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RS Aggarwal Solutions: Exercise 3B - Squares and Square Roots | Mathematics (Maths) Class 8 PDF Download

Q.1. Give reason to show that none of the numbers given below is a perfect square:
(i) 5372

(ii) 5963
(iii) 8457
(iv) 9468
(v) 360
(vi) 64000
(viii) 2500000
Ans. By observing the properties of square numbers, we can determine whether a given number is a square or not.
(i) 5372
A number that ends with 2 is not a perfect square.
Thus, the given number is not a perfect square.
(ii) 5963
A number that ends with 3 is not a perfect square.
Thus, the given number is not a perfect square.
(iii) 8457
A number that ends with 7 is not a perfect square.
Thus, the given number is not a perfect square.
(iv) 9468
A number ending with 8 is not a perfect square.
Thus, the given number is not a perfect square.
(v) 360
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.
(vi) 64000
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.
(vii) 2500000
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

Q.2. Which of the following are squares of even numbers?
(i) 196
(ii) 441
(iii) 900
(iv) 625
(v) 324
Ans.
The square of an even number is always even.
Thus, even numbers in the given list of squares will be squares of even numbers.
(i) 196
This is an even number. Thus, it must be a square of an even number.
(ii) 441
This is an odd number. Thus, it is not a square of an even number.
(iii) 900
This is an even number. Thus, it must be a square of an even number.
(iv) 625
This is an odd number. Thus, it is not a square of an even number.
(v) 324
This is an even number. Thus, it is a square of an even number.

Q.3. Which of the following are squares of odd numbers?
(i) 484
(ii) 961
(iii) 7396
(iv) 8649
(v) 4225
Ans. 
According to the property of squares, the square of an odd number is also an odd number.
Using this property, we will determine which of the numbers in the given list of squares is a square of an odd number.
(i) 484.
This is an even number. Thus, it is not a square of an odd number.
(ii) 961
This is an odd number. Thus, it is a square of an odd number.
(iii) 7396
This is an even number. Thus, it is not a square of an odd number.
(iv) 8649
This is an odd number. Thus, it is a square of an odd number.
(v) 4225
This is an odd number. Thus, it is a square of an odd number.

Q.4. Without adding, find the sum:
(i) (1 + 3 + 5 + 7 + 9 + 11 + 13)
(ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)
(iii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
Ans.
Sum of first n odd numbers = n2
(i) (1+3+5+7+9+11+13) = 7= 49
(ii) (1+3+5+7+9+11+13+15+17+19)=102=100
(iii) (1+3+5+7+9+11+13+15+17+19+21+23) = 122 = 144

Q.5. (i) Express 81 as the sum of 9 odd numbers.
(ii) Express 100 as the sum of 10 odd numbers.
Ans.
Sum of first n odd natural numbers = n2
(i) Expressing 81 as a sum of 9 odd numbers:
81=(9)2
n=9
81=1+3+5+7+9+11+13+15+17
(ii) Expressing 100 as a sum of 10 odd numbers:
100=(10)2
n=10
100=1+3+5+7+9+11+13+15+17+19

Q.6. Write a pythagorean triplet whose smallest member is

(i) 6
(ii) 14
(iii) 16
(iv) 20
Ans.
For every number m > 1, the Pythagorean triplet is (2m, m2−1, m2+1)
Using the above result:
(i) 2m=6
m=3, m2=9
m2−1=9−1=8
m2+1=9+1=10
Thus, the Pythagorean triplet is [6,8,10]6,8,10.
(ii) 2m=14
m=7, m2=49
m2−1=49−1=48
m2+1=49+1=50
Thus, the Pythagorean triplet is [14,48,50]14,48,50.
(iii) 2m=16
m=8, m2=64
m2−1=64−1=63
m2+1=64+1=65
Thus, the Pythagorean triplet is: [16,63,65]16,63,65
(iv) 2m=20
m=10, m2=100
m2−1=100−1=99
m2+1=100+1=101
Thus, the Pythagorean triplet is [20,99,101].

Q.7. Evaluate:
(i) (38)2 − (37)2
(ii) (75)2 − (74)2
(iii) (92)2 − (91)2
(iv) (105)2 − (104)2
(v) (141)2 − (140)2
(vi) (218)2 − (217)2
Ans. Given: [(n+1)2−n2] = (n+1)+n
(i) (38)2−(37)= 38+37 = 75
(ii) (75)2−(74)= 75+74 = 149
(iii) (92)2−(91)= 92+91 = 183
(iv) (105)2−(104)= 105+104 = 209
(v) (141)2−(140)= 141+140 = 281
(vi) (218)2−(217)= 218+217 = 435

Q.8. Using the formula (a + b)2 = (a2 + 2ab + b2), evaluate:
(i) (310)2
(ii) (508)2
(iii) (630)2
Ans.
(i) 310= (300+10)2= 3002+2(300×10)+102) = 90000+6000+100 = 96100
(ii) 508= (500+8)= (5002+2(500×8)+82) = 250000+8000+64 = 258064
(iii) 630= (600+30)= (6002+2(600×30)+302) = 360000+36000+900 = 396900

Q.9. Using the formula (a − b)2 = (a2 − 2ab + b2), evaluate:
(i) (196)2
(ii) (689)2
(iii) (891)2
Ans. (i) (196)2=(200−4)2=2002−2(200×4)+42=40000−1600+16=38416
(ii) (689)2=(700−11)2=7002−2(700×11)+112=490000−15400+121=474721
(iii) (891)2=(900−9)2=9002−2(900×9)+92=810000−16200+81=793881

Q.10. Evaluate:

(i) 69 × 71
(ii) 94 × 106
Ans. 
(i) 69×71=(70−1)×(70+1)=(702−12)=4900−1=4899
(ii) 94×106=(100−6)×(100+6)=(1002−62)=10000−36=9964

Q.11. Evaluate:
(i) 88 × 92
(ii) 78 × 82
Ans.
(i) 88×92=(90−2)×(90+2)=(902−22)=8100−4=8096
(ii) 78×82=(80−2)×(80+2)=(802−22)=6400−4=6396

Q.12. Fill in the blanks:
(i) The square of an even number is .........
(ii) The square of an odd number is .........
(iii) The square of a proper fraction is ......... than the given fraction.
(iv) n2 = the sum of first n ......... natural numbers.
Ans.
(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 smaller  than the given fraction.
(iv) n2 = the sum of first n odd natural numbers.

Q.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.
(ii) The square of a prime number is prime.
(iii) The sum of two perfect squares is a perfect square.
(iv) The difference of two perfect squares is a perfect square.
(v) The product of two perfect squares is a perfect square.
Ans.
(i) F
The number of digits in a square can also be odd. For example: 121
(ii) F
A prime number is one that is not divisible by any other number, except by itself and 1. Thus, square of any number cannot be a prime number.
(iii) F
Example: 4+9=134+9=13
4 and 9 are perfect squares of 2 and 3, respectively. Their sum (13) is not a perfect square.
(iv) F
Example: 36−25=1136-25=11
36 and 25 are perfect squares. Their difference is 11, which is not a perfect square.
(v) T

The document RS Aggarwal Solutions: Exercise 3B - Squares and Square Roots | Mathematics (Maths) Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8.
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