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

Exercise 3.2 | Set 1 of Chapter 3 (Squares and Square Roots) in RD Sharma's Class 8 mathematics textbook focuses on finding square roots using various methods. This exercise introduces students to techniques such as the division method, estimation, and the use of patterns to calculate square roots. The problems in this set are designed to enhance students' computational skills and deepen their understanding of the relationship between squares and square roots. This exercise builds upon the concepts learned in Exercise 3.1 and prepares students for more advanced applications of square roots in algebra and geometry.

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

Question 5. Observe the following pattern
1 + 3 = 2²

1 + 3 + 5 = 3²

1 + 3 + 5 + 7 = 4²
And 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).

Question 6. Observe the following pattern

2² - 1² = 2 + 1

3² - 2² = 3 + 2

4² - 3² = 4 + 3

5² - 4² = 5 + 4

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 of 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 of 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.

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

1² = 1/6 (1×(1+1)×(2×1+1))

1²+2² = 1/6 (2×(2+1)×(2×2+1)))

1²+2²+3² = 1/6 (3×(3+1)×(2×3+1)))

1²+2²+3²+4² = 1/6 (4×(4+1)×(2×4+1)))

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

Summary

Exercise 3.2 | Set 1 of Chapter 3 in RD Sharma's Class 8 mathematics textbook delves deeper into the methods of finding square roots. It introduces students to systematic approaches for calculating square roots, including the long division method for both whole numbers and decimals. The exercise also covers estimation techniques and methods to determine the number of digits in a square root without actual calculation. Through a variety of problems, students learn to apply these methods to find exact and approximate square roots, compare square roots, and solve equations involving square roots. This set of problems is crucial in developing students' computational skills and logical thinking, preparing them for more complex mathematical concepts in higher grades. By mastering these techniques, students gain a solid foundation in working with irrational numbers and develop the ability to solve real-world problems involving square roots.