6 . Squares and Square Roots

Chapter 6 . Squares and Square Roots

6. Squares and Square Roots

Exercise 6.1

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: (i) The unit digit of the square of the number 81 is

(ii) The unit digit of the square of the number 272 is

(iii) The unit digit of the square of the number 799 is [9² = 81]

(iv) The unit digit of the square of the number 3853 is

(v) The unit digit of the square of the number 1234 is [ ]

(vi) The unit digit of the square of the number 26387 is []

(vii) The unit digit of the square of the number 52698 is []

(viii) The unit digit of the square of the number 99880 is

(ix) The unit digit of the square of the number 12796 is []

(x) The unit digit of the square of the number 55555 is []

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: (i) The number 1057 ends with unit place 7 . So, 1057 is not the perfect number .

(ii) The number 23453 ends with unit place 3 . So, 23453 is not the perfect number .

(iii) The number 7928 ends with unit place 8 . So, 7928 is not the perfect number .

(iv) The number 222222 ends with unit place 2 . So, 222222 is not the perfect number .

(v) The number 64000 ends with unit place 0 . So, 64000 is not the perfect number .

(vi) The number 89722 ends with unit place 2 . So, 89722 is not the perfect number .

(vii) The number 222000 ends with unit place 0 . So, 222000 is not the perfect number .

(viii) The number 505050 ends with unit place 0 . So, 505050 is not the perfect number .

3. The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Solution: (i) 431² is an odd number . (i.e., 1² = 1)

(ii) 2826² is an even number (i.e., 6² = 36 ) .

(iii) 7779² is an odd number (i.e., 9² = 81)

(iv) 82004² is an even number (i.e., 4² =16)

4. Observe the following pattern and find the missing digits.

Solution: We have,

5. Observe the following pattern and supply the missing numbers.

Solution: We have ,

6. Using the given pattern, find the missing numbers.

Solution: We have,

[Note: (i)

(ii)

(iii)

and ]

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: (i) We have, 1 + 3 + 5 + 7 + 9
= 5² = 5 × 5 = 25
(ii) We have, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

= 10² = 10 × 10 = 100
(iii) We have, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 +21 +23

= 12² = 12 × 12 = 144

8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

Solution: (i) We have , 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) We have, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Solution: (i) 12 and 13

We know that , the squares of the numbers and is .

Therefore, the squares of the numbers and is .

(ii) 25 and 26

We know that , the squares of the numbers and is .

Therefore, the squares of the numbers and is .

(iii) 99 and 100

We know that , the squares of the numbers and is .

Therefore, , the squares of the numbers and is .

Exercise 6.2

1. Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

2. Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18

[Note : For any natural number ,

We have, . So, the Pythagorean triplets are and ]

Solution: (i) 6

We know that , the Pythagorean triplets are and

We take,

So,

Then

Then the value of will not be integer .

We take,

Thus, and

Thus , the triplet are 6 , 8 and 10 .

(ii) 14

We know that , the Pythagorean triplets are and

We take ,

So ,

Then

Then the value of will not be integer .

We take,

Then

Again , and

Thus , the triplet are 14 , 48 and 50 .

(iii) 16

We know that, the Pythagorean triplets are and

We take

So ,

Then

Then the value of will not be integer .

We take,

Again, will not give an integer value for .

We take,

Thus, and

Therefore , the triplet are 16 , 63 and 65 .

(iv) 18

We know that , the Pythagorean triplets are and

We take

So ,

Then

Then the value of will not be integer .

We take,

Again, will not give an integer value for .

We take,

Thus, and

Therefore , the triplet are 18 , 80 and 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: (i) 9801

Therefore, the possible ‘one’s’ digits of the square root of 9801 are 1 and 9

(ii) 99856

Therefore, the possible ‘one’s’ digits of the square root of 99856 are 4 and 6 .

(iii) 998001

Therefore, the possible ‘one’s’ digits of the square root of 998001 are 1 and 9 .

(iv) 657666025

Therefore, the possible ‘one’s’ digits of the square root of 657666025 is 5 .

2. Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153 (ii) 257 (iii) 408 (iv) 441

Solution: Since, the unit’s place of the digits with 2 , 3 , 7 or 8 are not a perfect square number .

(i) The number 153 is the unit’s place of the digits with 3 . So, 153 is not a perfect square number .

(ii) The number 257 is the unit’s place of the digits with 7 . So, 257 is not a perfect square number .

(iii) The number 408 is the unit’s place of the digits with 8 . So, 408 is not a perfect square number .

(iv) The number 441 is the unit’s place of the digits with 1 . So, 441 is a perfect square number .

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

Solution: (i) We have , 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

We know that, the sum of the first odd natural numbers is .There are 10 odd numbers for 100 .

Therefore, the square roots of 100 is 10 .

(ii) We have , 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 , 68 – 21 = 48 , 48 – 23 = 25 , 25 – 25 = 0

We know that, the sum of the first odd natural numbers is . There are 13 odd numbers for 169 .

Therefore, the square roots of 169 is 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

Using prime factorisation Method , we have

729 = 3×3×3×3×3×3 = 3²×3²×3²

So,

(ii) 400

Using prime factorisation Method , we have

400 = 2×2×2×2×5×5 = 2²×2²×5²

So,

(iii) 1764

Using prime factorisation Method , we have

1764 = 2×2×3×3×7×7 = 2²×3²×7²

So,

(iv) 4096
Using prime factorisation Method , we have

4096 = 2×2×2×2×2×2×2×2×2×2×2×2 = 2² × 2² × 2² × 2² × 2² × 2²

So,

(v) 7744

Using prime factorisation Method , we have

7744 = 2×2×2×2×2×2×11×11 = 2² × 2² × 2² × 11²

So,

(vi) 9604

Using prime factorisation Method , we have

9604 = 2×2×7×7×7×7 = 2²×7²×7²

So,

(vii) 5929

Using prime factorisation method , we have

5929 = 7×7×11×11 = 7² × 11²

So,

(viii) 9216
Using prime factorisation method , we have

9216 = 2×2×2×2×2×2×2×2×2×2×3×3 = 2² × 2² × 2² × 2² × 2² × 3²

So,

(ix) 529

Using prime factorisation method , we have

529 = 23×23 = 23²

So,

(x) 8100

Using prime factorisation method , we have

8100 = 2×2×3×3×3×3×5×5 = 2² × 3² × 3² × 5²

So,

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

The prime factors 7 has no pair , so 252 is not a perfect square .

Now, we multiply 252 by 7

Therefore, 252 × 7 = 2×2×3×3×7×7 is a perfect square .

Thus,

(ii) We have,

180 = 2×2×3×3×5

The prime factors 5 has no pair , so 180 is not a perfect square .

Now, we multiply 180 by 5

Therefore, 180 × 5 = 2×2×3×3×5×5 is a perfect square .

Thus,

(iii) We have,

1008 = 2×2×2×2×3×3×7

The prime factors 7 has no pair , so 1008 is not a perfect square .

Now, we multiply 1008 by 7

Therefore, 1008 × 7 = 2×2×2×2×3×3×7×7 is a perfect square .

Thus ,

(iv) We have,

2028 = 2×2×13×13×3

The prime factors 3 has no pair , so 2028 is not a perfect square .

Now, we multiply 2028 by 3

Therefore, 2028 × 3 = 2×2×13×13×3×3 is a perfect square .

Thus ,

(v) We have,

1458 = 2×3×3×3×3×3×3

The prime factors 2 has no pair , so 1458 is not a perfect square .

Now, we multiply 1458 by 2

Therefore, 1458 × 2 = 2×2×3×3×3×3×3×3 is a perfect square .

Thus,

(vi) We have,

768 = 2×2×2×2×2×2×2×2×3

The prime factors 3 has no pair , so 768 is not a perfect square .

Now, we multiply 768 by 3

Therefore, 768 × 3 = 2×2×2×2×2×2×2×2×3×3 is a perfect square .

Thus,

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: (i) We have,

252 = 2×2×3×3×7

Now, we divide 252 by the factor 7 , then 252 7 = 2×2×3×3 is a perfect square number .

Therefore, the required smallest number is 7 .

So,

(ii) We have,

2925 =3×3×5×5×13

Now, we divide 2925 by the factor 13,then 2925 13 = 3×3×5×5 is a perfect square number .

Therefore, the required smallest number is 13 .

So,

(iii) We have,

396 = 2×2×3×3×11

Now , we divide 396 by the factor 11 , then 396 11 = 2×2×3×3 is a perfect square number .

Therefore, the required smallest number is 11 .

So,

(iv) We have,

2645 = 5×23×23
Now , we divide 2645 by the factor 5 , then 2645 ÷ 5 = 23×23 is a perfect square number .

Therefore, the required smallest number is 5 .

So,

(v) We have,

2800 = 2×2×2×2×5×5×7

Now, we divide 2800 by the factor 7 , then 2800 ÷ 7 = 2×2×2×2×5×5 is a perfect square number .

Therefore, the required smallest number is 7 .

So,

(vi) We have,

1620 = 2×2×3×3×3×3×5

Now, we divide 1620 by the factor 5 , then 1620 5 = 2×2×3×3×3×3 is a perfect square number .

Therefore, the required smallest number is 5 .

So,

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: The total amount of donation = Rs 2401 .

We have,

2401 = 7×7×7×7 = 7² × 7²

Therefore, the number of students in the class

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 = 2025 .

Here, the number of rows = the number of plants in each row .

We have,

2025 = 3×3×3×3×5×5 = 3² × 3² × 5²

So, the number of rows

Therefore, 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: The least number divisible by each one of 4 , 9 and 10 is their LCM .

We have, 4 = 2 × 2 , 9 = 3 × 3 , 10 = 2 × 5

So, the LCM of 8 , 15 and 20 is 2 × 2 × 3 × 3 × 5 = 180

The prime factorization of 180 is 180 = 2 × 2 × 3 × 3 × 5 is not perfect square .

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: The least number divisible by each one of 8 , 15 and 20 is their LCM .

We have, 8 = 2 × 2 × 2 , 15 = 3 × 5 , 20 = 2 × 2 × 5

So, the LCM of 8 , 15 and 20 is 2 × 2 × 2 × 3 × 5 = 120

The prime factorization of 120 is 120 = 2 × 2 × 2 × 3 × 5 is not perfect square .

Therefore, 120 should be multiplied by 2 × 3 × 5 = 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

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

[Note : If a perfect square is of digits, then its square root will have digits if is even or if is odd ]

Solution: (i) 64

Here, (even)