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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