CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Evaluate:

Question 1.
Solution:

Question 2.
Solution:

\(\sqrt { 33.64 } \) = 5.8

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

\(\sqrt { 10.0489 } \) = 3.17

Question 7.
Solution:

\(\sqrt { 1.0816 } \) = 1.04

Question 8.
Solution:

\(\sqrt { 0.2916 } \) = 0.54

Question 9.
Solution:
\(\sqrt { 3 } \) = 1.73

Question 10.
Solution:
\(\sqrt { 2.8 } \) = 1.6733 = 1.67

Question 11.
Solution:
\(\sqrt { 0.9 } \) = 0.948
= 0.95

Question 12.
Solution:
Length of rectangle (l) = 13.6 m
and width (b) = 3.4 m

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Evaluate:

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Finding the square root of 2509 by division we find that 9 is left as remainder

9 must be subtracted to get the perfect square 100.
Least number to be subtracted = 9

Question 14.
Solution:
Finding the square root of 7581 by division method, we find that 12 is left as remainder.
12 must be subtracted from 7581 to get a perfect square i.e., 7581 – 12 = 7569

(i) The least number to be subtracted = 12
(ii) Perfect square = 7569
(iii) and square root = 87 Ans.

Question 15.
Solution:
Finding the square root of 6203 by division method, we find that 38 is to be added to get a perfect square.
(i) Least number to be added = 38
(ii) Perfect square = 6241
(iii) Square root = 79 Ans.

Question 16.
Solution:
Finding the square root of 8400 by long division method, we find that 64 is to be added to 8400,
We, get 8400 + 64 = 8464

Least number to be added = 64
Perfect square = 8464
Square root = 92 Ans.

Question 17.
Solution:
Least four-digit number = 1000

Finding the square root of 1000 by the division method, we find that 24 must be added to get a perfect square of 4 digits.
Perfect square = 1000 + 24 = 1024 Ans.
square root of 1024 = 32

Question 18.
Solution:
Greatest number of five digits = 99999
Finding the square root of 99999
We get remainder = 143

Required perfect square = 99999 – 143 = 99856
and square root = 316 Ans

Question 19.
Solution:
Area of a square field = 60025 m²
Let its side = a

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Find the square root of each of the following numbers by using the method of prime factorization:

Question 1.
Solution:
225 = 3 x 3 x 5 x 5

Question 2.
Solution:
441 = 3 x 3 x 7 x 7

Question 3.
Solution:
729 =

Question 4.
Solution:
1296 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3

Question 5.
Solution:
2025 =

Question 6.
Solution:
4096 =

Question 7.
Solution:
7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7

Question 8.
Solution:
8100 =

Question 9.
Solution:
9216 =

Question 10.
Solution:
11025 =

Question 11.
Solution:
15876 =

Question 12.
Solution:
17424 =

Question 13.
Solution:
Factorizing 252, we get :
252 = \(\overline { 2X2 } X\overline { 3X3 } X7\)

To make it is a perfect square it must by multiplied by 7.
Square root of 252 x 7 = 1764
= 2 x 3 x 7 = 42 Ans.

Question 14.
Solution:
Factorizing 2925, we get :
2925 = \(\overline { 3X3 } X\overline { 5X5 } X13\)

To make it a perfect square it must be divided by 13.
2925 ÷ 13 = 225
Square root of 225 = 3 x 5 = 15 Ans.

Question 15.
Solution:
Let no. of rows = x
Then the number of plants in each row = x

Hence no. of rows = 35
and no. of plants in each row = 35

Question 16.
Solution:
Let no. of students in the class = x
Then contribution of each student = x

No. of students of the class = 34 Ans.

Question 17.
Solution:
L.C.M. of 6, 9, 15 and 20
= 2 x 3 x 5 x 2 x 3 = 180
180 = \(\overline { 2X2 } X\overline { 3X3 } X5\)

To make it a perfect square, it must be multiplied by 5
Product = 180 x 5 = 900
Hence, the least square number = 900 Ans.

Question 18.
Solution:
L.C.M. of 8, 12, 15, 20 = 2 x 2 x 2 x 3 x 5= 120
120 = \(\overline { 2X2 }\)X2X3X5
To make it a perfect square it must be multiplied by 2 x 3 x 5 = 30
Then least perfect square = 120 x 30 = 3600

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Question 1.
Solution:
(23)² = 529

Question 2.
Solution:
(35)² = 1225

Question 3.
Solution:
(52)² = 2704

Question 4.
Solution:
(96)² = 9216

Find the value of each of the following using the diagonal method :

Question 5.
Solution:
(67)² = 4489 Ans.

Question 6.
Solution:
(86)² = 7396 Ans

Question 7.
Solution:
(137)² = 18769 Ans

Question 8.
Solution:
(256)² = 65536 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Question 1.
Solution:
We know that a number ending in 2, 3, 7 or 8 is never a perfect square. So, (i) 5372, (ii) 5963, (iii) 8457, (iv) 9468 cannot be perfect square
Again number ending in an odd number of zeros is also never a perfect square.
Among (v) 360, (vi) 64000, (vii) 2500000 each one has odd zeros at its end. So, there cannot be a perfect square.

Question 2.
Solution:
We know that the square of an even number is also an even number.
(i) 196 (iii) 900, (v) 324 are the squares of even numbers.

Question 3.
Solution:
We know that square.of an odd number is alway is an odd number and square of an even number is always an even number.
Therefore the (ii) 961, (iv) 8649 (v) 4225 are squares of odd numbers.

Question 4.
Solution:
We know that sum of the first n odd natural numbers = n² Therefore.
(i) ∵ It ends with 13
and 1 + 3 + 5 + 7 + 9 + 11 + 13 is the sum of first 7 odd numbers
∵Its sum = (7)² = 49
(ii) 1 + 3 + 5 + 7 + 9+ 11 + 13 + 15 + 17 + 19
Here, n = 10
∵Sum = n² = (10)² = 100
(iii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
Here, n = 12
Sum = n² = (12)² = 144 Ans.

Question 5.
Solution:
(i) 81 = (9)² = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 (Sum of first 9 odd numbers)
(ii) 100 = (10)² =1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 (Sum of first 10 odd numbers)

Question 6.
Solution:
We know that 2m, m² – 1 and m² + 1 is a Pythagorean triplet where m > 1
(i) One number = 6
∵ 2m = 6 => m = 3
∵ Other members of triplet will be
m² – 1 = (3)² – 1 = 9 – 1 = 8
and m² + 1 = (3)² + 1 = 9 + 1 = 10
∵ Pythagorean triplet is 6, 8, 10
(ii) Let 2m = 14 = m = \(\\ \frac { 14 }{ 2 } \) = 7
m² – 1 = (7)² – 1 = 49 – 1 = 48
and m² + 1 = (7)² + 1 = 49 + 1 = 50
∵Pythagorean triplet = 14, 48, 50
(iii) Let 2m = 16 => m = 8
m² – 1 = (8)² – 1 = 64 – 1 = 63
and m² + 1 = (8)² + 1 = 64 + 1 = 65
∵Pythagorean triplet =16, 63, 65
(iv) Let 2m = 20 => m = 10
∵m² – 1 = (10)² – 1 = 100 – 1 = 99
and m² + 1 = (10)² + 1 = 100 + 1 = 101
∵Pythagorean triplet = 20, 99, 101 Ans.

Question 7.
Solution:
We know that :
(n + 1)² – n² = {(n + 1) + n}
Therefore :
(i) (38)² – (37)² = 38 + 37 = 75
(ii) (75)² – (74)² = 75 + 74 = 149
(iii) (92)² – (91)² = 92 + 91 = 183
(iv) (105)² – (104)² = 105 + 104 = 209
(v) (141)² – (140)² = 141 + 140 = 281
(vi) (218)² – (217)² = 218 + 217 = 435

Question 8.
Solution:
We know that (a + b)² = a² + 2ab + b²
(i) (310)² = (300 + 10)²
= (300)² + 2 x 300 x 10 + (10)²
= 90000 + 6000 + 100 = 96100
(ii) (508)² = (500 + 8)²
= (500)² + 2 x 500 x 8 + (8)²
= 250000 + 8000 + 64 = 258064
(iii) (630)² = (600 + 30)²
= (600)² + 2 x 600 x 30 + (30)²
= 360000 + 36000 + 900 = 396900

Question 9.
Solution:
We know that (a – b)² = a² – 2ab + b²
(i) (196)² = (200 – 4)²
= (200)² – 2 x 200 x 4 + (4)²
= 40000 – 1600 + 16
= 40016 – 1600 = 38416
(ii) (689)² = (700 – 11)²
= (700)² – 2 x 700 x 11 +(11)²
= 490000 – 15400 + 121
= 490121 – 15400 = 474721
(iii) (891)² = (900 – 9)²
= (900)² – 2 x 900 x 9 + (9)²
= 810000 – 16200 + 81
= 810081 – 16200 = 793881

Question 10.
Solution:
Using (a – b) (a + b) = a² – b²
(i) 69 x 71 = (70 – 1) (70 + 1)
= (70)² – (1)² = 4900 – 1
= 4899
(ii) 94 x 106 = (100 – 6) (100 + 6)
= (100)² – (6)²
= 10000 – 36 = 9964

Question 11.
Solution:
Using (a – b) (a + b) – a² – b²
(i) 88 x 92 = (90 – 2) (90 + 2)
= (90)² – (2)²
= 8100 – 4 = 8096
(ii) 78 x 82 = (80 – 2) (80 + 2)
= (80)² – (2)²
= 6400 – 4 = 6396

Question 12.
Solution:
(i) The square of an even number is even
(ii) The square of an odd number is odd
(iii) The square of a proper fraction is less than the given fraction.
(iv) n² = the sum of first n odd natural numbers. Ans.

Question 13.
Solution:
(i) False: No. of digits of a perfect square can be even or odd.
(ii) False: Square of a prime number is not a prime number.
(iii) False: It is not always possible.
(iv) False: It is not always possible.
(v) True: The product of two squares is always a perfect square.

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.