Math, asked by manashvi2006, 9 months ago

*MATHS*
CLASS - 8
CHAPTER 6
SQUARE AND SQUARE ROOTS

Q5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so
obtained:
i) 252
iii) 1008
v) 1458

Q6:For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also find the square root of the square number so obtained:

i) 2925
ii) 2645
iv) 1620

Q1: Find the square root of each of the following numbers by Division method:

i) 2304
ii) 3481
vi) 5776
viii) 1024

PLEASE GIVE CORRECT ANSWERS AND ACCORDING TO THE GIVEN METHOD GIVEN IN THE QUESTION . I WILL MARK BRAINLIEST.​

Attachments:

Answers

Answered by ritika123475
0

Answer:

answer is 7 for question no 1. 1

Attachments:
Answered by jatinraghav123
0

Q.5 Solution:

i.

NCERT Solution For Class 8 Maths Chapter 6 Image 10

252 = 2×2×3×3×7

= (2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will multiply 252 by 7 to get perfect square.

New number = 252 × 7 = 1764

NCERT Solution For Class 8 Maths Chapter 6 Image 11

1764 = 2×2×3×3×7×7

⇒ 1764 = (2×2)×(3×3)×(7×7)

⇒ 1764 = 22×32×72

⇒ 1764 = (2×3×7)2

⇒ √1764 = 2×3×7 = 42

ii.

NCERT Solution For Class 8 Maths Chapter 6 Image 12

180 = 2×2×3×3×5

= (2×2)×(3×3)×5

Here, 5 cannot be paired.

∴ We will multiply 180 by 5 to get perfect square.

New number = 180 × 5 = 900

NCERT Solution For Class 8 Maths Chapter 6 Image 13

900 = 2×2×3×3×5×5×1

⇒ 900 = (2×2)×(3×3)×(5×5)

⇒ 900 = 22×32×52

⇒ 900 = (2×3×5)2

⇒ √900 = 2×3×5 = 30

iii.

NCERT Solution For Class 8 Maths Chapter 6 Image 14

1008 = 2×2×2×2×3×3×7

= (2×2)×(2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will multiply 1008 by 7 to get perfect square.

New number = 1008×7 = 7056

NCERT Solution For Class 8 Maths Chapter 6 Image 15

7056 = 2×2×2×2×3×3×7×7

⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)

⇒ 7056 = 22×22×32×72

⇒ 7056 = (2×2×3×7)2

⇒ √7056 = 2×2×3×7 = 84

iv.

NCERT Solution For Class 8 Maths Chapter 6 Image 16

2028 = 2×2×3×13×13

= (2×2)×(13×13)×3

Here, 3 cannot be paired.

∴ We will multiply 2028 by 3 to get perfect square. New number = 2028×3 = 6084

NCERT Solution For Class 8 Maths Chapter 6 Image 17

6084 = 2×2×3×3×13×13

⇒ 6084 = (2×2)×(3×3)×(13×13)

⇒ 6084 = 22×32×132

⇒ 6084 = (2×3×13)2

⇒ √6084 = 2×3×13 = 78

v.

NCERT Solution For Class 8 Maths Chapter 6 Image 18

1458 = 2×3×3×3×3×3×3

= (3×3)×(3×3)×(3×3)×2

Here, 2 cannot be paired.

∴ We will multiply 1458 by 2 to get perfect square. New number = 1458 × 2 = 2916

NCERT Solution For Class 8 Maths Chapter 6 Image 19

2916 = 2×2×3×3×3×3×3×3

⇒ 2916 = (3×3)×(3×3)×(3×3)×(2×2)

⇒ 2916 = 32×32×32×22

⇒ 2916 = (3×3×3×2)2

⇒ √2916 = 3×3×3×2 = 54

vi.

NCERT Solution For Class 8 Maths Chapter 6 Image 20

768 = 2×2×2×2×2×2×2×2×3

= (2×2)×(2×2)×(2×2)×(2×2)×3

Here, 3 cannot be paired.

∴ We will multiply 768 by 3 to get perfect square.

New number = 768×3 = 2304

NCERT Solution For Class 8 Maths Chapter 6 Image 21

2304 = 2×2×2×2×2×2×2×2×3×3

⇒ 2304 = (2×2)×(2×2)×(2×2)×(2×2)×(3×3)

⇒ 2304 = 22×22×22×22×32

⇒ 2304 = (2×2×2×2×3)2

⇒ √2304 = 2×2×2×2×3 = 48

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