NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 8 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 8 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 8 Mathematics are an important part of exams for Class 8 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 8 Mathematics and also download more latest study material for all subjects. Chapter 6 Squares And Square Roots is an important topic in Class 8, please refer to answers provided below to help you score better in exams
Chapter 6 Squares And Square Roots Class 8 Mathematics NCERT Solutions
Class 8 Mathematics students should refer to the following NCERT questions with answers for Chapter 6 Squares And Square Roots in Class 8. These NCERT Solutions with answers for Class 8 Mathematics will come in exams and help you to score good marks
Chapter 6 Squares And Square Roots NCERT Solutions Class 8 Mathematics
Exercise 6.1
Q.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
Sol.1) (i) The number 81 contains its unit’s place digit 1. So, square of 1 is 1. Hence, unit’s digit of square of 81 is 1.
(ii) The number 272 contains its unit’s place digit 2. So, square of 2 is 4. Hence, unit’s digit of square of 272 is 4.
(iii) The number 799 contains its unit’s place digit 9. So, square of 9 is 81. Hence, unit’s digit of square of 799 is 1.
(iv) The number 3853 contains its unit’s place digit 3. So, square of 3 is 9. Hence, unit’s digit of square of 3853 is 9.
(v) The number 1234 contains its unit’s place digit 4. So, square of 4 is 16. Hence, unit’s digit of square of 1234 is 6.
(vi) The number 26387 contains its unit’s place digit 7. So, square of 7 is 49. Hence, unit’s digit of square of 26387 is 9.
(vii) The number 52698 contains its unit’s place digit 8. So, square of 8 is 64. Hence, unit’s digit of square of 52698 is 4.
(viii) The number 99880 contains its unit’s place digit 0. So, square of 0 is 0. Hence, unit’s digit of square of 99880 is 0.
(ix) The number 12796 contains its unit’s place digit 6. So, square of 6 is 36. Hence, unit’s digit of square of 12796 is 6.
(x) The number 55555 contains its unit’s place digit 5. So, square of 5 is 25. Hence, unit’s digit of square of 55555 is 5.
Q.2) The following numbers are obviously not perfect squares. Give reasons.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222
(v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
Sol.2) (i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 505050 is not a perfect square because its unit’s place digit is 0.
Q.3) The squares of which of the following would be odd number:
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Sol.3) (i) 431 – Unit’s digit of given number is 1 and square of 1 is 1.
Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36.
Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81.
Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16.
Therefore, square of 82004 would not be an odd number.
Q.4) Observe the following pattern and find the missing digits:
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 … … .2 … … .1
10000002 = 1 … … … … … … … …
Sol.4) 112 = 121
1012= 10201
10012 = 1002001
1000012 = 10000200001
10000002 = 100000020000001
Q.5) Observe the following pattern and supply the missing numbers:
112 = 121
1012 = 10201
101012 = 102030201
10101012 = … … … … … … … … …
… … … … … … … … … .2 = 10203040504030201
Sol.5) 112 = 121
1012 = 10201
101012 = 102030201
10101012 = 1020304030201
1010101012 = 10203040504030201
Q.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
Sol.6) 12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432
Q.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
Sol.7) (i) Here, there are five odd numbers.
Therefore, square of 5 is 25.
∴ 1 + 3 + 5 + 7 + 9 = 52 = 25
(ii) Here, there are ten odd numbers.
Therefore, square of 10 is 100.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100
(iii) Here, there are twelve odd numbers.
Therefore, square of 12 is 144.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144
Q.8) (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers
Sol.8) (i) 49 is the square of 7.
Therefore, it is the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 is the square of 11. Therefore, it is the sum of 11 odd numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Q.9) How many numbers lie between squares of the following numbers:
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Sol.9) (i) Since, non-perfect square numbers between 𝑛2 and (𝑛 + 1)2 are 2𝑛.
Here, 𝑛 = 12
Therefore, non-perfect square numbers between 12 and 13 = 2n = 2 x 12 = 24
(ii) Since, non-perfect square numbers between 𝑛2 and (𝑛 + 1)2 are 2𝑛.
Here, n = 25
Therefore, non-perfect square numbers between 25 and 26 = 2𝑛 = 2 × 25 = 50
(iii) Since, non-perfect square numbers between 𝑛2 and (𝑛 + 1)2 are 2𝑛
Here, n = 99
Therefore, non-perfect square numbers between 99 and 100 = 2𝑛 = 2 × 99 = 198
Exercise 6.2
Q.1) Find the squares of the following numbers:
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Sol.1) i) (32)2 = (30 + 2)2 = (30)2 + 2 × 30 × 2 + (2)2 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2]
= 900 + 120 + 4 = 1024
ii) (35)2 = (30 + 5)2 = (30)2 + 2 × 30 × 5 + (5)2 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2]
= 900 + 300 + 25 = 1225
iii) (86)2 = (80 + 6)2 = (80)2 + 2 × 80 × 6 + (6)2 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2]
= 1600 + 960 + 36 = 7386
iv) (93)2 = (90 + 3)2 = (90)2 + 2 × 90 × 3 + (3)2 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2]
= 8100 + 540 + 9 = 8649
v) (71)2 = (70 + 1)2 = (70)2 + 2 × 70 × 1 + (1)2 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2]
= 4900 + 140 + 1 = 5041
vi) (46)2 = (40 + 6)2 = (40)2 + 2 × 40 × 6 + (6)2 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2]
= 1600 + 480 + 36 = 2116
Q.2) Write a Pythagoras triplet whose one member is:
(i) 6 (ii) 14 (iii) 16 (iv) 18
Sol.2) i) There are three numbers 2𝑚, 𝑚2 − 1 and 𝑚2 + 1 in a Pythagorean Triplet.
Here, 2𝑚 = 6 ⇒ 𝑚 = 6/2 = 3
Therefore,
Second number 𝑚2 − 1 = (3)2 − 1 = 9 − 1 = 8
Third number 𝑚2 + 1 = (3)2 + 1 = 9 + 1 = 10
Hence, Pythagorean triplet is (6, 8, 10).
(ii) There are three numbers 2𝑚, 𝑚2 − 1 and 𝑚2 + 1 in a Pythagorean Triplet.
Here, 2𝑚 = 14
⇒ 𝑚 = 14/2 = 7
Therefore,
Second number (𝑚 − 1)2 = (7)2 − 1 = 49 − 1 = 48
Third number (𝑚 + 2)2 = (7)2 + 1 = 49 + 1 = 50
Hence, Pythagorean triplet is (14, 48, 50).
(iii) There are three numbers 2𝑚, 𝑚2 − 1 and 𝑚2 + 1 in a Pythagorean Triplet.
Here, 2𝑚 = 18
⇒ 𝑚 = 16/2 = 8
Therefore,
Second number (𝑚 − 1)2 = (8)2 − 1 = 64 − 1 = 48
Third number (𝑚 + 2)2 = (8)2 + 1 = 64 + 1 = 65
Hence, Pythagorean triplet is (16,63,65).
(iv) There are three numbers 2𝑚, 𝑚2 − 1 and 𝑚2 + 1 in a Pythagorean Triplet.
Here, 2𝑚 = 18
⇒ 𝑚 = 18/2 = 9
Therefore,
Second number (𝑚 − 1)2 = (9)2 − 1 = 81 − 1 = 80
Third number (𝑚 + 2)2 = (9)2 + 1 = 81 + 1 = 82
Hence, Pythagorean triplet is (18, 80, 82).
Exercise 6.3
Q.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
Sol.1) Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:
(i) 1 (ii) 6 (iii) 1 (iv) 5
Q.2) Without doing any calculation, find the numbers which are surely not perfect squares:
(i) 153 (ii) 257 (iii) 408 (iv) 441
Sol.2) Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number
Q.3) Find the square roots of 100 and 169 by the method of repeated subtraction.
Sol.3) By successive subtracting odd natural numbers from 100,
100 – 1 = 99 99 – 3 = 96 96 – 5 = 91 91 – 7 = 84
84 – 9 = 75 75 – 11 = 64 64 – 13 = 51 51 – 15 = 36
36 – 17 = 19 19 – 19 = 0
This successive subtraction is completed in 10 steps.
Therefore, √100 = 10
By successive subtracting odd natural numbers from 169,
169 – 1 = 168 168 – 3 = 165 165 – 5 = 160 160 – 7 = 153
153 – 9 = 144 144 – 11 = 133 133 – 13 = 120 120 – 15 = 105
105 – 17 = 88 88 – 19 = 69 69 – 21 = 48 48 – 23 = 25
25 – 25 = 0
This successive subtraction is completed in 13 steps.
Therefore, √169 = 13
Q.4) Find the square roots of the following numbers by the Prime Factorization method:
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100
Sol.4)
Q.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
Sol.5) (i) 252 = 2 × 2 × 3 × 3 × 7
Here, prime factor 7 has no pair.
Therefore 252 must be multiplied by 7 to
make it a perfect square.
∴ 252 × 7 = 1764 and
√1764 = 2 × 3 × 7 = 42
ii) 180 = 2 × 2 × 3 × 3 × 5
Here, prime factor 5 has no pair.
Therefore 180 must be multiplied by 5 to
make it a perfect square.
∴ 180 × 5 = 900
And 900 = 2 × 3 × 5 = 30
(iii) 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, prime factor 7 has no pair.
Therefore 1008 must be multiplied by 7 to make it a perfect square.
∴ 1008 x 7 = 7056
And √7056 = 2 × 2 × 3 × 7 = 84
(iv) 2028 = 2 × 2 × 3 × 13 × 13
Here, prime factor 3 has no pair.
Therefore 2028 must be multiplied by 3 to make it a perfect square.
∴ 2028 × 3 = 6084
And √6084 = 2 × 2 × 3 × 3 × 13 × 13 = 78
(v) 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, prime factor 2 has no pair.
Therefore 1458 must be multiplied by 2 to make it a perfect square.
∴ 1458 × 2 = 2916
And √2916 = 2 × 3 × 3 × 3 = 54
(vi) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, prime factor 3 has no pair.
Therefore 768 must be multiplied by 3 to make it a perfect square.
∴ 768 × 3 = 2304 And
2304 = 2 × 2 × 2 × 2 × 3 = 48
Q.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
Sol.6) (i) 252 = 2 × 2 × 3 × 3 × 7
Here, prime factor 7 has no pair.
Therefore 252 must be divided by 7 to make it a perfect square.
∴ 252 ÷ 7 = 36 And
36 = 2 × 3 = 6
(ii) 2925 = 3 × 3 × 5 × 5 × 13
Here, prime factor 13 has no pair.
Therefore 2925 must be divided by 13 to make it a perfect square.
∴ 2925 ÷ 13 = 225 And
225 = 3 × 5 = 15
(iii) 396 = 2 × 2 × 3 × 3 × 11
Here, prime factor 11 has no pair.
Therefore 396 must be divided by 11 to make it a perfect square.
∴ 396 ÷ 11 = 36 And
36 = 2 × 3 = 6
(iv) 2645 = 5 × 23 × 23
Here, prime factor 5 has no pair.
Therefore 2645 must be divided by 5 to make it a perfect square.
∴ 2645 ÷ 5 = 529 And
529 = 23 × 23 = 23
(v) 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, prime factor 7 has no pair.
Therefore 2800 must be divided by 7 to make it a perfect square.
∴ 2800 ÷ 7 = 400 And
400 = 2 × 2 × 5 = 20
(vi) 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, prime factor 5 has no pair.
Therefore 1620 must be divided by 5 to make it a perfect square.
∴ 1620 ÷ 5 = 324 And
324 = 2 × 3 × 3 = 18
Q.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.
Sol.7) Here, Donated money = 2401
Let the number of students be 𝑥.
Therefore donated money 𝑥 × 𝑥
According to question, 𝑥2 = 2401
⇒ 𝑥 = √2401 = √7 × 7 × 7 × 7
⇒ 𝑥 = 7 × 7 = 49
Hence, the number of students is 49.
Q.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
Sol.8) Here, Number of plants = 2025
Let the number of rows of planted plants be 𝑥.
And each row contains number of plants = 𝑥
According to question,
𝑥2 = 2025
Hence, each row contains 45 plants.
Q.9) Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Sol.9) L.C.M. of 4, 9 and 10 is 180. Prime factors of
180 = 2 × 2 × 3 × 3 × 5
Here, prime factor 5 has no pair.
Therefore 180 must be multiplied by 5 to make it a perfect
square.
∴ 180 x 5 = 900
Hence, the smallest square number
which is divisible by 4, 9 and 10 is 900.
Q.10) Find the smallest square number that is divisible by each of the numbers 8, 15 and 20
Sol.10) L.C.M. of 8, 15 and 20 is 120. Prime factors of
120 = 2 × 2 × 2 × 3 × 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.
∴ 120 x 2 x 3 x 5 = 3600
Hence, the smallest square number which is
divisible by 8, 15 and 20 is 3600
Exercise 6.4
Q.1) 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
Sol.1)
Q.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
Sol.2) (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
(iv) 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
Q.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
Sol.3) (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.
Q.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
Sol.4) (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.
Q.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
Sol.5) (i) 525
Since remainder is 41.
Therefore 222 < 525
Next perfect square number 232 = 529
Hence, number to be added
= 529 – 525 = 4
∴ 525 + 4 = 529
Hence, the square root of 529 is 23.
(ii) 1750
Since remainder is 69.
Therefore 412 < 1750
Next perfect square number 422 = 1764
Hence, number to be added
= 1764 – 1750 = 14
∴ 1750 + 14 = 1764
Hence, the square root of 1764 is 42
(iii) 252
Since remainder is 27.
Therefore 152 < 252
Next perfect square number 162 = 256
Hence, number to be added
= 256 – 252 = 4
∴ 252 + 4 = 256
Hence, the square root of 256 is 16.
(iv) 1825
Since remainder is 61.
Therefore 422 < 1825
Next perfect square number 432 = 1849
Hence, number to be added
= 1849 – 1825 = 24
∴ 1825 + 24 = 1849
Hence, the square root of 1849 is 43.
(v) 6412
Since remainder is 12.
Therefore 802 < 6412
Next perfect square number 812 = 6561
Hence, number to be added
= 6561 – 6412 = 149
∴ 6412 + 149 = 6561
Hence, the square root of 6561 is 81.
Q.6) Find the length of the side of a square whose area is 441 𝑚2?
Sol.6) Let the length of side of a square be x meter.
Area of square = (𝑠𝑖𝑑𝑒)2 = 𝑥2
According to question, 𝑥2 = 441
Q.7) In a right triangle ABC, ∠ 𝐵 = 90 .
(i) If AB = 6 cm, BC = 8 cm, find AC. (ii) If AC = 13 cm, BC = 5 cm, find AB.
Sol.7) (i) Using Pythagoras theorem,
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
⇒ 𝐴𝐶2 = (6)2 + (8)2
⇒ 𝐴𝐶2 = 36 + 84 = 100
⇒ 𝐴𝐶 = 10 𝑐𝑚
(ii) Using Pythagoras theorem,
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
⇒ (13)2 = 𝐴𝐵2 + (5)2
⇒ 169 = 𝐴𝐵2 + 25
⇒ 𝐴𝐵2 = 169 – 25
⇒ 𝐴𝐵2 = 144
⇒ 𝐴𝐵 = 12 𝑐𝑚
Q.8) A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and number of columns remain same. Find the minimum number of plants he needs more for this.
Sol.8) Here, plants = 1000
Since remainder is 39.
Therefore 312 < 1000
Next perfect square number 322 = 1024
Hence, number to be added = 1024 – 1000 = 24
∴ 1000 + 24 = 1024
Hence, the gardener required 24 more plants.
Q.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?
Sol.9) Here, Number of children = 500
By getting the square root of this number,
we get,
In each row, the number of children is 22.
And left out children are 16.
| NCERT Solutions Class 8 Mathematics Chapter 1 Rational Numbers |
| NCERT Solutions Class 8 Mathematics Chapter 2 Linear Equations In One Variable |
| NCERT Solutions Class 8 Mathematics Chapter 3 Understanding Quadrilaterals |
| NCERT Solutions Class 8 Mathematics Chapter 4 Practical Geometry |
| NCERT Solutions Class 8 Mathematics Chapter 5 Data Handling |
| NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots |
| NCERT Solutions Class 8 Mathematics Chapter 7 Cube and Cube Roots |
| NCERT Solutions Class 8 Mathematics Chapter 8 Comparing Quantities |
| NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities |
| NCERT Solutions Class 8 Mathematics Chapter 10 Visualising Solid Shapes |
| NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration |
| NCERT Solutions Class 8 Mathematics Chapter 12 Exponents And Powers |
| NCERT Solutions Class 8 Mathematics Chapter 13 Direct and Inverse Proportions |
| NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation |
| NCERT Solutions Class 8 Mathematics Chapter 15 Introduction To Graphs |
| NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers |
NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots
The above provided NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 8 Mathematics textbook online or you can easily download them in pdf. The answers to each question in Chapter 6 Squares And Square Roots of Mathematics Class 8 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 6 Squares And Square Roots Class 8 chapter of Mathematics so that it can be easier for students to understand all answers. These solutions of Chapter 6 Squares And Square Roots NCERT Questions given in your textbook for Class 8 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 8.
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All questions given in the end of the chapter Chapter 6 Squares And Square Roots have been answered by our teachers