184‾‾‾‾‾√=72×72‾‾‾‾‾‾‾√ = 72

∴ s = 72

∴ Perimeter of the square field = 4 × s

= 4 × 72

= 288 m

According to the sum,

Perimeter of a rectangular field

= Perimeter of a square field = 288 m

Let the breadth of a rectangular field

= x m say

∴ Length = 2 × x = 2 × m

∴ Perimeter of the rectangular field

= 2 (1 + b)

= 2 (2x + x)

= 2 × 3x

= 6x

∴ 6x = 288 .

x = 2886

x = 48

∴ Breadth of the rectangular field

= x = 48 m

Length of the rectangular field = 2x

= 2 × 48

=96m

∴ Area of the rectangular field

= l × b

= 96 × 48

= 4608 m^{2}

## AP Board Textbook Solutions PDF for Class 8th Maths

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