ML Aggarwal CBSE Solutions Class 8 Math Third Chapter Squares and Square Roots Exercise 3.2

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

= 8² =64

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

= (15)² = 225

(7) (i) 64= 8² =1+3+5+7+9+11+13+15

(8) (i) 15² =225

Therefore, so first number = 15²-1/2

= 225-1/2 = 224/2

= 112

Therefore, 2^nd number = 15²+1/2 = 225+1/2 = 226/2

= 113

Therefore, 15² = 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, m²-1 = m²-1 = 4²-1 = 16-1 =15

Let, m²+1 = 4²+1 = 16+1 = 17

Therefore, we know the triplet is (2m, m²-1, m²+1)

The triplet is (8, 15, 17)

(ii) 15

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

Again let –

M²-1 = 15

Or, m² =15+1 =16

Or, m= 4

Therefore, 2m =2×4 =8

M²+1 = 4²+1

= 16+1 =17

Therefore, We know the triplet is (2m, m²-1, m²+1)

The triplet is (8, 15, 17)

(iii) 63

= Let, we know that,

The triplet is (2m, m²-1, m²+1)

Let, m²-1 =63

Or, m² = 63+1 = 64

Or, m = 8

Therefore, 2m = 2×8 = 16

M²+1 = 8²+1 = 64+1 =65

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

(iv) 80

= Let, we know that

The triplet is (2m, m²-1, m²+1)

Let, m²-1 = 80

Or, m² = 80+1 =81

Or, m = 9

Therefore, 2m = 2×9 = 18

M²+1 = 9²+1 = 81+1 = 82

Therefore, the triplet is 18, 80, 82