**ML Aggarwal CBSE Solutions Class 8 Math 3rd Chapter Squares and Square Roots Exercise 3.2**

**(1)** 237, 353, 247, 478, 378

**(2)** (i) 1

(ii) 2

(iii) 9

(iv) 3

(v) 4

(vi) 5

(vii) 7

(viii) 8

(ix) 0

(x) 6

**(3)** (i) 4567 – unit digit of this num is 7

∴ It is not perfect square

(ii) 2463 – init digit of this number is 3

∴ It is not perfect square.

(iii) 6298 – unit digit of this number is 8

∴ It is not a perfect square

(iv) 47292 – unit digit of this number is 2.

∴ It is not a perfect square.

(v) 64000 – unit digit of this number is 0

∴ It is not a perfect square.

**(5) (1)** 12 & 13

= 12= (12)^{2} =144

13= (13)^{2} = 169

The number between = {(13)^{2} – (12)^{2}} -1

= (169-144) -1

= 25-1 =24

**(ii)** 35 & 36

= 35 = (35)^{2} = 1,225

36 = (36)^{2} = 1,296

The number between = {(36)^{2} – (35)^{2}} -1

= (1,296 – 1,225) -1

= 71-1

= 70

**(6) (i) **1+3+5+7+9+11+13+15

= 8^{2} =64

**(ii) **1+3+5+7+9+11+13+15+17+19+21+23+25+27+29

= (15)^{2} = 225

**(7) (i) **64= 8^{2} =1+3+5+7+9+11+13+15

**(8) (i) **15^{2} =225

Therefore, so first number = 15^{2}-1/2

= 225-1/2 = 224/2

= 112

Therefore, 2^{nd} number = 15^{2}+1/2 = 225+1/2 = 226/2

= 113

Therefore, 15^{2} = 112+113 =225

**(ii) **(23)^{2} =529

Therefore, 1^{st} number = (23)^{2}-1/2 = 529-1/2 = 528/2

= 264

2^{nd} number = (23)^{2} +1/2 = 529+1/2 = 530/2

= 265

Therefore, (23)^{2} = 264+265 =529

**(iii) **(37)^{2} = 1369

Therefore, 1^{st} number = (37)^{2} -1/2 = 1369-1/2 = 1368/2 = 684

Therefore, 2^{nd} number = (37)^{2}+1/2 = 1369+1/2= 1370/2 = 685

Therefore, (37)^{2} = 1369 = 684+685

**(9) (i) **(25)^{2} = 625 = 600+25 =(2×3) hundred+25

**(ii) **(305)^{2} = 93, 025= 93000+25 =(3×31) thousand+25

**(iii) **(525)^{2} = 275, 625 = 275600+25=

**(10) (i)** 8

= Let, 2m =8

Or, m= 8/2 =4

Therefore, m^{2}-1 = m^{2}-1 = 4^{2}-1 = 16-1 =15

Let, m^{2}+1 = 4^{2}+1 = 16+1 = 17

Therefore, we know the triplet is (2m, m^{2}-1, m^{2}+1)

The triplet is (8, 15, 17)

**(ii) **15

= Let, 2m =15 it is not natural number.

Again let –

M^{2}-1 = 15

Or, m^{2} =15+1 =16

Or, m= 4

Therefore, 2m =2×4 =8

M^{2}+1 = 4^{2}+1

= 16+1 =17

Therefore, We know the triplet is (2m, m^{2}-1, m^{2}+1)

The triplet is (8, 15, 17)

**(iii) **63

= Let, we know that,

The triplet is (2m, m^{2}-1, m^{2}+1)

Let, m^{2}-1 =63

Or, m^{2} = 63+1 = 64

Or, m = 8

Therefore, 2m = 2×8 = 16

M^{2}+1 = 8^{2}+1 = 64+1 =65

Therefore, the triplet is (16, 63, 65)

**(iv) **80

= Let, we know that

The triplet is (2m, m^{2}-1, m^{2}+1)

Let, m^{2}-1 = 80

Or, m^{2} = 80+1 =81

Or, m = 9

Therefore, 2m = 2×9 = 18

M^{2}+1 = 9^{2}+1 = 81+1 = 82

Therefore, the triplet is 18, 80, 82

Q 11 , 12

Q. 11,12?