ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.2

 Exercise 3.2

Question 1

Sol :
(i) 2

(ii) 13

(iii) 27

(iv) 88

(v) 243


Question 2 

Sol :

(i) 1

(ii) 4

(iii) 1

(iv) 9 

(v) 6

(vi) 5

(vii) 9

(viii) 4

(ix) 0

(x) 6


Question 3 

Sol :
(i) 567

567 has '7' in its unit's place. a perfect square 

Should have 1,4,5,6,9,0 in it's unit's place.

so 567 is not a perfect square.


(ii) 2453

2453 has '3' in it's unit's place. But a perfect square

should have 0,1,4,5,6,9 in it's unit's place.

So 2453 is not a perfect square.


(iii) 5298

5298 has 8 in it's unit's place. But a perfect square

should have 0,1,4,5,6,9 in it's unit's place.

so 5298 is not a perfect square.


(iv)4692

46292 has 2 in it's unit's place. But a perfect square

Should have 0,1,4,5,6,9 in it's unit's place

so 46292 is not a perfect square.


(v) 74000

74000 has 0 in it's unit's place but it has

odd no.of zero's and 740  is not a perfect square

so 74000 is not a perfect square.

Question 4 

Sol :
(i) 573
square of 573 is a odd number because ,If a number  has 3 in the units place , then its square and in '9'


(ii) 4096

Square of 4096 is a even number because, If a number has ' 6 ' in the units place, Then its square ends in ' 6 '


(iii) 8267

Square of 8267 is a odd number because, If a number has 7 in the Units place, Then its square ends in ' 9


(iv) 37916

square of 37916 is a even number because if a number has ' 6 ' in the Units place, then it square ends in ' 6 '

Question 5

Sol :
(i) 12 and 13

There are 2 non-square numbers between the squares of

two consecutive numbers n and n+1

∴ natural numbers between 12 and (12+1)=2×12=24

hence , there are 24 natural number between 122 and 132 


(ii) 90 and 91

There are 2 non-square numbers between the Squares of two consecutive numbers n and n+1

∴ Natural numbers between 90 and 91=2 \times 90= 180

Hence, There are 180 natural numbers between 902 and 912

Question 6

Sol :
(i) 1+3+5+7+9+11+13=72=49

(ii) 1+3+5+7+9+11+13+15+17+...+29=152=225

sum of first 'n' odd numbers = n2


Question 7

Sol :
(i) 64

641=63;633=60;605=55

557=48:489=39;3911=28

2813=15;1515=0

64=1+3+5+7+9+11+13+15=82


(ii) 121 

1211=120;1203=117;1175=112;1127=105;

1059=96;9611=85;8513=72;7215=57;

5717=40;4019=21;2121=0

121=1+3+5+7+9+11+13+15+17+19+21=112

Question 8

Sol :

(i) 192=361

"we can express the square of any odd number greater

than 1 as the sum of two consecutive natural numbers."

First number =19212=180

Second number =192+12=181

192=361=180+181


(ii) 332=1089

First number =33212=544

Second number =332+12=545

332=1089=544+545


(iii) 472=2209

First number =47212=1104

Second number =472+12=1105

472=2209=1104+1105

Question  9

Sol :

(i) 312=(30+1)2=(30+1)(30+1)

=30(30+1)+1(30+1)

=900+30+30+1

312=961


(ii) 422=(40+2)2=(40+2)(40+2)

=40(40+2)+2(40+2)

=1600+80+80+4

422=1764


(iii) 862=(80+6)2=(80+6)(80+6)

=80(80+6)+6(80+6)

=6400+480+480+36

862=7396


(iv) 942=(90+4)2=(90+4)(90+4)

=90(90+4)+4(90+4)

=8100+360+360+16

942=8836

Question 10

Sol :

(i) 45

Comparing with a5 where a = 4 

41(a5)2=a(a+1) hundreds +25

45r=4(4+1) hundreds +25

=20 hundreds +25
 
452=2025


(ii) 305

Comparing with a5 where a =30

(a5)2=a(a+1) hundreds +25

(305)2=30(30+1) hundreds +25

 ⇒930 hundred +25

(305)2= 93025


(iii) 525 

Comparing with a5 where a = 52

(a5)2=a(a+1) hundreds +25

(525)2=52(52+1) hundreds +25

⇒ 2756 hundreds +25 

(525)2=275625

Question 11 

Sol :

(i) 8 

Given number = 8 

let us assume m21=8

m2=9

⇒m = 3

Remaining two numbers of Pythagorean triplet are

m2+1,2m

32+1,2×3


10 , 6

The required triplet (6,8,10) with one number 


(ii) 15

Given number =15

Let us assume m21=15

m2=16

m = 4

Remaining two numbers of Pythagorean triplet are

m2+1,2m

16+1,2×4

17,8

∴ The required triplet (8,15,17) with one number as 15 


(iii) 63 

Given number 63

Let us assume m21=63

m2=64

m=8

Remaining two numbers of Pythagorean triplet are

m2+1,2m

82+12×8

65-16

∴ We required triplet (16,63,65) with one number '63'


(iv) 80 

given number 80 

let us assume m21=80m2=81

m=9

Remaining two numbers of Pythagorean triplet are

m2+1,2m

q2+1,2×9

82,18

∴ The required triplet (18,80,82) wits one number '80'


Question 12 

Sol :

212= 441

2012= 40401

20012= 4004001

200012= 40004001

2000012= 4000400001


Question 14

Sol :

72= 49

672= 4489

6672= 444889

66672= 44448889

666672= 4444488889

6666672= 444444888889

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