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Class 8 Maths Notes Squares and Square Roots

⏹️ Chapter 6: Squares and Square Roots (Class 8) √

Hi Math Magicians! Aapne numbers ko multiply karna seekha hai. Jab hum ek number ko usi se multiply karte hain (like 3 × 3), toh humein uska **Square** milta hai. Is chapter mein hum squares aur unke opposite operation – **Square Roots** ke baare mein detail mein padhenge.

Hi Math Magicians! Aapne numbers ko multiply karna seekha hai. Jab hum ek number ko usi se multiply karte hain (jaise 3 × 3), toh humein uska **Varg (Square)** milta hai. Is chapter mein hum squares aur unke ulte operation – **Vargmool (Square Roots)** ke baare mein detail mein padhenge.

🔢 Squares and Square Numbers

Square of a Number: The number obtained when a number is multiplied by itself.

If ‘n’ is a number, its square is n × n = n².

Ek Sankhya ka Varg: Vah sankhya jo kisi sankhya ko usi se guna karne par milti hai. Agar ‘n’ ek sankhya hai, toh uska varg hai n × n = n².

Example: Square of 3 is 3 × 3 = 9 (written as 3² = 9).

Example: Square of 7 is 7 × 7 = 49 (written as 7² = 49).

Example: Square of 10 is 10 × 10 = 100 (written as 10² = 100).

Perfect Square (or Square Number): A natural number that is the square of some natural number.

Poorna Varg (Perfect Square/Square Number): Ek prakritik sankhya jo kisi prakritik sankhya ka varg ho.

Examples: 1 (1²), 4 (2²), 9 (3²), 16 (4²), 25 (5²), 36 (6²), … are perfect squares.

Numbers like 2, 3, 5, 6, 7, 8, 10 are NOT perfect squares.

⭐ Properties of Square Numbers (Varg Sankhyaon Ke Gun)

Perfect squares have some interesting properties related to their ending digits:

Poorna varg sankhyaon ke antim ank (ending digits) se judi kuch dilchasp properties hoti hain:
Ending Digit Properties:
  • A number ending in 2, 3, 7, or 8 is NEVER a perfect square. ❌
    • Jo number 2, 3, 7, ya 8 par khatm hota hai, vah kabhi bhi perfect square nahi hota.
  • A number ending in 0, 1, 4, 5, 6, or 9 MAY OR MAY NOT be a perfect square.
    • Jo number 0, 1, 4, 5, 6, ya 9 par khatm hota hai, vah perfect square ho bhi sakta hai aur nahi bhi.
  • Square numbers can only end with an EVEN number of zeroes. (e.g., 100 (2 zeros), 90000 (4 zeros) are squares, but 10, 1000 are not).
    • Square numbers ke end mein hamesha even number of zeroes (0, 2, 4, 6…) hote hain.
  • If a number ends in 1 or 9, its square ends in 1. (1²=1, 9²=81, 11²=121, 19²=361)
  • If a number ends in 2 or 8, its square ends in 4. (2²=4, 8²=64, 12²=144, 18²=324)
  • If a number ends in 3 or 7, its square ends in 9. (3²=9, 7²=49, 13²=169, 17²=289)
  • If a number ends in 4 or 6, its square ends in 6. (4²=16, 6²=36, 14²=196, 16²=256)
  • If a number ends in 5, its square ends in 5 (specifically ends in 25). (5²=25, 15²=225, 25²=625)
Other Interesting Patterns:
  • There are 2n non-perfect square numbers between the squares of two consecutive numbers n and (n+1).
    • Do lagatar numbers n aur (n+1) ke squares ke beech mein 2n non-perfect square numbers hote hain.
    Ex: Between 3²(9) and 4²(16), there are 2n=2×3=6 numbers (10, 11, 12, 13, 14, 15).
  • The sum of the first n odd natural numbers is n².
    • Pehle n visham prakritik sankhyaon (odd natural numbers) ka yog n² hota hai.
    Ex: 1 = 1²
    1 + 3 = 4 = 2²
    1 + 3 + 5 = 9 = 3²
    1 + 3 + 5 + 7 = 16 = 4²
  • A perfect square can be expressed as the sum of consecutive odd numbers starting from 1.
  • The square of any odd number can be expressed as the sum of two consecutive positive integers. Formula: n² = n²-12 + n²+12 (where n is odd).
    • Kisi bhi visham sankhya (odd number) ke square ko do lagatar positive integers ke sum ke roop mein likha ja sakta hai.
    Ex: 3²=9 = 4+5; 5²=25 = 12+13; 7²=49 = 24+25.

✖️ Finding the Square of a Number

The basic way is direct multiplication. For slightly larger numbers, we can use identities.

Sabse basic tareeka seedha multiply karna hai. Thode bade numbers ke liye, hum identities use kar sakte hain.
  • Direct Multiplication: Find 23². Simply multiply 23 × 23 = 529.
  • Using Identity (a + b)² = a² + 2ab + b²:
    Find 42²:

    Write 42 as (40 + 2).

    (40 + 2)² = 40² + 2(40)(2) + 2²

    = 1600 + 160 + 4 = 1764

  • Using Identity (a - b)² = a² - 2ab + b²:
    Find 98²:

    Write 98 as (100 – 2).

    (100 - 2)² = 100² - 2(100)(2) + 2²

    = 10000 - 400 + 4 = 9604

  • Special Case for numbers ending in 5: Square of number ending in 5 (say ‘a5’). Result ends in 25. The part before 25 is a × (a + 1). 5 par khatm hone wale number (maano ‘a5’) ka square. Result 25 par khatm hoga. 25 se pehle ka hissa a × (a + 1) hai.
    Find 35²:

    Ends in 5. Here a = 3.

    a × (a+1) = 3 × (3+1) = 3 × 4 = 12.

    Result: Put 12 before 25 => 1225.

    Find 75²:

    Ends in 5. Here a = 7.

    a × (a+1) = 7 × (7+1) = 7 × 8 = 56.

    Result: Put 56 before 25 => 5625.

📐 Pythagorean Triplets

Pythagorean Triplet: A set of three positive integers a, b, c such that a² + b² = c². These numbers can form the sides of a right-angled triangle 📐.

Pythagorean Trik: Teen positive integers a, b, c ka ek set jiske liye a² + b² = c² ho. Yeh sankhyaen ek samkon tribhuj (right-angled triangle) ki bhujayen ban sakti hain.
The simplest triplet is (3, 4, 5): 3² + 4² = 9 + 16 = 25 = 5².

For any natural number m > 1, we have the Pythagorean triplet: (2m, m² - 1, m² + 1)

Kisi bhi natural number m > 1 ke liye, humare paas Pythagorean triplet hai: (2m, m² – 1, m² + 1)
Finding a Pythagorean Triplet when one member is given:
  1. Let the given number be equal to one form: 2m OR m² - 1 OR m² + 1.
  2. Solve for m. Make sure m is a natural number greater than 1.
  3. Usually, if the given number is EVEN, try setting it equal to 2m first. If ODD, try setting it equal to m² - 1 or m² + 1.
    4. Agar diya gaya number EVEN hai, toh use 2m ke barabar rakh kar try karo. Agar ODD hai, toh use m² – 1 ya m² + 1 ke barabar rakh kar try karo.
  4. Once you find a valid m, calculate the other two members using the formulas: 2m, m² - 1, m² + 1.
Example 1: Find a Pythagorean triplet whose one member is 6.

Try setting 2m = 6.

Then m = 3 (which is > 1, so valid).

Other members are:

m² - 1 = 3² - 1 = 9 - 1 = 8

m² + 1 = 3² + 1 = 9 + 1 = 10

Answer: The triplet is (6, 8, 10). Check: 6² + 8² = 36 + 64 = 100 = 10². ✅

Example 2: Find a Pythagorean triplet whose one member is 12.

Try 2m = 12.

m = 6 (valid).

Other members:

m² - 1 = 6² - 1 = 36 - 1 = 35

m² + 1 = 6² + 1 = 36 + 1 = 37

Answer: The triplet is (12, 35, 37).

Example 3: Find a Pythagorean triplet whose smallest member is 8.

Try 2m = 8 => m = 4.

Members are: 2m = 8, m²-1 = 4²-1 = 15, m²+1 = 4²+1 = 17.

Triplet is (8, 15, 17). Here 8 is the smallest.

*(Let’s check other options)*

Try m²-1 = 8 => m²=9 => m=3.

Members are: m²-1 = 8, 2m=2(3)=6, m²+1 = 3²+1 = 10.

Triplet is (6, 8, 10). Here 8 is NOT the smallest. So (8, 15, 17) is the required triplet.

Answer: The triplet is (8, 15, 17).

√ Square Roots (Vargmool)

Square Root: The square root of a number ‘x’ is the number which, when multiplied by itself, gives ‘x’. It’s the inverse operation of squaring.

Symbol: √ (radical sign)

Vargmool: Ek sankhya ‘x’ ka vargmool vah sankhya hoti hai jise khud se guna karne par ‘x’ milta hai. Yeh squaring ka ulta operation hai. Chinh: √.

Example: Since 3² = 9, the square root of 9 is 3 (√9 = 3).

Example: Since 8² = 64, the square root of 64 is 8 (√64 = 8).

A perfect square has exactly two square roots: one positive and one negative (e.g., both 3 and -3 are square roots of 9 because 3²=9 and (-3)²=9). However, the symbol √ usually denotes the positive square root (principal square root).

Ek perfect square ke theek do vargmool hote hain: ek positive aur ek negative. Lekin √ chinh aam taur par positive vargmool ko darshata hai.

🧩 Finding Square Root through Prime Factorisation (Abhajya Gunankhandan Dwara Vargmool)

This method works well for perfect squares.

Yeh method perfect squares ke liye achha kaam karta hai.
Steps:
  1. Find the prime factors of the given number.
    2. Diye gaye number ke abhajya gunankhand (prime factors) nikalo.
  2. Make pairs of identical prime factors.
    3. Ek jaise prime factors ke jode (pairs) banao.
  3. Take one factor from each pair.
    4. Har jode (pair) se ek factor lo.
  4. Multiply these chosen factors together.
    5. Chune gaye factors ko aapas mein multiply karo.
  5. The product obtained is the square root.
Example 1: Find √729 by prime factorisation.

Solution:

Prime factorise 729:

  • 3 | 729
  • 3 | 243
  • 3 | 81
  • 3 | 27
  • 3 | 9
  • 3 | 3
  • | 1

729 = 3 × 3 × 3 × 3 × 3 × 3

Make pairs: (3 × 3) × (3 × 3) × (3 × 3)

Take one factor from each pair: 3 × 3 × 3

Multiply: 3 × 3 × 3 = 27

Answer: √729 = 27.

Example 2: Find √1764 by prime factorisation.

Solution:

Prime factorise 1764:

  • 2 | 1764
  • 2 | 882
  • 3 | 441
  • 3 | 147
  • 7 | 49
  • 7 | 7
  • | 1

1764 = 2 × 2 × 3 × 3 × 7 × 7

Pairs: (2 × 2) × (3 × 3) × (7 × 7)

Take one from each pair: 2 × 3 × 7

Multiply: 2 × 3 × 7 = 42

Answer: √1764 = 42.

Example 3: Is 2352 a perfect square? If not, find the smallest whole number by which it must be multiplied to get a perfect square.

Solution:

Prime factorise 2352:

  • 2 | 2352
  • 2 | 1176
  • 2 | 588
  • 2 | 294
  • 3 | 147
  • 7 | 49
  • 7 | 7
  • | 1

2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7

Pairs: (2 × 2) × (2 × 2) × (7 × 7) × 3

The prime factor 3 does not have a pair.

Answer (Part 1): No, 2352 is not a perfect square.

To make it a perfect square, we need to complete the pair for 3. So, we must multiply by 3.

Answer (Part 2): Smallest whole number is 3.

(New number = 2352 × 3 = 7056. √7056 = 2 × 2 × 7 × 3 = 84).

(Note: Repeated Subtraction method also exists but is only practical for small numbers).

➗ Finding Square Root by Division Method (Bhaag Vidhi Dwara Vargmool)

This method works for large numbers and non-perfect squares (to find approximate value or decimals). Very useful!

Yeh method bade numbers aur non-perfect squares ke liye kaam karta hai. Bahut upyogi hai!
Steps (for Perfect Squares like √529):
  1. Pair Digits: Place bars over pairs of digits starting from the unit’s place (right). If number of digits is odd, the leftmost single digit will also have a bar. Ex: 5 29.
  2. Find Divisor/Quotient: Find the largest number whose square is less than or equal to the number under the leftmost bar (here, 5). That number is 2 (since 2²=4 < 5). Write 2 as divisor and quotient. Subtract the square (4) from 5. Remainder = 1.
    3.       2     
       2|5 29 
        -4    
          1
  3. Bring Down Next Pair: Bring down the next pair of digits (29) to the right of the remainder. New dividend is 129.
    4.       2     
       2|5 29 
        -4    
          1 29
  4. Double Quotient, Find New Digit: Double the current quotient (2) -> 4. Find a digit (say ‘y’) such that when placed next to 4 (making it ‘4y’), multiplying ‘4y’ by ‘y’ gives a product ≤ 129. Try 41×1=41, 42×2=84, 43×3=129. The digit is 3. Write 3 next to the quotient (making it 23) and next to the new divisor (making it 43).
    5.       2 3   
       2|5 29 
        -4 ↓
      43|1 29
  5. Multiply and Subtract: Multiply the new divisor (43) by the new digit in quotient (3): 43 × 3 = 129. Subtract this from the dividend (129). Remainder is 0.
    6.       2 3   
       2|5 29 
        -4 ↓
      43|1 29 
        -1 29
          0
  6. Since remainder is 0 and no pairs left, the quotient (23) is the square root.
Example 1: Find √529 using division method.

Solution: (Steps shown above).

Answer: √529 = 23.

Example 2: Find √4096 using division method.

Solution:

       6  4
   6|40 96
    -36  ↓
 124|  4 96
     -4 96
          0

Answer: √4096 = 64.

Example 3: Find the least number to be subtracted from 5607 to make it a perfect square. Also find the square root of the perfect square.

Solution: Use long division on 5607.

       7  4
   7|56 07
    -49  ↓
 144|  7 07
     -5 76
       1 31 --> Remainder

We find 74² < 5607. The remainder is 131.

This means if we subtract the remainder (131) from 5607, we get a perfect square.

Least number to be subtracted = 131.

Perfect square = 5607 – 131 = 5476.

Square root of the perfect square (5476) is the quotient we found = 74.

Answer: Subtract 131; √5476 = 74.

.√ Square Roots of Decimals (Dashamlav Ka Vargmool)

Use the long division method with slight modification for pairing.

Long division method istemal karo, pairing mein thoda sa badlav karke.
Steps:
  1. Pair Digits: Place bars on the integral part starting from unit digit. Place bars on the decimal part starting from the first decimal place. Add zeroes at the end if needed to complete pairs. Ex: 17.64, 12.25, 2.56, 6.4000
  2. Perform Long Division: Follow the same steps as for whole numbers.
  3. Place Decimal Point: Put the decimal point in the quotient as soon as the integral part is exhausted (when you bring down the first pair after the decimal point).
    4. Jaise hi integer wala hissa khatam ho jaye (decimal ke baad wala pehla pair neeche lao), quotient mein decimal point laga do.
Example 1: Find √17.64

Solution:

       4. 2
   4|17.64
    -16  ↓↓ <-- Decimal in quotient
  82|  1 64
     -1 64
          0

Answer: √17.64 = 4.2

Example 2: Find √12.25

Solution:

       3. 5
   3|12.25
     -9  ↓↓
  65|  3 25
     -3 25
          0

Answer: √12.25 = 3.5

Example 3: Find √2 (approx to 2 decimal places)

Solution: Write 2 as 2.00 00 00

       1. 4  1  4
   1| 2.00 00 00
     -1  ↓↓
  24| 1 00
      - 96 ↓↓
 281|   4 00
       - 2 81 ↓↓
2824|   1 19 00
        -1 12 96
            6 04

Answer: √2 ≈ 1.414 (Approximation needed)

≈ Estimating Square Root (Vargmool Ka Anuman)

We can estimate the square root of non-perfect square numbers.

Hum non-perfect square numbers ka vargmool anumaan laga sakte hain.
Steps (Example: Estimate √80):
  1. Find the perfect squares just less than and just greater than the number. For 80, they are 64 (8²) and 81 (9²).
    2. Number se theek chhota aur theek bada perfect square dhoondho. 80 ke liye, 64 aur 81 hain.
  2. So, 64 < 80 < 81, which means √64 < √80 < √81.
  3. Therefore, 8 < √80 < 9. The square root lies between 8 and 9.
  4. See which perfect square is closer. 80 is closer to 81 (distance 1) than to 64 (distance 16).
    5. Dekho kaunsa perfect square zyada paas hai. 80, 81 ke zyada paas hai.
  5. So, √80 is approximately closer to 9 than 8. A reasonable estimate might be 8.9.
    6. Toh √80 lagbhag 9 ke zyada paas hoga 8 se. Ek anumaan 8.9 ho sakta hai.
Example 1: Estimate √10

Perfect squares around 10 are 9 (3²) and 16 (4²).

9 < 10 < 16 => 3 < √10 < 4.

10 is closer to 9 (distance 1) than 16 (distance 6).

So, √10 is slightly greater than 3. Estimate ≈ 3.1 or 3.2.

Example 2: Estimate √410

We know 20² = 400 and 21² = 441.

400 < 410 < 441 => 20 < √410 < 21.

410 is closer to 400 (distance 10) than 441 (distance 31).

So, √410 is slightly greater than 20. Estimate ≈ 20.2.

Example 3: Estimate √90

Perfect squares around 90 are 81 (9²) and 100 (10²).

81 < 90 < 100 => 9 < √90 < 10.

Distance from 81 is 9. Distance from 100 is 10. It's slightly closer to 81.

So, √90 is roughly in the middle, slightly closer to 9. Estimate ≈ 9.5 (or maybe 9.4).

❓ Sawal Jawab (Questions & Answers)

🤏 Very Short Answer Questions

1. What is the square of 6?

36.

2. What is the square root of 81?

9.

3. Is 50 a perfect square?

No.

4. What are the possible ending digits of a perfect square?

0, 1, 4, 5, 6, 9.

5. If a number ends in 3, can it be a perfect square?

No.

6. If a number ends in 4, what can be the ending digit of its square root?

2 or 8.

7. How many zeroes can a perfect square end with?

An even number of zeroes.

8. How many non-square numbers are between 5² and 6²?

2n = 2 × 5 = 10.

9. What is 1 + 3 + 5 + 7 + 9?

Sum of first 5 odd numbers = 5² = 25.

10. Express 49 as sum of odd numbers.

1 + 3 + 5 + 7 + 9 + 11 + 13.

11. Write a Pythagorean triplet using the formula for m=4.

2m=8, m²-1=15, m²+1=17. Triplet is (8, 15, 17).

12. Find the square root of 64 by repeated subtraction.

64-1=63, 63-3=60, 60-5=55, 55-7=48, 48-9=39, 39-11=28, 28-13=15, 15-15=0. (Subtracted 8 odd numbers). Ans: 8.

13. Find √100 using prime factorisation.

100 = 2×2×5×5 = (2×2)×(5×5). √100 = 2×5 = 10.

14. In long division of √289, what is the first divisor?

Pairing is 2 89. Largest square ≤ 2 is 1²=1. Divisor is 1.

15. Estimate √50.

Between √49=7 and √64=8. Closer to 7. Approx 7.1.

16. Find √2.25

Pairing 2.25. √225 = 15. One pair after decimal -> One decimal place in answer. Ans: 1.5.

17. Is (6, 8, 10) a Pythagorean triplet?

Yes (6²+8²=36+64=100=10²).

📝 Short Answer Questions

1. Write down the squares of numbers from 1 to 10.

  • 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36, 7²=49, 8²=64, 9²=81, 10²=100.

2. Why do numbers ending in 2, 3, 7, 8 never form a perfect square?

  • Observe the unit digits of squares of numbers 1-10 (1, 4, 9, 6, 5, 6, 9, 4, 1, 0).
  • The ending digits of perfect squares can only be 0, 1, 4, 5, 6, or 9.
  • Digits 2, 3, 7, 8 never appear as the unit digit of a perfect square.

3. Find the square of 43 using the identity (a+b)².

  • 43 = 40 + 3.
  • (40+3)² = 40² + 2(40)(3) + 3²
  • = 1600 + 240 + 9
  • = 1849.

4. Find the square of 58 using the identity (a-b)².

  • 58 = 60 - 2.
  • (60-2)² = 60² - 2(60)(2) + 2²
  • = 3600 - 240 + 4
  • = 3364.

5. Find the square of 65 using the pattern for numbers ending in 5.

  • Number is 'a5' where a=6.
  • Calculate a × (a+1) = 6 × (6+1) = 6 × 7 = 42.
  • Put '42' before '25'.
  • Result: 4225.

6. Write a Pythagorean triplet whose one member is 14.

  • Try 2m = 14 => m = 7 (valid).
  • Other members:
  • m²-1 = 7²-1 = 49-1 = 48.
  • m²+1 = 7²+1 = 49+1 = 50.
  • Triplet is (14, 48, 50).

7. Find the square root of 1296 by prime factorisation.

  • 1296 = 2 | 648 -> 2 | 324 -> 2 | 162 -> 2 | 81
  • 81 = 3 | 27 -> 3 | 9 -> 3 | 3 -> 3 | 1
  • 1296 = (2×2) × (2×2) × (3×3) × (3×3)
  • Take one from each pair: 2 × 2 × 3 × 3
  • Result = 4 × 9 = 36. √1296 = 36.

8. Find the square root of 17.64 by long division.

  • Pairing: 17 . 64
  • Largest square ≤ 17 is 4²=16. (Quotient=4, Divisor=4). Subtract 16, Rem=1.
  • Bring down 64. Dividend=164.
  • Double quotient: 2×4=8. New divisor looks like 8y.
  • Try y=2. 82 × 2 = 164.
  • Put decimal in quotient. Subtract 164, Rem=0.
  • √17.64 = 4.2.
   4. 2 
 4|17.64
  -16
 ---
 82| 1 64
   -1 64
   -----
      0

9. Estimate √150.

  • Squares around 150: 12² = 144, 13² = 169.
  • 144 < 150 < 169 => 12 < √150 < 13.
  • Distance from 144 is 6. Distance from 169 is 19.
  • 150 is much closer to 144.
  • Estimate: Slightly more than 12. Approx 12.2 or 12.3.

10. Find the smallest square number divisible by each of 8, 15, 20.

  • Find LCM of 8, 15, 20.
  • 8=2×2×2; 15=3×5; 20=2×2×5.
  • LCM = 2×2×2 × 3 × 5 = 8 × 15 = 120.
  • Prime factors of LCM: 120 = 2 × 2 × 2 × 3 × 5.
  • Pairs: (2×2). Unpaired: 2, 3, 5.
  • To make it a perfect square, multiply by unpaired factors: 2 × 3 × 5 = 30.
  • Smallest square number = 120 × 30 = 3600.

11. 2025 plants are to be planted in a garden such that each row contains as many plants as the number of rows. Find the number of rows and plants in each row.

  • Let number of rows = x.
  • Number of plants in each row = x.
  • Total plants = (Number of rows) × (Plants per row) = x × x = x².
  • Given: x² = 2025.
  • We need to find x = √2025.
  • Using prime factorisation or long division: √2025 = 45.
  • Answer: Number of rows = 45, Plants in each row = 45.

📜 Long Answer Questions

1. Explain the properties related to the ending digits of square numbers.

  • Numbers ending in 2, 3, 7, 8 are NEVER perfect squares.
  • Numbers ending in 0, 1, 4, 5, 6, 9 MAY be perfect squares.
  • Numbers ending in 1 or 9 have squares ending in 1.
  • Numbers ending in 2 or 8 have squares ending in 4.
  • Numbers ending in 3 or 7 have squares ending in 9.
  • Numbers ending in 4 or 6 have squares ending in 6.
  • Numbers ending in 5 have squares ending in 5 (specifically 25).
  • Square numbers end with an EVEN number of zeroes.

2. Explain the method of finding square root using Prime Factorisation with an example (e.g., √1764).

This method uses the prime factors of the number.

Steps:
  1. Find prime factors of the number.
  2. Group the factors in pairs of identical factors.
  3. Take one factor from each pair.
  4. Multiply these chosen factors. The product is the square root.
Example: √1764
  • Factorisation: 1764 = 2×2 × 3×3 × 7×7.
  • Pairs: (2×2) × (3×3) × (7×7).
  • Take one from each pair: 2, 3, 7.
  • Multiply: 2 × 3 × 7 = 42.
  • So, √1764 = 42.

3. Explain the method of finding square root using Long Division with an example (e.g., √729).

This method is suitable for large numbers.

Steps (Example √729):
  1. Pairing: Pair digits from right: 7 29.
  2. First Divisor/Quotient: Largest square ≤ 7 is 2²=4. Divisor=2, Quotient=2. Subtract 4 from 7. Rem=3.
  3. Bring Down Pair: Bring down next pair (29). New dividend=329.
  4. New Divisor: Double current quotient (2) -> 4. Find digit 'y' such that 4y × y ≤ 329. Try 47×7=329. So y=7.
  5. Update Quotient/Divisor: New digit in quotient is 7 (making it 27). New divisor is 47.
  6. Multiply & Subtract: 47 × 7 = 329. Subtract from dividend. Rem=0.
    2  7
 2| 7 29
  -4 ↓
 47| 3 29
   -3 29
     0

Since remainder is 0, √729 = 27.

4. Find the square root of 6.4009 by Long Division method.

Number: 6.4009

  • Step 1: Pairing: Pair integral part from right (6). Pair decimal part from left (.40 09).
  • Step 2: First Digit: Largest square ≤ 6 is 2²=4. Quotient=2, Divisor=2. Subtract 4. Rem=2.
  • Step 3: Place Decimal, Bring Down Pair: Place decimal in quotient. Bring down 40. Dividend=240.
  • Step 4: New Divisor: Double quotient (2) -> 4. Find 'y' so 4y × y ≤ 240. Try 45×5=225. (46×6=276 > 240). So y=5. Quotient=2.5, Divisor=45. Subtract 225. Rem=15.
  • Step 5: Bring Down Pair: Bring down 09. Dividend=1509.
  • Step 6: New Divisor: Double current quotient (25 ignore decimal) -> 50. Find 'y' so 50y × y ≤ 1509. Try 503×3=1509. So y=3. Quotient=2.53, Divisor=503.
  • Step 7: Multiply & Subtract: 503 × 3 = 1509. Subtract. Rem=0.
      2. 5  3
  2| 6.40 09
    -4  ↓↓
  45| 2 40
    -2 25 ↓↓
 503|  15 09
      -15 09
          0

Answer: √6.4009 = 2.53

5. Find the least number that must be subtracted from 402 to obtain a perfect square. Also find the square root of the perfect square obtained.

  • Step 1: Find square root of 402 by division. Pairing: 4 02.
  • First step: 2²=4. Quotient=2, Divisor=2. Subtract 4, Rem=0.
  • Bring down 02. Dividend=02.
  • New divisor: Double quotient (2) -> 4. Find 'y' so 4y × y ≤ 02. Only possible y=0. 40×0=0. Quotient=20, Divisor=40.
  • Subtract 0 from 02. Rem=2.
    •      2  0
      2| 4 02
        -4 ↓
      40|  0 02
          -  0
             2 --> Remainder
  • Step 2: Interpret remainder. The remainder is 2. This means 20² < 402. If we subtract the remainder (2) from 402, the result will be a perfect square (20²).
  • Least number to be subtracted = Remainder = 2.
  • Perfect square = 402 - 2 = 400.
  • Square root of the perfect square = Quotient = 20.

Answer: Subtract 2; √400 = 20.

6. Find the least number that must be added to 1300 to obtain a perfect square. Also find the square root of the perfect square obtained.

  • Step 1: Find square root of 1300 by division. Pairing: 13 00.
  • First step: 3²=9. Quotient=3, Divisor=3. Subtract 9. Rem=4.
  • Bring down 00. Dividend=400.
  • New divisor: Double quotient (3) -> 6. Find 'y' so 6y × y ≤ 400. Try 66×6=396. (67×7=469 > 400). So y=6. Quotient=36.
    •       3  6
      3| 13 00
        - 9 ↓
      66|  4 00
          -3 96
              4 --> Remainder
  • Step 2: Interpret result. This shows 36² < 1300. The next perfect square is 37².
  • Calculate the next square: 37² = 1369.
  • Step 3: Find number to add. Number to be added = (Next perfect square) - (Given number).
    Number to add = 37² - 1300 = 1369 - 1300 = 69.
  • The perfect square obtained is 1369.
  • Its square root is 37.

Answer: Add 69; √1369 = 37.

7. Write a Pythagorean triplet where one number is 16.

We look for triplet (2m, m²-1, m²+1) with one member = 16.

  • Option 1: Try 2m = 16.
    Then m = 16 / 2 = 8. (Valid since m > 1).
    Other members: m²-1 = 8²-1 = 64-1 = 63. m²+1 = 8²+1 = 64+1 = 65.
    Triplet is (16, 63, 65).
  • Option 2: Try m²-1 = 16.
    m² = 17. Since 17 is not a perfect square, m is not an integer. Reject this option.
  • Option 3: Try m²+1 = 16.
    m² = 15. Since 15 is not a perfect square, m is not an integer. Reject this option.

Answer: The required Pythagorean triplet is (16, 63, 65).

8. Find the square root of 90 by estimation.

Estimating √90:

  • Step 1: Find surrounding perfect squares.
    We know 9² = 81 and 10² = 100.
  • Step 2: Place the number between them.
    81 < 90 < 100.
  • Step 3: Take square roots.
    √81 < √90 < √100
    9 < √90 < 10.
  • Step 4: Check closeness.
    Distance from lower square: 90 - 81 = 9.
    Distance from upper square: 100 - 90 = 10.
  • Step 5: Estimate. 90 is slightly closer to 81 than to 100. Therefore, √90 will be slightly closer to 9 than to 10 (but almost halfway). A good estimate is around 9.4 or 9.5.

(Using calculator, √90 ≈ 9.4868...)

9. Find the square root of 7.29 by division method.

  • Step 1: Pairing. Pair integral part from right (7). Pair decimal part from left (.29). Pairing: 7 . 29.
  • Step 2: First Digit. Largest square ≤ 7 is 2²=4. Quotient=2, Divisor=2. Subtract 4. Rem=3.
  • Step 3: Place Decimal, Bring Down Pair. Put decimal in quotient. Bring down 29. Dividend=329.
  • Step 4: New Divisor. Double quotient (2) -> 4. Find 'y' so 4y × y ≤ 329. Try 47×7=329. So y=7. Quotient=2.7, Divisor=47.
  • Step 5: Multiply & Subtract. 47 × 7 = 329. Subtract from dividend. Rem=0.
      2. 7
  2| 7.29
    -4  ↓↓
  47| 3 29
    -3 29
      0

Answer: √7.29 = 2.7

10. Find the smallest number by which 9408 must be divided to get a perfect square. Also find the square root of the quotient.

  • Step 1: Prime Factorise 9408.
    9408 = 2 | 4704 -> 2 | 2352 -> 2 | 1176 -> 2 | 588 -> 2 | 294 -> 2 | 147
    147 = 3 | 49 -> 7 | 7 -> 7 | 1.
    So, 9408 = (2×2) × (2×2) × (2×2) × (7×7) × 3.
  • Step 2: Identify unpaired factor. The prime factor 3 is left unpaired.
  • Step 3: Find number to divide by. To make it a perfect square, we must divide by the unpaired factor, which is 3.
  • Step 4: Find the quotient (perfect square).
    Quotient = 9408 / 3 = 3136.
  • Step 5: Find square root of quotient.
    3136 = (2×2) × (2×2) × (2×2) × (7×7).
    √3136 = 2 × 2 × 2 × 7 = 8 × 7 = 56.

Answer: Smallest number to divide by is 3. Square root of quotient (3136) is 56.

🤔 Check Your Understanding! (Quiz Time!)

1. The square of 8 is:

2. Which of the following is a perfect square?

3. A number ending in which digit can NEVER be a perfect square?

4. The square of an even number is always:

5. Sum of first 4 odd numbers (1+3+5+7) is:

6. How many non-square numbers lie between 7² and 8²?

7. Which forms a Pythagorean triplet with 2m = 10?

8. The symbol √ represents:

9. Square root of 144 by prime factorization is:

10. The number of digits in the square root of 900 is:

11. In long division of √2304, what's the first digit of the root?

12. Square root of 1.21 is:

13. Which perfect square is closest to 55?

14. Estimate √20.

15. Which number must be multiplied by 72 to make it a perfect square?

16. Square of 0.3 is:

17. Which triplet is NOT Pythagorean?

18. The square root sign is √. What is ⁴√ (index 4)?

19. If a number ends in 6, its square root could end in:

20. Find the smallest number to divide 98 to get a perfect square.

21. Find √0.09

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