RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.2) Part - 1, Class 8 Math Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

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Class 8 : RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.2) Part - 1, Class 8 Math Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.2) Part - 1, Class 8 Math Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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Question 1: The following numbers are not perfect squares. Give reason.
(i) 1547
(ii) 45743
(iii) 8948
(iv) 333333

Answer 1: A number ending with 2, 3, 7 or 8 cannot be a perfect square.
(i) Its last digit is 7. Hence, 1547 cannot be a perfect square.
(ii) Its last digit is 3. Hence, 45743 cannot be a perfect square.
(iii) Its last digit is 8. Hence, 8948 cannot be a perfect square.
(iv) Its last digit is 3. Hence, 333333 cannot be a perfect square. 

Question 2: Show that the following numbers are not perfect squares:
(i) 9327
(ii) 4058
(iii) 22453
(iv) 743522

Answer 2: A number ending with 2, 3, 7 or 8 cannot be a perfect square.
(i) Its last digit is 7. Hence, 9327 is not a perfect square.
(ii) Its last digit is 8. Hence, 4058 is not a perfect square.
(iii) Its last digit is 3. Hence, 22453 is not a perfect square.
(iv) Its last digit is 2. Hence, 743522 is not a perfect square. 

Question 3: The square of which of the following numbers would be an odd number?
(i) 731
(ii) 3456
(iii) 5559
(iv) 42008

Answer 3: The square of an odd number is always odd.
(i) 731 is an odd number. Hence, its square will be an odd number.
(ii) 3456 is an even number. Hence, its square will not be an odd number.
(iii) 5559 is an odd number. Hence, its square will not be an odd number.
(iv) 42008 is an even number. Hence, its square will not be an odd number.
Hence, only the squares of 731 and 5559 will be odd numbers.

Question 4: What will be the units digit of the squares of the following numbers?
(i) 52
(ii) 977
(iii) 4583
(iv) 78367
(v) 52698
(vi) 99880
(vii) 12796
(viii) 55555
(ix) 53924

Answer 4: The units digit is affected only by the last digit of the number. Hence, for each question, we only need to examine the square of its last digit.
(i) Its last digit is 2. Hence, the units digit is 22, which is equal to 4.
(ii) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 72), which is 9.
(iii) Its last digit is 3. Hence, the units digit is 32, which is equal to 9.
(iv) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 72), which is 9.
(v) Its last digit is 8. Hence, the units digit is the last digit of 64 (64 = 82), which is 4.
(vi) Its last digit is 0. Hence, the units digit is 02, which is equal to 0.
(vii) Its last digit is 6. Hence, the units digit is the last digit of 36 (36 = 62), which is 6.
(viii) Its last digit is 5. Hence, the units digit is the last digit of 25 (25 = 52), which is 5.
(ix) Its last digit is 4. Hence, the units digit is the last digit of 16 (16 = 42), which is 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 + ... upto n terms.

Answer 5: From the pattern, we can say that the sum of the first n positive odd numbers is equal to the square of the n-th positive number. Putting that into formula:
1 + 3 + 5 + 7 + ...  n =  n2, where the left hand side consists of 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
(ii) 1112 − 1092
(iii) 992 − 962

Answer 6: From the pattern, we can say that the difference between the squares of two consecutive numbers is the sum of the numbers itself.
In a formula:
 (n+1)2  (n)2 = (n+1) + nn+12 - n2 = n+1 + n
Using this formula, we get:
(i) 1002 − 992   = (99 + 1) + 99
= 199
(ii) 1112 − 1092 = 1112 − 110+ 1102 − 1092
= (111 + 110) + (110 + 109)
= 440
(iii) 992 − 962 = 992 − 98+ 98− 972 + 97− 96
= 99 + 98 + 98 + 97 + 97 + 96
= 585 

Question 7: Which of the following triplets are pythagorean?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(v) (16, 63, 65)
(vi) (12, 35, 38)

Answer 7: Only (i), (ii), (iv) and (v) are Pythagorean triplets.
A triplet (abc) is called Pythagorean if the sum of the squares of the two smallest numbers is equal to the square of the biggest number.
(i) The two smallest numbers are 8 and 15. The sum of their squares is:
82 + 152 = 289 = 172
Hence, (8, 15, 17) is a Pythagorean triplet.
(ii) The two smallest numbers are 18 and 80. The sum of their squares is:
182 + 802 = 6724 = 822
Hence, (18, 80, 82) is a Pythagorean triplet.
(iii) The two smallest numbers are 14 and 48. The sum of their squares is:
142 + 482 = 2500, which is not equal to 512 = 2601
Hence, (14, 48, 51) is not a Pythagorean triplet.
(iv) The two smallest numbers are 10 and 24. The sum of their squares is:
102 + 242 = 676 = 262
Hence, (10, 24, 26) is a Pythagorean triplet.
(v) The two smallest numbers are 16 and 63. The sum of their squares is:
162 + 632 = 4225 = 652
Hence, (16, 63, 65) is a Pythagorean triplet.
(vi) The two smallest numbers are 12 and 35. The sum of their squares is:
122 + 352 = 1369, which is not equal to 382 = 1444
Hence, (12, 35, 38) is not a Pythagorean triplet.

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