(i) 402 We know that, if we subtract the remainder from the number, we get a perfect square. Here, we get remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square. ∴ 402 – 2 = 400 Hence, the square root of 400 is 20. (ii) 1989 We know that, if we subtract the remainder from the nRead more
(i) 402
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 2.
Therefore 2 must be subtracted from 402 to get a
perfect square.
∴ 402 – 2 = 400
Hence, the square root of 400 is 20.
(ii) 1989
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 53. Therefore 53 must be
subtracted from 1989 to get a perfect square.
∴ 1989 – 53 = 1936
Hence, the square root of 1936 is 44.
(iii) 3250
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 1. Therefore 1 must be
subtracted from 3250 to get a perfect square.
∴ 3250 – 1 = 3249
Hence, the square root of 3249 is 57.
(iv) 825
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 41. Therefore 41 must be
subtracted from 825 to get a perfect square.
∴ 825 – 41 = 784
Hence, the square root of 784 is 28.
(v) 4000
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 31. Therefore 31 must be
subtracted from 4000 to get a perfect square.
∴ 4000 – 31 = 3969
Hence, the square root of 3969 is 63.
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
(i) 2.56 Hence, the square root of 2.56 is 1.6. (ii) 7.29 Hence, the square root of 7.29 is 2.7. (iii) 51.84 Hence, the square root of 51.84 is 7.2. (iv) 42.25 Hence, the square root of 42.25 is 6.5. (v) 31.36 Hence, the square root of 31.36 is 5.6. Class 8 Maths Chapter 6 Exercise 6.4 Solution in VRead more
(i) 2.56
Hence, the square root of 2.56 is 1.6.
(ii) 7.29
Hence, the square root of 7.29 is 2.7.
(iii) 51.84
Hence, the square root of 51.84 is 7.2.
(iv) 42.25
Hence, the square root of 42.25 is 6.5.
(v) 31.36
Hence, the square root of 31.36 is 5.6.
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
(i) Here, 64 contains two digits which is even. Therefore, number of digits in square root = n/2 = 2/2 = 1 (ii) Here, 144 contains three digits which is odd. Therefore, number of digits in square root = n+1/2 = 3+1/2 = 4/2 = 2 (iii) Here, 4489 contains four digits which is even. Therefore, number ofRead more
(i) Here, 64 contains two digits which is even.
Therefore, number of digits in square root = n/2 = 2/2 = 1
(ii) Here, 144 contains three digits which is odd.
Therefore, number of digits in square root = n+1/2 = 3+1/2 = 4/2 = 2
(iii) Here, 4489 contains four digits which is even.
Therefore, number of digits in square root = n/2=4/2=2
(v) Here, 390625 contains six digits which is even.
Therefore, number of digits in square root = n/2=6/2=3
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
(i) 2304 Hence, the square root of 2304 is 48. (ii) 4489 Hence, the square root of 4489 is 67. (iii) 3481 Hence, the square root of 3481 is 59. (iv) 529 Hence, the square root of 529 is 23. (v) 3249 Hence, the square root of 3249 is 57. (vi) 1369 Hence, the square root of 1369 is 37. (vii) 5776 HencRead more
(i) 2304
Hence, the square root of 2304 is 48.
(ii) 4489
Hence, the square root of 4489 is 67.
(iii) 3481
Hence, the square root of 3481 is 59.
(iv) 529
Hence, the square root of 529 is 23.
(v) 3249
Hence, the square root of 3249 is 57.
(vi) 1369
Hence, the square root of 1369 is 37.
(vii) 5776
Hence, the square root of 5776 is 76.
(viii) 7921
Hence, the square root of 7921 is 89.
(ix) 576
Hence, the square root of 576 is 24.
(x) 1024
Hence, the square root of 1024 is 32.
(xi) 3136
Hence, the square root of 3136 is 56.
(vi) 1369
Hence, the square root of 1369 is 37.
(vii) 5776
Hence, the square root of 5776 is 76.
(viii) 7921
Hence, the square root of 7921 is 89.
(ix) 576
Hence, the square root of 576 is 24.
(x) 1024
Hence, the square root of 1024 is 32.
(xi) 3136
Hence, the square root of 3136 is 56.
(xii) 900
Hence, the square root of 900 is 30.
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
L.C.M. of 8, 15 and 20 is 120. Prime factors of 120 = 2 x 2 x 2 x 3 x 5 Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by 2 x 3 x 5 to make it a perfect square. ∴ 120x2x3x5 = 3600 Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600. Class 8 MathsRead more
L.C.M. of 8, 15 and 20 is 120.
Prime factors of 120 = 2 x 2 x 2 x 3 x 5
Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be
multiplied by
2 x 3 x 5 to make it a perfect square.
∴ 120x2x3x5 = 3600
Hence, the smallest square number which is divisible by 8, 15 and 20
is 3600.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
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
(i) 402 We know that, if we subtract the remainder from the number, we get a perfect square. Here, we get remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square. ∴ 402 – 2 = 400 Hence, the square root of 400 is 20. (ii) 1989 We know that, if we subtract the remainder from the nRead more
(i) 402
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 2.
Therefore 2 must be subtracted from 402 to get a
perfect square.
∴ 402 – 2 = 400
Hence, the square root of 400 is 20.
(ii) 1989
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 53. Therefore 53 must be
subtracted from 1989 to get a perfect square.
∴ 1989 – 53 = 1936
Hence, the square root of 1936 is 44.
(iii) 3250
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 1. Therefore 1 must be
subtracted from 3250 to get a perfect square.
∴ 3250 – 1 = 3249
Hence, the square root of 3249 is 57.
(iv) 825
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 41. Therefore 41 must be
subtracted from 825 to get a perfect square.
∴ 825 – 41 = 784
Hence, the square root of 784 is 28.
(v) 4000
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 31. Therefore 31 must be
subtracted from 4000 to get a perfect square.
∴ 4000 – 31 = 3969
Hence, the square root of 3969 is 63.
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Find the square root of the following decimal numbers: (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
(i) 2.56 Hence, the square root of 2.56 is 1.6. (ii) 7.29 Hence, the square root of 7.29 is 2.7. (iii) 51.84 Hence, the square root of 51.84 is 7.2. (iv) 42.25 Hence, the square root of 42.25 is 6.5. (v) 31.36 Hence, the square root of 31.36 is 5.6. Class 8 Maths Chapter 6 Exercise 6.4 Solution in VRead more
(i) 2.56
Hence, the square root of 2.56 is 1.6.
(ii) 7.29
Hence, the square root of 7.29 is 2.7.
(iii) 51.84
Hence, the square root of 51.84 is 7.2.
(iv) 42.25
Hence, the square root of 42.25 is 6.5.
(v) 31.36
Hence, the square root of 31.36 is 5.6.
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
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
(i) Here, 64 contains two digits which is even. Therefore, number of digits in square root = n/2 = 2/2 = 1 (ii) Here, 144 contains three digits which is odd. Therefore, number of digits in square root = n+1/2 = 3+1/2 = 4/2 = 2 (iii) Here, 4489 contains four digits which is even. Therefore, number ofRead more
(i) Here, 64 contains two digits which is even.
Therefore, number of digits in square root = n/2 = 2/2 = 1
(ii) Here, 144 contains three digits which is odd.
Therefore, number of digits in square root = n+1/2 = 3+1/2 = 4/2 = 2
(iii) Here, 4489 contains four digits which is even.
Therefore, number of digits in square root = n/2=4/2=2
(v) Here, 390625 contains six digits which is even.
Therefore, number of digits in square root = n/2=6/2=3
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Find the square roots 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
(i) 2304 Hence, the square root of 2304 is 48. (ii) 4489 Hence, the square root of 4489 is 67. (iii) 3481 Hence, the square root of 3481 is 59. (iv) 529 Hence, the square root of 529 is 23. (v) 3249 Hence, the square root of 3249 is 57. (vi) 1369 Hence, the square root of 1369 is 37. (vii) 5776 HencRead more
(i) 2304
Hence, the square root of 2304 is 48.
(ii) 4489
Hence, the square root of 4489 is 67.
(iii) 3481
Hence, the square root of 3481 is 59.
(iv) 529
Hence, the square root of 529 is 23.
(v) 3249
Hence, the square root of 3249 is 57.
(vi) 1369
Hence, the square root of 1369 is 37.
(vii) 5776
Hence, the square root of 5776 is 76.
(viii) 7921
Hence, the square root of 7921 is 89.
(ix) 576
Hence, the square root of 576 is 24.
(x) 1024
Hence, the square root of 1024 is 32.
(xi) 3136
Hence, the square root of 3136 is 56.
(vi) 1369
Hence, the square root of 1369 is 37.
(vii) 5776
Hence, the square root of 5776 is 76.
(viii) 7921
Hence, the square root of 7921 is 89.
(ix) 576
Hence, the square root of 576 is 24.
(x) 1024
Hence, the square root of 1024 is 32.
(xi) 3136
Hence, the square root of 3136 is 56.
(xii) 900
Hence, the square root of 900 is 30.
Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/
Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
L.C.M. of 8, 15 and 20 is 120. Prime factors of 120 = 2 x 2 x 2 x 3 x 5 Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by 2 x 3 x 5 to make it a perfect square. ∴ 120x2x3x5 = 3600 Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600. Class 8 MathsRead more
L.C.M. of 8, 15 and 20 is 120.
Prime factors of 120 = 2 x 2 x 2 x 3 x 5
Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be
multiplied by
2 x 3 x 5 to make it a perfect square.
∴ 120x2x3x5 = 3600
Hence, the smallest square number which is divisible by 8, 15 and 20
is 3600.
Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/