RS Aggarwal Class 8 Maths Solutions

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.

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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.

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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.

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

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 3A.

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:
(i) 441
= 3 x 3 x 7 x 7

Question 2.
Solution:
(i) 1225
= 5 x 5 x 7 x 7

Question 3.
Solution:
(i) Factors of 3675
3 x 5 x 5 x 7 x 7

Question 4.
Solution:
(i) 1575
= 3 x 3 x 5 x 5 x 7

Question 5.
Solution:
The largest two digit number = 99

Finding the square root of 99, we get remainder = 18
∴The greatest two digit number which is a perfect square will be = 99 – 18 = 81

Question 6.
Solution:
The largest 3 digit number = 999

Finding the square root of 999, we get remainder = 38
∴The greatest 3 digit number which is a perfect square = 999 – 38 = 961

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

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RS Aggarwal Class 8 Solutions Chapter 2 Exponents Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2A
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2B
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C

Objective Questions :
Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (c)

Question 2.
Solution:
Answer = (d)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Answer = (b)

Question 6.
Solution:

Question 7.
Solution:
Answer = (a)

Question 8.
Solution:
Answer = (a)

Question 9.
Solution:
Answer = (d)

Question 10.
Solution:
Answer = (d)

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
3670000 = 3.670000 x 1000000
= 3.67 x 10⁶ (c)

Question 16.
Solution:

Question 17.
Solution:

Hope given RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C are helpful to complete your math homework.

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