Class 8 - Maths - Squares and Square Roots

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

(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.

Question 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

Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers

of 0.

1. 1057 is not a perfect square because its unit’s place digit is 7.
2. 23453 is not a perfect square because its unit’s place digit is 3.
• 7928 is not a perfect square because its unit’s place digit is 8.
1. 222222 is not a perfect square because its unit’s place digit is 2.
2. 64000 is not a perfect square because its unit’s place digit is single 0.
3. 89722 is not a perfect square because its unit’s place digit is 2.
• 222000 is not a perfect square because its unit’s place digit is triple 0.
• 505050 is not a perfect square because its unit’s place digit is 0.

Question 3:

The squares of which of the following would be odd number:

(i) 431                (ii) 2826            (iii) 7779            (iv) 82004

(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.

Question 4:

Observe the following pattern and find the missing digits:

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1……2……..1

100000012 = 1……………….

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 10000200001

100000012 = 100000020000001

Question 5:

Observe the following pattern and supply the missing numbers:

112 = 121

1012 = 10201

101012 = 102030201

10101012 = ………………….

……………….2 = 1020304050403020101

112 = 121

1012 = 10201

101012 = 102030201

10101012 = 1020304030201

1010101012 = 1020304050403020101

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

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

Question 7:

(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

(i) Here, there are five odd numbers. Therefore square of 5 is 25.

So, 1 + 3 + 5 + 7 + 9 = 52 = 25

(ii) Here, there are ten odd numbers. Therefore square of 10 is 100.

So, 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.

So, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144

Question 8:

(i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

(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

Question 9:

How many numbers lie between squares of the following numbers:

(i) 12 and 13                          (ii) 25 and 26                      (iii) 99 and 100

(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 * 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

Question 1:

Find the squares of the following numbers:

(i) 32               (ii) 35            (iii) 86            (iv) 93            (v) 71              (vi) 46

(i) (32)2 = (30 + 2)2 = (30)2 + 2 * 30 * 2 + 22      [(a + b)2 = a2 + 2 * a * b + b2]

= 900 + 120 + 4

= 1024

(ii) (35)2 = (30 + 5)2 = (30)2 + 2 * 30 * 5 + 52      [(a + b)2 = a2 + 2 * a * b + b2]

= 900 + 300 + 25

= 1225

(iii) (86)2 = (80 + 6)2 = (60)2 + 2 * 80 * 6 + 62      [(a + b)2 = a2 + 2 * a * b + b2]

= 1600 + 960 + 36

= 7386

(iv) (93)2 = (90 + 3)2 = (90)2 + 2 * 90 * 2 + 32      [(a + b)2 = a2 + 2 * a * b + b2]

= 8100 + 540 + 9

= 8649

(v) (71)2 = (70 + 1)2 = (70)2 + 2 * 70 * 1 + 12      [(a + b)2 = a2 + 2 * a * b + b2]

= 4900 + 140 + 1

= 5041

(vi) (46)2 = (40 + 6)2 = (40)2 + 2 * 40 * 6 + 62      [(a + b)2 = a2 + 2 * a * b + b2]

= 1600 + 480 + 36

= 2116

Question 2:

Write a Pythagoras triplet whose one member is:

(i) 6                                  (ii) 14                               (iii) 16                                 (iv) 18

(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 is:

m2 – 1 = 32 – 1 = 9 – 1 = 8

Third Number is:

m2 + 1 = 32 + 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 is:

m2 – 1 = 72 – 1 = 49 – 1 = 48

Third Number is:

m2 + 1 = 72 + 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 = 16 => m = 16/2 = 8

Therefore, second number is:

m2 – 1 = 82 – 1 = 64 – 1 = 63

Third Number is:

m2 + 1 = 82 + 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 is:

m2 – 1 = 92 – 1 = 81 – 1 = 80

Third Number is:

m2 + 1 = 92 + 1 = 81 + 1 = 82

Hence, Pythagorean triplet is (9, 80, 82)

Exercise 6.3

Question 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

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

Question 2:

Without doing any calculation, find the numbers which are surely not perfect squares:

(i) 153                       (ii) 257                      (iii) 408                        (iv) 441

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

Question 3:

Find the square roots of 100 and 169 by the method of repeated subtraction.

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

Question 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

(i) 729

√729 = √(3 * 3 * 3 * 3 * 3 * 3)

= 3 * 3 * 3

= 27

(ii) 400

√400 = √(2 * 2 * 2 * 2 * 5 * 5)

= 2 * 2 * 5

= 20

(iii) 1764

√1764 = √(2 * 2 * 3 * 3 * 7 * 7)

= 2 * 3 * 7

= 42

(iv) 4096

√4096 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)

= 2 * 2 * 2 * 2 * 2 * 2

= 64

(v) 7744

√7744 = √(2 * 2 * 2 * 2 * 2 * 2 * 11 * 11)

= 2 * 2 * 2 * 11

= 88

(vi) 9608

√9608 = √(2 * 2 * 7 * 7 * 7 * 7)

= 2 * 7 * 7

= 98

(vii) 5929

√5929 = √(7 * 7 * 11 * 11)

= 7 * 11

= 77

(viii) 9216

√9216 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3)

= 2 * 2 * 2 * 2 * 2 * 3

= 96

(ix) 529

√529 = √(23 *23)

= 23

(x) 8100

√8100 = √(2 * 2 * 3 * 3 * 3 * 3 * 5 * 5)

= 2 * 3 * 3 * 5

= 90

Question 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

(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.

So, 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.

So, 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.

So, 1008 * 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.

So, 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.

So, 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.

So, 768 * 3 = 2304

And 2304 = 2 * 2 * 2 * 2 * 3 = 48

Question 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

(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.

So, 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.

So, 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.

So, 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.

So, 2645/5 = 529

And √529 = 23 x 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.

So, 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.

So, 1620/5 = 324

And 324 = 2 * 3 * 3 = 18

Question 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.

Here, Donated money = 2401

Let the number of students be x.

Therefore donated money = x * x

According to question,

x2 = 2401

=> x = √2401

=> x = √(7 * 7 * 7 * 7)

=> x = 7 * 7 = 49

Hence, the number of students is 49.

Question 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.

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

=> x = √2025

=> x = √(3 * 3 * 3 * 3 * 5 * 5)

=> x = 3 * 3 * 5 = 45

Hence, each row contains 45 plants.

Question 9:

Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

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.

So, 180 * 5 = 900

Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.

Question 10:

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 * 2 * 2 * 3 * 5

Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be

multiplied by 2 * 3 * 5 to make it a perfect square.

So, 120 * 2 * 3 * 5 = 3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.

Exercise 6.4

Question 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

(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.

(xii) 900

Hence, the square root of 900 is 30.

Question 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

(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 = 4/2 = 4/2 = 2

(iv) Here, 27225 contains five digits which is odd.

Therefore, number of digits in square root = (n + 1)2 = (5 + 1)/2 = 6/2 = 3

(v) Here, 390625 contains six digits which is even.

Therefore, number of digits in square root = n/2 = 6/2 = 3

Question 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

(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

Question 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

(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.

So, 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.

So, 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.

So, 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.

So, 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.

So, 4000 – 31 = 3969

Hence, the square root of 3969 is 63.

Question 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

(i) 525

Since remainder is 41. Therefore 222 < 525

Next perfect square number 232 = 529

Hence, number to be added = 529 – 525 = 4

So, 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

So, 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

So, 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

So, 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

So, 6412 + 149 = 6561

Hence, the square root of 6561 is 81.

Question 6:

Find the length of the side of a square whose area is 441 m2?

Let the length of side of a square be x meter.

Area of square = (side)2 = x2

According to question,

x2 = 441

=> x = √441

=> x = √(3 * 3 * 7 * 7)

=> x = 3 * 7  => x = 21 m

Hence, the length of side of a square is 21 m.

Question 7:

In a right triangle ABC, ÐB = 900.

(i) If AB = 6 cm, BC = 8 cm, find AC.

(ii) If AC = 13 cm, BC = 5 cm, find AB.

(i) Using Pythagoras theorem,

AC2 = AB2 + BC2

=> AC2 = 62 + 82

=> AC2 = 36 + 84 = 100

=> AC = √100

=> AC = 10 cm

(ii) Using Pythagoras theorem,

AC2 = AB2 + BC2

=> 132 = AB2 + 52

=> 169 = AB2 + 25

=> AB2 = 169 – 25

=> AB2 = 144

=> AB = √144

=> AB = 12 cm

Question 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.

Here, plants = 1000

Since remainder is 39. Therefore 312 < 1000

Next perfect square number 322 = 1024

Hence, number to be added = 1024 – 1000 = 24

So, 1000 + 24 = 1024

Hence, the gardener required 24 more plants.

Question 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?

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.