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 $(12+1)=2 \times 12=24$

hence , there are 24 natural number between $12^{2}$ and $13^{2}$ 


(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 $90^{2}$ and $91^{2}$

Question 6

Sol :
(i) $1+3+5+7+9+11+13=7^{2}=49$

(ii) $1+3+5+7+9+11+13+15+17+...+29=15^{2}=225$

sum of first 'n' odd numbers = $n^{2}$


Question 7

Sol :
(i) 64

$64-1=63 ; 63-3=60 ; 60-5=55$

$55-7=48: \quad 48-9=39 ; \quad 39-11=28$

$28-13=15 ; \quad 15-15=0$

∴ $64=1+3+5+7+9+11+13+15=8^{2}$


(ii) 121 

$121-1=120 ; 120-3=117 ; \quad 117-5=112 ; 112-7=105 ;$

$105-9=96 ; 96-11=85 ; 85-13=72 ; 72-15=57 ;$

$57-17=40 ; 40-19=21 ; \quad 21-21=0$

∴ $121=1+3+5+7+9+11+13+15+17+19+21=11^{2}$

Question 8

Sol :

(i) $19^{2}=361$

"we can express the square of any odd number greater

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

First number $=\frac{19^{2}-1}{2}=180$

Second number $=\frac{19^{2}+1}{2}=181$

$19^{2}=361=180+181$


(ii) $33^{2}=1089$

First number $=\frac{33^{2}-1}{2}=544$

Second number $=\frac{33^{2}+1}{2}=545$

$33^{2}=1089=544+545$


(iii) $47^{2}=2209$

First number $=\frac{47^{2}-1}{2}=1104$

Second number $=\frac{47^{2}+1}{2}=1105$

$47^{2}=2209=1104+1105$

Question  9

Sol :

(i) $31^{2}=(30+1)^{2}=(30+1)(30+1)$

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

$=900+30+30+1$

$31^{2}=961$


(ii) $42^{2}=(40+2)^{2}=(40+2)(40+2)$

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

$=1600+80+80+4$

$42^{2}=1764$


(iii) $86^{2}=(80+6)^{2}=(80+6)(80+6)$

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

$=6400+480+480+36$

$86^{2}=7396$


(iv) $94^{2}=(90+4)^{2}=(90+4)(90+4)$

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

$=8100+360+360+16$

$94^{2}=8836$

Question 10

Sol :

(i) 45

Comparing with a5 where a = 4 

$4^{-1}(a 5)^{2}=a(a+1)$ hundreds +25

$45^{r}=4(4+1)$ hundreds +25

$=20$ hundreds +25
 
$45^{2}=2025$


(ii) 305

Comparing with a5 where a =30

$\left(a_{5}\right)^{2}=a(a+1)$ hundreds +25

$(305)^{2}=30(30+1)$ hundreds +25

 ⇒930 hundred +25

$(305)^{2}$= 93025


(iii) 525 

Comparing with a5 where a = 52

$(a 5)^{2}=a(a+1)$ 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 $m^{2}-1=8$

⇒$m^{2}=9$

⇒m = 3

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$3^{2}+1, 2 \times 3$


10 , 6

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


(ii) 15

Given number =15

Let us assume $m^{2}-1=15$

$m^{2}=16$

m = 4

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$16+1 \quad ,2 \times 4$

$17 \quad ,  8$

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


(iii) 63 

Given number 63

Let us assume $m^{2}-1=63$

$m^{2}=64$

m=8

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$8^{2}+1 \quad 2 \times 8$

65-16

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


(iv) 80 

given number 80 

let us assume $m^{2}-1=80 \Rightarrow m^{2}=81$

m=9

Remaining two numbers of Pythagorean triplet are

$m^{2}+1,2 m$

$q^{2}+1,2 \times 9$

82,18

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


Question 12 

Sol :

$21^{2}=$ 441

$201^{2}=$ 40401

$2001^{2}=$ 4004001

$20001^{2}=$ 40004001

$200001^{2}=$ 4000400001


Question 14

Sol :

$7^{2}=$ 49

$67^{2}=$ 4489

$667^{2}=$ 444889

$6667^{2}=$ 44448889

$66667^{2}=$ 4444488889

$666667^{2}=$ 444444888889

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