Class 8 RD Sharma Solutions Chapter 3 Squares and Square Roots Exercise 3.2 | Set 1 (original) (raw)
Last Updated : 11 Sep, 2024
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 = 2 2
**1 + 3 + 5 = 3 2
**1 + 3 + 5 + 7 = 4 2
**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
**2 2 - 1 2 = 2 + 1
**3 2 - 2 2 = 3 + 2
**4 2 - 3 2 = 4 + 3
**5 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, 1002 -992 =100 + 99 = 199
****(ii)111** 2 - 109 2
**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) 99** 2 - 96 2
**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
**1 2 = 1/6 (1×(1+1)×(2×1+1))
**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)))
**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)]
12 + 22 + 32 + 42 + … + 102 = 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)]
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.