RS Aggarwal Class 8 Math Third Chapter Squares and Square Roots Exercise 3B Solution

RS Aggarwal Class 8 Math Third Chapter Squares and Square Roots Exercise 3B Solution

EXERCISE 3B

(1) Give reason to show that none of the numbers given below is a perfect square:

(i) 5372

Solution: 5372 is not a perfect square, because the numbers end with 2, which is not a perfect square.

(ii) 5963

Solution: 5963 is not a perfect square, because the numbers end with 3, which is not a perfect square.

(iii) 8457

Solution: 8457 is not a perfect square, because the numbers end with 7, which is not a perfect square.

(iv) 9468

Solution: 9468 is not a perfect square, because the numbers end with 8, which is not a perfect square.

(v) 360

Solution: 360 is not a perfect square, because the numbers ending in an odd number of zeros, which is never a perfect square.

(vi) 64000

Solution: 64000 is not a perfect square, because the numbers ending in an odd number of zeros, which is never a perfect square.

(vii) 2500000

Solution: 2500000 is not a perfect square, because the numbers ending in an odd number of zeros, which is never a perfect square.

(2) Which of the following are squares of even numbers?

(i) 196

(ii) 441

(iii) 900

(iv) 625

(v) 324

Solution: We know that the square of an odd number is odd and square of an even number is even.

(i) 196 is even ⇒ (196)2 is even.

(ii) 441 is odd ⇒ (441)2 is odd.

(iii) 900 is even ⇒ (900)2 is even.

(iv) 625 is odd ⇒ (625)2 is odd.

(v) 324 is even ⇒ (324)2 is even.

(3) Which of the following are squares of odd numbers?

(i) 484

(ii) 961

(iii) 7396

(iv) 8649

(v) 4225

Solution: We know that the square of an odd number is odd and square of an even number is even.

(ii) 961 is odd ⇒ (961)2 is odd

(iv) 8649 is odd ⇒ (8649)2 is odd

(v) 4225  is odd ⇒ (4225)2 is odd.

(4) Without adding, find the sum:

(i) (1 + 3 + 5 + 7 + 9 + 11 + 13)

Solution: Sum of first 6 odd numbers = 72 = 49.

(ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)

Solution: Sum of first 10 odd numbers = 102 = 100.

(iii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)

Solution: Sum of first 12 odd numbers = 122 = 144.

(5) (i) Express 81 as the sum of 9 odd numbers.

Solution: We know that n2 is equal to the sum of first n odd numbers.

81 = 92 = Sum of 9 odd numbers = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17)

(ii) Express 100 as the sum of 10 odd numbers.

Solution: We know that n2 is equal to the sum of first n odd numbers.

100 = 102 = Sum of 10 odd numbers = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)

(6) Write a Pythagorean triplet whose smallest member is

(i) 6

Solution: For every natural number m > 1, (2m, m2 -1, m2 + 1) is a Pythagorean triplet.

Putting 2m = 6, we get m = 3

Thus, we get the triplet (6, 8, 10).

(ii) 14

Solution: For every natural number m > 1, (2m, m2 -1, m2 + 1) is a Pythagorean triplet.

Putting 2m = 14, we get m = 7

Thus, we get the triplet (14, 48, 50).

(iii) 16

Solution: For every natural number m > 1, (2m, m2 -1, m2 + 1) is a Pythagorean triplet.

Putting 2m = 16, we get m = 8

Thus, we get the triplet (16, 63, 65).

(iv) 20

Solution: For every natural number m > 1, (2m, m2 -1, m2 + 1) is a Pythagorean triplet.

Putting 2m = 20, we get m = 10

Thus, we get the triplet (20, 99, 101).

(7) Evaluate:

(i) (38)2 – (37)2

Solution: We have (n + 1)2 – n2 = (n + 1) + n.

Taking n = 37 and (n + 1) = 38, we get

(38)2 – (37)2 = (38 + 37) = 75.

(ii) (75)2 – (74)2

Solution: We have (n + 1)2 – n2 = (n + 1) + n.

Taking n = 74 and (n + 1) = 75, we get

(75)2 – (74)2 = (75 + 74) = 149.

(iii) (92)2 – (91)2

Solution: We have (n + 1)2 – n2 = (n + 1) + n.

Taking n = 91 and (n + 1) = 92, we get

(92)2 – (91)2 = (92 + 91) = 183.

(iv) (105)2 – (104)2

Solution: We have (n + 1)2 – n2 = (n + 1) + n.

Taking n = 104 and (n + 1) = 105, we get

(105)2 – (104)2 = (105 + 104) = 209.

(v) (141)2 – (140)2

Solution: We have (n + 1)2 – n2 = (n + 1) + n.

Taking n = 140 and (n + 1) = 141, we get

(141)2 – (140)2 = (141 + 140) = 281.

(vi) (218)2 – (217)2

Solution: We have (n + 1)2 – n2 = (n + 1) + n.

Taking n = 217 and (n + 1) = 218, we get

(218)2 – (217)2 = (218 + 217) = 435.

(8) Using the formula (a + b)2 = (a2 + 2ab + b2), evaluate:

(i) (310)2 = (300 + 10)2 = (300)2 + 2 × 300 × 10 + (10)2

= (90000 + 6000 + 100) = 96100.

(ii) (508)2 = (500 + 8)2 = (500)2 + 2 × 500 × 8 + (8)2

= (250000 + 8000 + 64) = 258064.

(iii) (630)2 = (600 + 30)2 = (600)2 + 2 × 600 × 30 + (30)2

= (360000 + 36000 + 900) = 396900.

(9) Using the formula (a – b)2 = (a2 – 2ab + b2), evaluate:

(i) (196)2 = (200 – 4)2 = (200)2 – 2 × 200 × 4 + (4)2

= 40000 – 1600 + 16 = 38416.

(ii) (689)2 = (700 – 11)2 = (700)2 – 2 × 700 × 11 + (11)2

= 490000 – 15400 + 121 = 474721.

(iii) (891)2 = (900 – 9)2 = (900)2 – 2 × 900 × 9 + (9)2

= 810000 – 16200 + 81 = 793881.

(10) Evaluate:

(i) 69 × 71

= (70 – 1) × (70 + 1)

= [(70)2 – (1)2]

= (4900 – 1) = 4899.

(ii) 94 × 106

= (100 – 6) × (100 + 6)

= [(100)2 – (6)2]

= (10000 – 36) = 9964.

(11) Evaluate:

(i) 88 × 92

= (90 – 2) × (90 + 2)

= [(90)2 – (2)2]

= (8100 – 4) = 8096.

(ii) 78 × 82

= (80 – 2) × (80 + 2)

= [(80)2 – (2)2]

= (6400 – 4) = 6396.

(12) Fill in the blanks:

(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.

(13) Write (T) for true and (F) for false for each of the statements given below:

(i) The number of digits in a perfect square is even. = F

(ii) The square of a prime number is prime. = F

(iii) The sum of two perfect squares is perfect square. = F

(iv) The difference of two perfect squares is a perfect square. = F

(v) The product of two perfect squares is a perfect square. = T


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