# Finding the Square of a Numbers

Solved exercise of square numbers

# Summary

To find the square of any number we needed to divide the number into two parts then we can solve it easily. If number is ‘x’ then x = (p + q) and x^{2} = (p + q)^{2}

You can also use the formula (p + q)^{2} = p^{2} + 2pq + q^{2}I

**Example:** Find the square of 53.

**Solution:** Divide the number in two parts.

53 = 50 + 3

53^{2} = (50 + 3)^{2}

= (50 + 3) (50 + 3)

= 50(50 + 3) +3(50 + 3)

= 2500 + 150 + 150 + 9

= 2809

**1. Other pattern for the number ending with 5**

For numbers ending with 5 we can use the pattern

(a5)^{2} = a × (a + 1)100 + 25

**Example**

25^{2} = 625 = (2 × 3) 100 + 25

45^{2} = 2025 = (4 × 5) 100 + 25

95^{2} = 9025 = (9 × 10) 100 + 25

125^{2 }= 15625 = (12 × 13) 100 + 25

**2. Pythagorean Triplets**

If the sum of two square numbers is also a square number, then these three numbers form a Pythagorean triplet.

For any natural number p >1, we have (2p) ^{2} + (p^{2} -1)^{2} = (p^{2} + 1)^{2}.

So, 2p, p^{2}-1 and p^{2}+1 forms a Pythagorean triplet.

**Example**

Write a Pythagorean triplet having 22 as one its member

**Solution**

Let 2p = 6

P = 3

p^{2} + 1 = 10

p^{2} – 1 = 8.

Thus, the Pythagorean triplet is 6, 8 and 10.

6^{2} + 8^{2} = 10^{2}

36 + 64 = 100

**Square Roots**

The square root is the inverse operation of squaring. To find the number with the given square is called the **Square Roots.**

2^{2} = 4, so the square root of 4 is 2

10^{2} = 100, therefore square root of 100 is 10

There are two square roots of any number. One is positive and other is negative.

The square root of 100 could be 10 or -10.

**Symbol of Positive Square Root**

**Finding Square Root**

**1. Through Repeated Subtraction**

As we know that every square number is the sum of consecutive odd natural numbers starting from 1,

so we can find the square root by doing opposite because root is the inverse of the square.We need to subtract the odd natural numbers starting from 1 from the given square number until the remainder is zero to get its square root.

The number of steps will be the square root of that square number.

**Example**: Calculate the square root of 64 by repeated subtraction.

**Solution:**

- 64 – 1 = 63
- 63 – 3 = 60
- 60 – 5 = 55
- 55 – 7 = 48
- 28 – 13 = 15
- 48 – 9 = 39
- 15 – 15 = 0
- 39 – 11 = 28

**2. Prime Factorization**

In this method, we need to list the prime factors of the given number and then make the pair of two same numbers.

Then write one number for each pair and multiply to find the square root.

**Example**

Calculate the square root of 784 using prime factorization method.

**Solution:**

List the prime factors of 784.

784 = 2 × 2 × 2 × 2 × 7 × 7

√784 = 2 × 2 × 7 = **28**

we know the prime factorization is 324 is 324=2x2x3x3x3x3

By pairing the prime factors we get

324=2x^{2}x3x^{3}x3x^{3}=2^{2}x3^{2}x3^{2}=(2x3x3)^{2}

Example : Find the square root of 640

solution 6400=2x2x2x2x2x2x2x2x2x5x5

**3. Division Method**

Steps to find the square root by division method

**Step 1: **First we have to start making the pair of digits starting from the right and if there are odd number of digits then the single digit left over at the left will also have bar .

**Step 2:** Take the largest possible number whose square is less than or equal to the number which is on the first bar from the left. Write the same number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide to get the remainder.

**Step 3: **Like a normal division process bring the digits in next bar down and write next to the remainder.

**Step 4: **In next part the quotient will get double and we will right in next line with a blank on its right.

**Step 5: **Now we have to take a number to fill the blank so that the if we take it as quotient then the product of the new divisor and the new digit in quotient is less than or equal to the dividend.

**Step 6:** If there are large number of digits then you can repeat the steps 3, 4, 5 until the remainder does not become 0.

**Example**

Calculate the square root of √729 using division method.

Thus, √729 = 27.

**Square Roots of Decimals**

To find the square root of a decimal number we have to put bars on the primary part of the number in the same manner as we did above. And for the digits on the right of the decimal we have to put bars starting from the first decimal place.

Rest of the method is same as above. We just need to put the decimal in between when the decimal will come in the division.

**Example**

Find √7.29 using division method.

**Solution:**

Thus, √7.29 = 2.7

**Remark:** To put the bar on a number like 174.241, we will put a bar on 74 and a bar on 1 as it is a single digit left. And in the numbers after decimal, we will put a bar on 24 and put zero after 1 to make it double-digit.

174. 24 10

**Estimating Square Root**

Sometimes we have to estimate the square root of a number if it’s not possible to calculate the exact square root.

**Example**

Estimate the square root of 300.

**Solution:** We know that, 300 comes between 100 and 400 i.e. 100 < 300 < 400.

Now, √100 = 10 and √400 = 20.

So, we can say that

10 < √300 < 20.

We can further estimate the numbers as we know that 17^{2} = 289 and 18^{2} = 324.

Thus, we can say that the square root of √300 = 17 as 289 is much closer to 300 than 324.

# Exercise

**Class 8 exercise 5.2**

1. Find the square of the following numbers.

(i) 32

32 = 30 + 2

(32)^{2} = (30 + 2)^{2}

= 30(30 + 2) + 2(30 + 2)

= 30^{2} + 30 × 2 + 2 × 30 + 2^{2}

= 900 + 60 + 60 + 4

= 1024

Thus (32)^{2} = 1024

ii) 35

35 = (30 + 5)

(35)^{2} = (30 + 5)^{2}

= 30(30 + 5) + 5(30 + 5)

= (30)^{2} + 30 × 5 + 5 × 30 + (5)^{2}

= 900 + 150 + 150 + 25

= 1225

Thus (35)^{2} = 1225

(iii) 86

86 = (80 + 6)

86^{2} = (80 + 6)^{2}

= 80(80 + 6) + 6(80 + 6)

= (80)^{2} + 80 × 6 + 6 × 80 + (6)^{2}

= 6400 + 480 + 480 + 36

= 7396

Thus (86)^{2} = 7396

(iv) 93

93 = (90+ 3)

93^{2} = (90 + 3)^{2}

= 90 (90 + 3) + 3(90 + 3)

= (90)^{2} + 90 × 3 + 3 × 90 + (3)^{2}

= 8100 + 270 + 270 + 9

= 8649

Thus (93)^{2} = 8649

(v) 71

71 = (70 + 1)

71^{2} = (70 + 1)^{2}

= 70 (70 + 1) + 1(70 + 1)

= (70)^{2} + 70 × 1 + 1 × 70 + (1)^{2}

= 4900 + 70 + 70 + 1

= 5041

Thus (71)^{2} = 5041

vi) 46

46 = (40+ 6)

46^{2} = (40 + 6)^{2}

= 40 (40 + 6) + 6(40 + 6)

= (40)^{2} + 40 × 6 + 6 × 40 + (6)^{2}

= 1600 + 240 + 240 + 36

= 2116

Thus (46)^{2} = 2116

2. Write a Pythagorean triplet whose one member is

Solution:

i) 6

Let m^{2} – 1 = 6

[Triplets are in the form 2m, m^{2} – 1, m^{2} + 1]

m^{2} = 6 + 1 =7

So, the value of m will not be an integer.

Now, let us try for m^{2} + 1 = 6

⇒ m^{2} = 6 – 1 = 5

Also, the value of m will not be an integer.

Now we let 2m = 6 ⇒ m = 3 which is an integer.

Other members are:

m^{2} – 1 = 3^{2} – 1 = 8 and m^{2} + 1 = 3^{2} + 1 = 10

Hence, the required triplets are 6, 8 and 10

(ii) 14

Let m^{2} – 1 = 14 ⇒ m^{2} = 1 + 14 = 15

The value of m will not be an integer.

Now take 2m = 14 ⇒ m = 7 which is an integer.

The member of triplets are 2m = 2 × 7 = 14

m^{2} – 1 = (7)^{2} – 1 = 49 – 1 = 48

and m^{2} + 1 = (7)^{2} + 1 = 49 + 1 = 50

i.e., (14, 48, 50)

(iii) 16

Let 2m = 16 m = 8

The required triplets are 2m = 2 × 8 = 16

m^{2} – 1 = (8)^{2} – 1 = 64 – 1 = 63

m^{2} + 1 = (8)^{2} + 1 = 64 + 1 = 65

i.e., (16, 63, 65)

(iv) 18

Let 2m = 18 ⇒ m = 9

Required triplets are:

2m = 2 × 9 = 18

m^{2} – 1 = (9)^{2} – 1 = 81 – 1 = 80

and m^{2 }+ 1 = (9)^{2} + 1 = 81 + 1 = 82

i.e., (18, 80, 82)

**Exercise 5.3**

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801

One’s digit in the square root of 9801 maybe 1 or 9

(ii) 99856

One’s digit in the square root of 99856 maybe 4 or 6

(iii) 998001

One’s digit in the square root of 998001 maybe 1 or 9

(iv) 657666025

One’s digit in the square root of 657666025 can be 5.

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

Solution:We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares

(i) 153

153 is not a perfect square number. (ending with 3)

(ii) 257

257 is not a perfect square number. (ending with 7)

(iii) 408

408 is not a perfect square number. (ending with 8)

(iv) 441

441 is a perfect square number.

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

Solution:Using the method of repeated subtraction of consecutive odd numbers, we have

(i)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 (Ten times repetition) Thus √100 = 10

ii) 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
- (Thirteen times repetition)

Thus √169 = 13

4. Find the square roots of the following numbers by the prime factorization Method.

(i) 729

Prime factors of 729

729 = 3 × 3 × 3 × 3 × 3 × 3 = 3^{2} × 3^{2} × 3^{2}

√729 = 3 × 3 × 3 = 27

(ii) 400

Prime factors of 400

400 = 2 × 2 × 2 × 2 × 5 × 5 = 2^{2} × 2^{2} × 5^{2}

√400 = 2 × 2 × 5 = 20

(iii) 1764

1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2^{2} × 3^{2} × 7^{2}

√1764 = 2 × 3 × 7 = 42

(iv) 4096

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

= 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2}

√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v) 7744

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

= 2^{2} × 2^{2} × 2^{2} × 11^{2}

√7744 = 2 × 2 × 2 × 11 = 88

(vi) 9604

Prime factorization of 9604 is

9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2^{2} × 7^{2} × 7^{2}

√9604 = 2 × 7 × 7 = 98

vii) 5929

Prime factorisation of 5929 is

5929 = 7 × 7 × 11 × 11 = 7^{2} × 11^{2}

√5929 = 7 × 11 = 77

(viii) 9216

Prime factorisation of 9216 is

9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3

= 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2} × 3^{2}

√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

ix) 529

Prime factorisation of 529 is

529 = 23 × 23 = 23^{2}

√529 = 23

(x) 8100

Prime factorisation of 8100 is

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2^{2} × 3^{2} × 3^{2} × 5^{2}

√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

Prime factorisation of 252 is

252 = 2 × 2 × 3 × 3 × 7

Here, the prime factorisation is not in pair. 7 has no pair.

Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.

The new square number is 252 × 7 = 1764

Square root of 1764 is

√1764 = 2 × 3 × 7 = 42

(ii) 180

Prime factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

New square number = 180 × 5 = 900

Here, 5 has no pair.

New square number = 180 × 5 = 900

Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

(iii) 1008

Prime factorisation of 1008 is

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, 7 has no pair.

New square number = 1008 × 7 = 7056

Thus, 7 is the required number.

Square root of 7056 is

√7056 = 2 × 2 × 3 × 7 = 84

(iv) 2028

Prime factorisation of 2028 is

2028 = 2 × 2 × 3 × 13 × 13

Here, 3 is not in pair.

Thus, 3 is the required smallest whole number.

New square number = 2028 × 3 = 6084

Square root of 6084 is

√6084 = 2 × 13 × 3 = 78

(v) 1458

Prime factorisation of 1458 is

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, 2 is not in pair.

Thus, 2 is the required smallest whole number.

New square number = 1458 × 2 = 2916

Square root of 1458 is

√2916 = 3 × 3 × 3 × 2 = 54

(vi) 768

Prime factorisation of 768 is

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, 3 is not in pair.

Thus, 3 is the required whole number.

New square number = 768 × 3 = 2304

Square root of 2304 is

√2304 = 2 × 2 × 2 × 2 × 3 = 48

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

Prime factorisation of 252 is

252 = 2 × 2 × 3 × 3 × 7

Here 7 has no pair.

7 is the smallest whole number by which 252 is divided to get a square number.

New square number = 252 ÷ 7 = 36

Thus, √36 = 6

(ii) 2925

Prime factorisation of 2925 is

2925 = 3 × 3 × 5 × 5 × 13

Here, 13 has no pair.

13 is the smallest whole number by which 2925 is divided to get a square number.

New square number = 2925 ÷ 13 = 225

Thus √225 = 15

(iii) 396

Prime factorisation of 396 is

396 = 2 × 2 × 3 × 3 × 11

Here 11 is not in pair.

11 is the required smallest whole number by which 396 is divided to get a square number.

New square number = 396 ÷ 11 = 36

Thus √36 = 6

iv) 2645

Prime factorisation of 2645 is

2645 = 5 × 23 × 23

Here, 5 is not in pair.

5 is the required smallest whole number.

By which 2645 is multiplied to get a square number

New square number = 2645 ÷ 5 = 529

Thus, √529 = 23

(v) 2800

Prime factorisation of 2800 is

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

Here, 7 is not in pair.

7 is the required smallest number.

By which 2800 is multiplied to get a square number.

New square number = 2800 ÷ 7 = 400

Thus √400 = 20

(vi) 1620

Prime factorisation of 1620 is

1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

Here, 5 is not in pair.

5 is the required smallest prime number.

By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324

Thus √324 = 18

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

Total amount of money donated = ₹ 2401

Total number of students in the class = √2401

= √ 7^2×7^2

= √7×7×7×7

= 7 × 7

= 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 rows = Total number of plants in each row = √2025

= √3×3×3×3×5×5

= √3^2×3^2×5^2

= 3 × 3 × 5

= 45

Thus the number of rows and plants = 45

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

Solution:LCM of 4, 9, 10 = 180

The least number divisible by 4, 9 and 10 = 180

Now prime factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

The required smallest square number = 180 × 5 = 900

10. Find the smallest number that is divisible by each of the numbers 8, 15 and 20.

Solution: The smallest number divisible by 8, 15 and 20 is equal to their LCM.

LCM = 2 × 2 × 2 × 3 × 5 = 120

Here, 2, 3 and 5 have no pair.

The required smallest square number = 120 × 2 × 3 × 5 = 120 × 30 = 3600