# 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 = 7^{2} = 49.

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

Solution: Sum of first 10 odd numbers = 10^{2} = 100.

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

Solution: Sum of first 12 odd numbers = 12^{2} = 144.

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

Solution: We know that n^{2} is equal to the sum of first n odd numbers.

81 = 9^{2} = 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 n^{2} is equal to the sum of first n odd numbers.

100 = 10^{2} = 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, m^{2} -1, m^{2 }+ 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, m^{2} -1, m^{2 }+ 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, m^{2} -1, m^{2 }+ 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, m^{2} -1, m^{2 }+ 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} – n^{2} = (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} – n^{2} = (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} – n^{2} = (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} – n^{2} = (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} – n^{2} = (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} – n^{2} = (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} = (a^{2} + 2ab + b^{2}), 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} = (a^{2} – 2ab + b^{2}), 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) n^{2} = 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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