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

**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}= (2^{2}) = 4

**(ii) 977**

**Solution:**

Unit digit is 7

Therefore, unit digit of (977)

^{2}= (7^{2}) = 49 = 9

**(iii) 4583**

**Solution:**

Unit digit is 3

Therefore, unit digit of (4583)

^{2}= (3^{2}) = 9

**(iv) 78367**

**Solution:**

Unit digit is 7

Therefore, unit digit of (78367)

^{2}= (7^{2}) = 49 = 9

**(v) 52698**

**Solution:**

Unit digit is 8

Therefore, unit digit of (52698)

^{2}= (8^{2}) = 64 = 4

**(vi) 99880**

**Solution:**

Unit digit is 0

Therefore, unit digit of (99880)

^{2}= (0^{2}) = 0

**(vii) 12796**

**Solution:**

Unit digit is 6

Therefore, unit digit of (12796)

^{2}=(6^{2}) = 36 = 6

**(viii) 55555**

**Solution:**

Unit digit is 5

Therefore, unit digit of (55555)

^{2}=(5^{2}) = 25 = 5

**(ix) 53924**

**Solution:**

Unit digit is 4

Therefore, unit digit of (53924)

^{2}=(4^{2}) = 16 = 6

**Question 5. Observe the following pattern**

**1 + 3 = 2**^{2}

^{2}

**1 + 3 + 5 = 3**^{2}

^{2}

**1 + 3 + 5 + 7 = 4**^{2}

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

^{2}

**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 = 2

^{2}Therefore, The value of 1 + 3 + 5 + 7 + 9 +……… up to n terms = n

^{2}(as there are only n terms).

**Question 6. Observe the following pattern**

**2**^{2} – 1^{2} = 2 + 1

^{2}– 1

^{2}= 2 + 1

**3**^{2} – 2^{2} = 3 + 2

^{2}– 2

^{2}= 3 + 2

**4**^{2} – 3^{2} = 4 + 3

^{2}– 3

^{2}= 4 + 3

**5**^{2} – 4^{2} = 5 + 4

^{2}– 4

^{2}= 5 + 4

**And find the value of**

**(i) 100 ^{2} – 99^{2}**

**Solution:**

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

Therefore, 100

^{2}-99^{2}=100 + 99 = 199

**(ii)111 ^{2} – 109^{2}**

**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,

= (111

^{2}– 110^{2}) + (110^{2}– 109^{2})= (111 + 110) + (100 + 109)

= 440

**(iii) 99 ^{2} – 96^{2}**

**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,

= 99

^{2}– 96^{2}= (99

^{2}– 98^{2}) + (98^{2}– 97^{2}) + (97^{2}– 96^{2})= (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 = 8

^{2}+ 15^{2}= 289

RHS = 17

^{2}= 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 = 18

^{2}+ 80^{2}= 6724

RHS = 82

^{2}= 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 = 14

^{2}+ 48^{2}= 2500

RHS = 51

^{2}= 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 = 10

^{2}+ 24^{2}= 676

RHS = 26

^{2}= 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 = 16

^{2}+ 63^{2}= 4225

RHS = 65

^{2}= 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 = 12

^{2}+ 35^{2}= 1369

RHS = 38

^{2}= 1444

LHS ≠RHS

Therefore, the given triplet is not a Pythagorean.

**Question 8. Observe the following pattern**

**(1×2) + (2×3) = (2×3×4)/3**

**(1×2) + (2×3) + (3×4) = (3×4×5)/3**

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

**And find the value of**

**(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

**Question 9. Observe the following pattern**

**1 = 1/2 (1×(1+1))**

**1+2 = 1/2 (2×(2+1))**

**1+2+3 = 1/2 (3×(3+1))**

**1+2+3+4 = 1/2 (4×(4+1))**

**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

**Question 10. Observe the following pattern**

**1**^{2} = 1/6 (1×(1+1)×(2×1+1))

^{2}= 1/6 (1×(1+1)×(2×1+1))

**1**^{2}+2^{2} = 1/6 (2×(2+1)×(2×2+1)))

^{2}+2

^{2}= 1/6 (2×(2+1)×(2×2+1)))

**1**^{2}+2^{2}+3^{2} = 1/6 (3×(3+1)×(2×3+1)))

^{2}+2

^{2}+3

^{2}= 1/6 (3×(3+1)×(2×3+1)))

**1**^{2}+2^{2}+3^{2}+4^{2} = 1/6 (4×(4+1)×(2×4+1)))

^{2}+2

^{2}+3

^{2}+4

^{2}= 1/6 (4×(4+1)×(2×4+1)))

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

**(i) 1 ^{2 }+ 2^{2 }+ 3^{2 }+ 4^{2 }+… + 10^{2}**

**Solution:**

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

1

^{2 }+ 2^{2 }+ 3^{2 }+ 4^{2 }+ … + 10^{2}= 1/6 (10 × (10 + 1) × (2 × 10 + 1))= 1/6 (2310)

= 385

**(ii) 5 ^{2 }+ 6^{2 }+ 7^{2 }+ 8^{2 }+ 9^{2 }+ 10^{2 }+ 11^{2 }+ 12^{2}**

**Solution:**

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

5

^{2 }+ 6^{2 }+ 7^{2 }+ 8^{2 }+ 9^{2 }+ 10^{2 }+ 11^{2 }+ 12^{2}= 1^{2 }+ 2^{2 }+ 3^{2 }+ … + 12^{2}– (1^{2}+2^{2}+3^{2}+4^{2})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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