# Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.2 | Set 1

• Last Updated : 06 Apr, 2021

### Question 1. The following numbers are not perfect squares. Give reason.

(i) 1547

Solution:

Number ending with 7 is not perfect square

(ii) 45743

Solution:

Number ending with 3 is not perfect square

(iii) 8948

Solution:

Number ending with 8 is not perfect square

(iv) 333333

Solution:

Number ending with 3 is not perfect square

### Question 2. Show that the following numbers are not, perfect squares:

(i) 9327

Solution:

Number ending with 7 is not perfect square.

(ii) 4058

Solution:

Number ending with 8 is not perfect square

(iii) 22453

Solution:

Number ending with 3 is not perfect square

(iv) 743522

Solution:

Number ending with 2 is not perfect square

### Question 3. The square of which of the following numbers would be an old number?

(i) 731

Solution:

Square of an odd number is odd

731 is an odd number. Therefore, square of 731 is an odd number.

(ii) 3456

Solution:

Square of an even number is even

3456 is an even number. Therefore, square of 3456 is an even number.

(iii)5559

Solution:

Square of an odd number is odd

5559 is an odd number. Therefore, square of 5559 is an odd number.

(iv) 42008

Solution:

Square of an even number is even

42008 is an even number. Therefore, square of 42008 is an even number.

### Question 4. What will be the unit’s digit of the squares of the following numbers?

(i) 52

Solution:

Unit digit is 2

Therefore, unit digit of (52)2 = (22) = 4

(ii) 977

Solution:

Unit digit is 7

Therefore, unit digit of (977)2 = (72) = 49 = 9

(iii) 4583

Solution:

Unit digit is 3

Therefore, unit digit of (4583)2 = (32) = 9

(iv) 78367

Solution:

Unit digit is 7

Therefore, unit digit of (78367)2 = (72) = 49 = 9

(v) 52698

Solution:

Unit digit is 8

Therefore, unit digit of (52698)2 = (82) = 64 = 4

(vi) 99880

Solution:

Unit digit is 0

Therefore, unit digit of (99880)2 = (02) = 0

(vii) 12796

Solution:

Unit digit is 6

Therefore, unit digit of (12796)2 =(62) = 36 = 6

(viii) 55555

Solution:

Unit digit is 5

Therefore, unit digit of (55555)2 =(52) = 25 = 5

(ix) 53924

Solution:

Unit digit is 4

Therefore, unit digit of (53924)2 =(42) = 16 = 6

### 1 + 3 + 5 + 7 = 42And write the value of 1 + 3 + 5 + 7 + 9 +……… up to n terms.

Solution:

Number on the right-hand side is square of the number of terms present on the left-hand side.

1 + 3, These are two terms So, 1 + 3 = 22

Therefore, The value of 1 + 3 + 5 + 7 + 9 +……… up to n terms = n2 (as there are only n terms).

### And find the value of

(i) 1002 – 992

Solution:

According to pattern right-hand side is the addition of two consecutive numbers on the left-hand side.

Therefore, 1002 -992 =100 + 99 = 199

(ii)1112 – 1092

Solution:

According to pattern right-hand side is the addition og two numbers on the left-hand side.

But these two numbers are not consecutive

Therefore,

= (1112 – 1102) + (1102 – 1092)

= (111 + 110) + (100 + 109)

= 440

(iii) 992 – 962

Solution:

According to pattern right-hand side is the addition og two numbers on the left-hand side.

But these two numbers are not consecutive

Therefore,

= 992 – 962

= (992 – 982) + (982 – 972) + (972 – 962)

= (99 + 98) + (98 + 97) + (97 + 96)

= 585

### Question 7. Which of the following triplets is Pythagorean?

(i) (8, 15, 17)

Solution:

(8, 15, 17)

As 17 is the largest number

LHS = 82 + 152

= 289

RHS = 172

= 289

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(ii) (18, 80, 82)

Solution:

(18, 80, 82)

As 82 is the largest number

LHS = 182 + 802

= 6724

RHS = 822

= 6724

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(iii) (14, 48, 51)

Solution:

(14, 48, 51)

As 51 is the largest number

LHS = 142 + 482

= 2500

RHS = 512

= 2601

LHS ≠ RHS

Therefore, the given triplet is not a Pythagorean.

(iv) (10, 24, 26)

Solution:

(10, 24, 26)

As 26 is the largest number

LHS = 102 + 242

= 676

RHS = 262

= 676

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(v) (16, 63, 65)

Solution:

(16, 63, 65)

As 65 is the largest number

LHS = 162 + 632

= 4225

RHS = 652

= 4225

LHS = RHS

Therefore, the given triplet is a Pythagorean.

(vi) (12, 35, 38)

Solution:

(12, 35, 38)

As 38 is the largest number

LHS = 122 + 352

= 1369

RHS = 382

= 1444

LHS ≠RHS

Therefore, the given triplet is not a Pythagorean.

### (1×2) + (2×3) + (3×4) + (4×5) + (5×6)

Solution:

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6) = (5 × 6 × 7)/3 = 70

### And find the values of each of the following:

(i) 1+2+3+4+5+…+50

Solution:

R.H.S = 1/2 [No. of terms in L.H.S × (No. of terms + 1)] (if only when L.H.S starts with 1)

1 + 2 + 3 + 4 + 5 + … + 50 = 1/2 (5 × (5 + 1))

25 × 51 = 1275

(ii) 31 + 32 + …. + 50

Solution:

R.H.S = 1/2 [No. of terms in L.H.S × (No. of terms + 1)] (if only when L.H.S starts with 1)

31 + 32 + …. + 50 = (1 + 2 + 3 + 4 + 5 + … + 50) – (1 + 2 + 3 + … + 30)

1275 – 1/2 (30 × (30 + 1))

1275 – 465

810

### And find the values of each of the following:

(i) 12 + 22 + 32 + 42 +… + 102

Solution:

RHS = 1/6 [(No. of terms in L.H.S) × (No. of terms + 1) × (2 × No. of terms + 1)]

12 + 22 + 32 + 42 + … + 102 = 1/6 (10 × (10 + 1) × (2 × 10 + 1))

= 1/6 (2310)

= 385

(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122

Solution:

RHS = 1/6 [(No. of terms in L.H.S) × (No. of terms + 1) × (2 × No. of terms + 1)]

52 + 62 + 72 + 82 + 92 + 102 + 112 + 122 = 12 + 22 + 32 + … + 122 – (12+22+32+42)

1/6 (12×(12+1)×(2×12+1)) — 1/6 (4×(4+1)×(2×4+1))

= 650 – 30

= 620

### Chapter 3 Squares and Square Roots – Exercise 3.2 | Set 2

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