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Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution
SQUARE AND SQUARE ROOTS (Chapter-6) SOLUTION
EXERCISE 6.1
Question 1. What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555
Solution- According to the chapter we read that when we find a square of a number than the unit digit of a number is multiply by the unit digit of same number,
the unit or one's digit of a square numberwhich numbers have 1 or 9 unit digit, square number have 1 unit digit.
which numbers have 2 or 8 unit digit, square number have 4 unit digit.
which numbers have 3 or 7 unit digit, square number have 9 unit digit.
which numbers have 4 or 6 unit digit, square number have 6 unit digit.
which numbers have 5 unit digit, square number have 5 unit digit.which numbers have 0 unit digit, square number have 0 unit digit.
(i) In 81, The unit digit is 1.
Then, 1 X 1 = 1
Hence, The unit digit of the square of the 81 is 1.
(ii) In 272, The unit digit is 2.
Then, 2 X 2 = 4
Hence, The unit digit of the square of the 272 is 4.
(iii) In 799, The unit digit is 9.
Then, 9 X 9 = 81
Hence, The unit digit of the square of the 799 is 1.
(iv) In 3853 The unit digit is 3.
Then, 3 X 3 = 9
Hence, The unit digit of the square of the 3853 is 9.
(v) In 1234 The unit digit is 4.
Then, 4 X 4 = 16
Hence, The unit digit of the square of the 1234 is 6.
(vi) In 26387 The unit digit is 7.
Then, 7 X 7 = 49
Hence, The unit digit of the square of the 26387 is 9.
(vii) In 52698 The unit digit is 8.
Then, 8 X 8 = 64
Hence, The unit digit of the square of the 52698 is 4
(viii) In 99880 The unit digit is 0.
Then, 0 X 0 = 0
Hence, The unit digit of the square of the 99880 is 0.
(ix) In 12796 The unit digit is 6.
Then, 6 X 6 = 36
Hence, The unit digit of the square of the 12796 is 6.
(x) In 55555 The unit digit is 5.
Then, 5 X 5 = 25
Hence, The unit digit of the square of the 55555 is 5.
Question 2. The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
Solution- All the square numbers end with 0, 1, 4, 5, 6 or 9 at units place. None of these end with 2, 3, 7 or 8 at unit’s place. So we Can say that if a number ends in 0, 1, 4, 5, 6 or 9 at units place, then it must be a square number. And if a number ends in 2, 3, 7 or 8 at units place, then it must be not a square number.
All the square numbers can only have even number of zeros at the end.(i) 1057 end with 7 at units place.
(ii) 23453 end with 3 at units place.
(iii) 7928 end with 8 at units place.
(iv) 222222 end with 2 at units place.
(v) 64000 have odd number of zeros at the end.
(vi) 89722 end with 2 at units place.
(vii) 222000 have odd number of zeros at the end.
(viii) 505050 have odd number of zeros at the end.
Question 3. The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Solution- As we know, when we multiply odd numbers with odd numbers product is always odd number, and when we multiply even numbers with even numbers product is always even number.
(i) 431, The unit digit of 431 is 1. 1 is a odd number. After square we get odd number.
(ii) 2826, The unit digit of 2826 is 6. 6 is a even number. After square we get even number.
(iii) 7779, The unit digit of 7779 is 9. 9 is a odd number. After square we get odd number.
(iv) 82004, The unit digit of 82004 is 4. 4 is a even number. After square we get even number.
Question 4. Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1…2…1
100000012 = ………
Solution- According to the above pattern, we have1000012 = 10000200001100000012 = 100000020000001
Question 5. Observe the following pattern and supply the missing numbers.
112=121
1012=10201
101012=102030201
10101012=..................
............2=10203040504030201
Solution- According to the observe above pattern we find the missing numbers.
10101012=1020304030201
1010101012=10203040504030201
Question 6. Using the given pattern, find the missing numbers.
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + ….2 = 212
52 + ….2 + 302 = 312
62 + 72 + …..2 = ……2
Solution- According to the observe above pattern we find the missing numbers.
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432
Question 7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution- According to chapter, We can also say that if a natural number can be expressed as a sum of successive odd natural numbers starting with 1, then it is a perfect square.
The sum of first n odd natural numbers is n2 .
So,
(i) 1 + 3 + 5 + 7 + 9,
Here n = 5,
Then n2 = 52 = 25
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
Here n = 10,
Then n2 = 102 = 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Here n = 12,
Then n2 = 122 = 144
Question 8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution- According to chapter, We can also say that if a natural number can be expressed as a sum of successive odd natural numbers starting with 1, then it is a perfect square.
n2 is the sum of first n odd natural numbers.
So,
(i) Express 49 as the sum of 7 odd numbers.
Here n2 = 49, and n = 7
Then, 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
(ii) Express 121 as the sum of 11 odd numbers.
Here n2 = 49, and n = 7
Then, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121
Question 9. How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Solution- Numbers lie between n2 and (n + 1)2 there are 2n numbers which is 1 less than the difference of two squares.
(i) 12 and 13
Here n = 12 and (n+1) = 13
132 - 122 = 169 - 144 = 25
25 - 1 = 24
or
2n = 2(12) = 24
Numbers lie between squares of the 12 and 13 is 24.
(ii) 25 and 26
Here n = 25 and (n+1) = 26
262 - 252 = 676 - 625 = 51
51 - 1 = 50
or
2n = 2(25) = 50
Numbers lie between squares of the 25 and 26 is 50.
(iii) 99 and 100
Here n = 99 and (n+1) = 100
1002 - 992 = 10000 - 9801 = 199
199 - 1 = 198
or
2n = 2(99) = 198
Numbers lie between squares of the 99 and 100 is 198.
EXERCISE 6.2
1. Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Solution- According to the chapter we know that When we multiply a number by same number. We get the square of a number,
(i) The square of 32 is 322,
322 = 32 X 32 = 1024
or
322 = (30 +2)2
= 302 + 2X30X2 + 22
= 900 + 120 + 4
= 1024
(ii) The square of 35 is 352,
352 = 35 X 35 = 1225
or
352 = (30 +5)2
= 302 + 2X30X5 + 52
= 900 + 300 + 25
= 1225
(iii) The square of 86 is 862,
862 = 86 X 86 = 7396
or
862 = (80 +6)2
= 802 + 2X80X6 + 62
= 6400 + 960 + 36
= 7396
(iv) 93 The square of 93 is 932,
932 = 93 X 93 = 8649
or
932 = (90 +3)2
= 902 + 2X90X3 + 32
= 8100 + 540 + 9
= 8649
(v) 71 The square of 71 is 712,
712 = 71 X 71 = 5041
or
712 = (70 + 1)2
= 702 + 2X70X1 + 12
= 4900 + 140 + 1
= 5041
(vi) 46 The square of 46 is 462,
462 = 46 X 46= 2116
or
462 = (40 +6)2
= 402 + 2X40X6 + 62
= 1600 + 480 + 36
= 2116
2. Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
Solution- According to the chapter we know that For any natural number m > 1, we have (2m) 2 + (m2 – 1)2 = (m2 + 1)2 . So, 2m, m2 – 1 and m2 + 1 forms a Pythagorean triplet.
To find Pythagorean triplet first we observe the this number if it is the nearest of the perfect square number then we assume m2 – 1 and m2 + 1 according to the conditions.
(i) 6 is the one of the element of Pythagorean triplet. 6 is not a nearest number of the perfect square so we can't assume 6 as m2 – 1 and m2 + 1.
let 2m = 6
m = 3
So, other element of Pythagorean triplet is-
m2 – 1 = 32 – 1 = 9 - 1 = 8
and
m2 + 1 = 32 + 1 = 9 + 1 = 10
Hence, Pythagorean triplet is (6, 8, 10)
(ii) 14 is the one of the element of Pythagorean triplet. 14 is not a nearest number of the perfect square so we can't assume 14 as m2 – 1 and m2 + 1.
let 2m = 14
m = 7
So, other element of Pythagorean triplet is-
m2 – 1 = 72 – 1 = 49 - 1 = 48
and
m2 + 1 = 72 + 1 = 49 + 1 = 50
Hence, Pythagorean triplet is (14, 48, 50)
(iii) 16 is the one of the element of Pythagorean triplet. 16 is not a nearest number of the perfect square so we can't assume 16 as m2 – 1 and m2 + 1.
let 2m = 16
m = 8
So, other element of Pythagorean triplet is-
m2 – 1 = 82 – 1 = 64 - 1 = 63
and
m2 + 1 = 82 + 1 = 64 + 1 = 65
Hence, Pythagorean triplet is (16, 63, 65)
(iv) 18 is the one of the element of Pythagorean triplet. 18 is not a nearest number of the perfect square so we can't assume 18 as m2 – 1 and m2 + 1.
let 2m = 18
m = 9
So, other element of Pythagorean triplet is-
m2 – 1 = 92 – 1 = 81 - 1 = 80
and
m2 + 1 = 92 + 1 = 81 + 1 = 82
Hence, Pythagorean triplet is (9, 80, 82)
EXERCISE 6.3
1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025
Solution- According to the chapter we know that when we find the unit or one's digit of a square number than we get-
which numbers have 1 or 9 unit digit, square number have 1 unit digit.
which numbers have 2 or 8 unit digit, square number have 4 unit digit.
which numbers have 3 or 7 unit digit, square number have 9 unit digit.
which numbers have 4 or 6 unit digit, square number have 6 unit digit.
which numbers have 5 unit digit, square number have 5 unit digit.
which numbers have 0 unit digit, square number have 0 unit digit.If process becomes opposite then-
If square numbers have 1 unit digit, square root will be 1 or 9 unit digit.
If square numbers have 4 unit digit, square root will be 2 or 8 unit digit.
If square numbers have 9 unit digit, square root will be 3 or 7 unit digit.
If square numbers have 6 unit digit, square root will be 4 or 6 unit digit.
If square numbers have 5 unit digit, square root will be 5 unit digit.
If square numbers have 0 unit digit, square root will be 0 unit digit.(i) 9801 have 1 unit digit, square root will be 1 or 9 unit digit.
(ii) 99856 have 6 unit digit, square root will be 4 or 6 unit digit.
(iii) 998001 have 1 unit digit, square root will be 1 or 9 unit digit.
(iv) 657666025 have 5 unit digit, square root will be 5 unit digit.
2. Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153 (ii) 257 (iii) 408 (iv) 441
Solution- All the square numbers end with 0, 1, 4, 5, 6 or 9 at units place. None of these end with 2, 3, 7 or 8 at unit’s place. So we Can say that if a number ends in 0, 1, 4, 5, 6 or 9 at units place, then it must be a square number. And if a number ends in 2, 3, 7 or 8 at units place, then it must be not a square number.
(i) 153 is not a perfect square because 153 end with 3.
(ii) 257 is not a perfect square because 257 end with 7.
(iii) 408 is not a perfect square because 408 end with 8.
(iv) 441 is a perfect square because 441 end with 1.
3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution- The sum of the first n odd natural numbers is n 2 ? That is, every square number can be expressed as a sum of successive odd natural numbers starting from 1.
√100
(i) 100 – 1 = 99 (ii) 99 – 3 = 96 (iii) 96 – 5 = 91 (iv) 91 – 7 = 84 (v) 84 – 9 = 75 (vi) 75 – 11 = 64 (vii) 64 – 13 = 51 (viii) 51 – 15 = 36 (ix) 36 – 17 = 19 (x) 19 – 19 = 0
From 100 we have subtracted successive odd numbers starting from 1 and obtained 0 at 10th step. Therefore √100= 10
√169
(i) 169 – 1 = 168 (ii) 168 – 3 = 165 (iii) 165 – 5 = 160 (iv) 160 – 7 = 153 (v) 153 – 9 = 144 (vi) 144 – 11 = 133 (vii) 133 – 13 = 120 (viii) 120 – 15 = 105 (ix) 105 – 17 = 88 (x) 88 – 19 = 69 (xi) 69 – 21= 48 (xii) 48 – 23 = 25 (xiii) 25 – 25 = 0
From 169 we have subtracted successive odd numbers starting from 1 and obtained 0 at 13th step. Therefore √169 = 13
4. Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100
Solution-
(i) 729
Prime factorisation of 729 is
729 = 3 × 3 × 3 × 3 × 3 × 3
By pairing the prime factors we get,
729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3 × 3)²
Therefore, √729 = 3 × 3 × 3 = 27
(ii) 400
Prime factorisation of 400 is
400 = 2 × 2 × 2 × 2 × 5 × 5
By pairing the prime factors we get,
400 = 2 × 2 × 2 × 2 × 5 × 5 = (2 × 2 × 5)²
Therefore, √400 = 2 × 2 × 5 = 20
(iii) 1764
Prime factorisation of 1764 is1764 = 2 × 2 × 3 × 3 × 7 × 7
By pairing the prime factors we get,
1764 = 2 × 2 × 3 × 3 × 7 × 7 = (2 × 3 × 7)²
Therefore, √1764 = 2 × 3 × 7 = 42
(iv) 4096Prime factorisation of 4096 is
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
By pairing the prime factors we get,
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2 × 2× 2 × 2)²
Therefore, √4096 = 2 × 2 × 2 × 2 × 2 × 2= 64
(v) 7744Prime factorisation of 7744 is
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
By pairing the prime factors we get,
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 = (2 × 2 × 2 × 11)²
Therefore, √7744 = 2 × 2 × 2 × 11 = 88
(vi) 9604Prime factorisation of 9604 is
9604 = 2 × 2 × 7 × 7 × 7 × 7
By pairing the prime factors we get,
9604 = 2 × 2 × 7 × 7 × 7 × 7 = (2 × 7 × 7)²
Therefore, √9604 = 2 × 7 × 7 = 98
(vii) 5929Prime factorisation of 5929 is
5929 = 7 × 7 × 11 × 11
By pairing the prime factors we get,
5929 = 7 × 7 × 11 × 11 = (7 × 11)²
Therefore, √5929 = 7 × 11 = 77
(viii) 9216Prime factorisation of 9216 is
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
By pairing the prime factors we get,
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = (2 × 2 × 2 × 2 × 2 × 3)²
Therefore, √9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96
(ix) 529
Prime factorisation of 529 is
529 = 23 × 23
By pairing the prime factors we get,
529 = 23 × 23 = (23)²
Therefore, √529 = 23
(x) 8100Prime factorisation of 8100 is
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
By pairing the prime factors we get,
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = (2 × 3 × 3 × 5)²
Therefore, √8100 = 2 × 3 × 3 × 5 = 90
5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768
Solution-
(i) We have 252 = 2 × 2 × 3 × 3 × 7
As the prime factor 7 has no pair, 252 is not a perfect square.
If 7 gets a pair then the number will become perfect square.
So, we multiply 252 by 7 to get,
252 × 7 = 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 252 × 7 = 1764 is a perfect square.
Thus the required smallest multiple of 252 is 1764 which is a perfect square.
And, √1764 = 2 × 3 × 7 = 42
(ii) 180 We have 180 = 2 × 2 × 3 × 3 × 5
As the prime factor 5 has no pair, 180 is not a perfect square.
If 5 gets a pair then the number will become perfect square.
So, we multiply 180 by 5 to get,
180 × 5 = 2 × 2 × 3 × 3 × 5 × 5
Now each prime factor is in a pair. Therefore, 180 × 5 = 900 is a perfect square.
Thus the required smallest multiple of 180 is 900 which is a perfect square.
And, √900 = 2 × 3 × 5 = 30
(iii) 1008 We have 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
As the prime factor 7 has no pair, 1008 is not a perfect square.
If 7 gets a pair then the number will become perfect square.
So, we multiply 1008 by 7 to get,
1008 × 7 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 1008 × 7 = 7056 is a perfect square.
Thus the required smallest multiple of 1008 is 7056 which is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(iv) We have 2028 = 2 × 2 × 3 × 13 × 13
As the prime factor 3 has no pair, 2028 is not a perfect square.
If 3 gets a pair then the number will become perfect square.
So, we multiply 2028 by 3 to get,
2028 × 3 = 2 × 2 × 3 × 3 × 13 × 13
Now each prime factor is in a pair. Therefore, 2028 × 3 = 6084 is a perfect square.
Thus the required smallest multiple of 2028 is 6084 which is a perfect square.
And, √6084 = 2 × 3 × 13 = 78
(v) We have 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
As the prime factor 2 has no pair, 1458 is not a perfect square.
If 2 gets a pair then the number will become perfect square.
So, we multiply 1458 by 2 to get,
1458 × 2 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Now each prime factor is in a pair. Therefore, 1458 × 2 = 2916 is a perfect square.
Thus the required smallest multiple of 1458 is 2916 which is a perfect square.
And, √2916 = 2 × 3 × 3 × 3 = 54
(vi) 768 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If 3 gets a pair then the number will become perfect square.
So, we multiply 2352 by 3 to get,
2352 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 × 3 = 7056 is a perfect square.
Thus the required smallest multiple of 2352 is 7056 which is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620
Solution-
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(ii) 2925 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(iii) 396 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(iv) 2645 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(v) 2800 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(vi) 1620 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
7. The students of Class VIII of a school donated ` 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution- According to the question,
The students donated Total `2401 rupees
The number of students in the class = x
Each student donated as many rupees =x
Then,
x² = 2401
x = √2401 = 49
Hence, The number of students in the class is 49.
8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution-
Total number of plants are to be planted in a garden = 2025
The number of rows = x
The number of rows and the number of plants in each row = x
Then,
x² = 2025
x = √2025= 45
Hence, The number of rows and the number of plants in each row is 45.
9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution-
First find the smallest common multiple and then find the square number needed.
The least number divisible by each one of 4, 9 and 10 is their LCM.
The LCM of 4, 9 and 10 is 2 × 2 × 3 × 3 × 5 = 180.
Prime factorisation of 180 is 180 = 2 × 2 × 3 × 3 × 5.
We see that prime factors 5 are not in pairs.
Therefore 180 is not a perfect square. In order to get a perfect square, each factor of 180 must be paired.
So we need to make pairs of 5.
Therefore, 180 should be multiplied by 5.
Hence, the required square number is 180 × 5 = 900.
10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution-
First find the smallest common multiple and then find the square number needed.
The least number divisible by each one of 8, 15 and 20 is their LCM.
The LCM of 8, 15 and 20 is 2 × 2 × 2 × 3 × 5 = 120.
Prime factorisation of 120 is 120 = 2 × 2 × 2 × 3 × 5.
We see that prime factors 2, 3 and 5 are not in pairs.
Therefore 120 is not a perfect square. In order to get a perfect square, each factor of 120 must be paired.
So we need to make pairs of 2, 3 and 5.
Therefore, 120 should be multiplied by 2 × 3 × 5, i.e., 30.
Hence, the required square number is 120 × 30 = 3600.
EXERCISE 6.4
1. Find the square root of each of the following numbers by Division method.
(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900
Solution-
2. Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625
Solution- A perfect square is of n-digits, then its square root will have
n2 digits if n is even orn+12 if n is odd.(i) In 64, n = 2 (even)
Number of digits in √64 =
22 = 1(ii) In 144, n = 3 (odd)
Number of digits in √144 =
42 = 2(iii) In 4489, n = 4 (even)
Number of digits in √4489 =
42 = 2(iv) In 27225, n = 5 (odd)
Number of digits in √27225 =
62 = 3(v) In 390625, n = 6 (even)
Number of digits in √390625 =
62 = 3
3. Find the square root of the following decimal numbers.
(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
Solution-
4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000
Solution-
(i) 402Here, the remainder is 02
New number = 402 – 02 = 400
Therefore,√400 = 20Hence, the least required number which should be subtracted from 402 to get a perfect square is 02.(ii) 1989Here, the remainder is 53
New number = 1989 – 53 = 1936
Therefore,√1936 = 44Hence, the least required number which should be subtracted from 1989 to get a perfect square is 53.(iii) 3250Here, the remainder is 01
New number = 3250– 01 = 3249
Therefore,√3249 = 57Hence, the least required number which should be subtracted from 3250 to get a perfect square is 01.(iv) 825Here, the remainder is 41
New number = 825 – 41 = 784
Therefore,√784 = 28Hence, the least required number which should be subtracted from 825 to get a perfect square is 41.(v) 4000Here, the remainder is 31
New number = 4000 – 31 = 3969
Therefore,√3969= 63Hence, the least required number which should be subtracted from 4000 to get a perfect square is 31.
5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412
Solution-
(i) 525
Here, the falling short is 04
New number = 525 + 04 = 529
Therefore,√529 = 23Hence, the least required number which should be added to 525 to get a perfect square is 04.(ii) 1750
Here, the falling short is 14
New number = 1750 + 14 = 1764
Therefore,√1764 = 42Hence, the least required number which should be added to 1750 to get a perfect square is 14.(iii) 252
Here, the falling short is 04
New number = 252 + 04 = 256
Therefore,√256 = 16Hence, the least required number which should be added to 252 to get a perfect square is 04.
(iv) 1825
Here, the falling short is 24
New number = 1825 + 24 = 1849
Therefore,√1849 = 43Hence, the least required number which should be added to 1825 to get a perfect square is 24.(v) 6412
Here, the falling short is 149
New number = 6412 + 149 = 6561
Therefore,√6561 = 81Hence, the least required number which should be added to 6412 to get a perfect square is 149.
6. Find the length of the side of a square whose area is 441 m².
Solution-
As we know that the length of all side of all square is equal.
Let the side of the square is x
The area of the square = side ✕ side = x ✕ x =x²
Then,
x² = 441
x = √441
x = 21
Hence, the length of the side of a square is 21.
7. In a right triangle ABC, ∠B = 90°. (a) If AB = 6 cm, BC = 8 cm, find AC (b) If AC = 13 cm, BC = 5 cm, find AB
Solution-
(a) If AB = 6 cm, BC = 8 cm
First we draw a figure of right triangle ABC
According to the Pythagoras theorem
we know that AC² = AB² + BC²
AC² = 6² + 8² = 36 + 64 = 100
AC = √100 = 10
Hence, AC is 10 cm.
(b) If AC = 13 cm, BC = 5 cm
According to the Pythagoras theorem
we know that AB² = AC² ー BC²
AB² =13² ー 5² = 169 ー 25 = 144
AB = √144 = 12
Hence, AB is 12 cm.
8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution -
Let the number of rows = x.
And the number of columns = x.
Total number of plants = x × x = x2We know that 1000 is not a perfect square so we need to add sum more plant to make a perfect square.
Then,
1000 + y = x2
And we also know that the perfect square number just greater than 1000 is 1024.
So,
1000 + y = 1024
x2 = 1024
x = √1024
x = 32
And
1000 + y = 1024
y = 1024 - 1000 = 24
Hence, the minimum number of plants he needs more for this is 24.
9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
Solution-
Let the number of children in a row = x.
And also that of in a column = x.
Total number of students = x × x = x2
We know that 500 is not a perfect square so we need to add sum more plant to make a perfect square.
Then,
1000 - y = x2
And we also know that the perfect square number just less than 500 is 484.
So,
500 - y = 484
x2 = 484
x = √484
x = 22
And
500 - y = 484
y = 500 - 484 = 16
Hence, to stand in such a manner that the number of rows is equal to number of columns. 16 children would be left out in this arrangement.
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