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, 2nd number = 152+1/2 = 225+1/2 = 226/2
= 113
Therefore, 152 = 112+113 =225
(ii) (23)2 =529
Therefore, 1st number = (23)2-1/2 = 529-1/2 = 528/2
= 264
2nd number = (23)2 +1/2 = 529+1/2 = 530/2
= 265
Therefore, (23)2 = 264+265 =529
(iii) (37)2 = 1369
Therefore, 1st number = (37)2 -1/2 = 1369-1/2 = 1368/2 = 684
Therefore, 2nd 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
Q 11 , 12
Q. 11,12?