Squares and Square Roots

Welcome dear students! Today we are going to learn about Squares and Square Roots from Class 8 Maths.

Section 5.1 Introduction. You know that the area of a square = side × side. Study the following table. When side is 1 cm, area is 1 × 1 = 1 = 1². When side is 2 cm, area is 2 × 2 = 4 = 2². For side 3, area is 3 × 3 = 9 = 3². For side 5, area is 5 × 5 = 25 = 5². For side 8, area is 8 × 8 = 64 = 8². For a general side a, area is a × a = a².

What is special about 4, 9, 25, 64? Since 4 = 2 × 2 = 2², and 9 = 3 × 3 = 3², all such numbers are the product of a number with itself. Such numbers like 1, 4, 9, 16, 25, ... are square numbers. In general, if a natural number m can be expressed as n², where n is also a natural number, then m is a square number. Is 32 a square number? We know 5² = 25 and 6² = 36. If 32 is a square number, it must be the square of a natural number between 5 and 6. But there is no natural number between 5 and 6. Therefore 32 is not a square number.

Consider the squares from 1 to 10: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100. The numbers 1, 4, 9, 16 ... are square numbers, also called perfect squares.

Try These 1. Find the perfect square numbers between (i) 30 and 40, and (ii) 50 and 60.

Section 5.2 Properties of Square Numbers. The table shows squares from 1 to 20: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100. 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225, 16² = 256, 17² = 289, 18² = 324, 19² = 361, 20² = 400.

Study the ending digits. All end with 0, 1, 4, 5, 6 or 9 at units place. None end with 2, 3, 7 or 8. Can we say that if a number ends in 0, 1, 4, 5, 6 or 9, it must be a square number? Think about it.

Try These 1. Can we say whether the following are perfect squares? (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 1069 (vi) 2061. Try These 2. Write five numbers you can decide by units digit are not square numbers. Write five numbers you cannot decide just by units digit.

Study the extended table. Square numbers ending in 1: 1² = 1, 9² = 81, 11² = 121, 19² = 361, 21² = 441. Try These: Which of 123², 77², 82², 161², 109² would end with digit 1? You will see that if a number has 1 or 9 in the units place, then its square ends in 1.

Consider square numbers ending in 6: 4² = 16, 6² = 36, 14² = 196, 16² = 256. Try These: Which of (i) 19² (ii) 24² (iii) 26² (iv) 36² (v) 34² would have digit 6 at unit place? When a square number ends in 6, the original number has 4 or 6 in unit place.

Consider numbers with zeros. 10² = 100, 20² = 400, 80² = 6400 (one zero). 100² = 10000, 200² = 40000, 700² = 490000, 900² = 810000 (two zeros). Square numbers can only have an even number of zeros at the end.

What about squares of even and odd numbers? Even² is even. Odd² is odd. Try These 1. The square of which of the following would be odd or even? (i) 727 (ii) 158 (iii) 269 (iv) 1980. Try These 2. Number of zeros in the square of (i) 60 (ii) 400.

Section 5.3 Some More Interesting Patterns. 1. Adding triangular numbers. Triangular numbers are 1, 3, 6, 10, 15. If we combine two consecutive triangular numbers, we get a square number: 1 + 3 = 4 = 2². 3 + 6 = 9 = 3². 6 + 10 = 16 = 4².

2. Numbers between square numbers. Between 1² and 2², there are 2 non-square numbers: 2, 3. Between 2² and 3², there are 4: 5, 6, 7, 8. Between 3² and 4², there are 6: 10 to 15. Between n² and (n + 1)², there are 2n non-perfect square numbers. Check for n = 5: between 5² and 6², there are 10 numbers.

Try These 1. How many natural numbers lie between 9² and 10²? Between 11² and 12²? Try These 2. How many non square numbers lie between (i) 100² and 101² (ii) 90² and 91² (iii) 1000² and 1001².

3. Adding odd numbers. 1 = 1². 1 + 3 = 4 = 2². 1 + 3 + 5 = 9 = 3². 1 + 3 + 5 + 7 = 16 = 4². 1 + 3 + 5 + 7 + 9 = 25 = 5². 1 + 3 + 5 + 7 + 9 + 11 = 36 = 6². The sum of first n odd natural numbers is n². If a number is a square number, it is the sum of successive odd numbers starting from 1. Non-perfect squares like 2, 3, 5, 6 cannot be expressed this way.

Consider 25. Successively subtract 1, 3, 5, 7, 9: 25 - 1 = 24, 24 - 3 = 21, 21 - 5 = 16, 16 - 7 = 9, 9 - 9 = 0. So 25 = 1 + 3 + 5 + 7 + 9. 25 is a perfect square. Consider 38. 38 - 1 = 37, 37 - 3 = 34, 34 - 5 = 29, 29 - 7 = 22, 22 - 9 = 13, 13 - 11 = 2, 2 - 13 = -11. We cannot express 38 as a sum of consecutive odd numbers starting with 1. 38 is not a perfect square.

Try These. Find whether each is a perfect square: (i) 121 (ii) 55 (iii) 81 (iv) 49 (v) 69.

4. A sum of consecutive natural numbers. 3² = 9 = 4 + 5. 5² = 25 = 12 + 13. 7² = 49 = 24 + 25. 9² = 81 = 40 + 41. 11² = 121 = 60 + 61. 15² = 225 = 112 + 113. The square of any odd number can be expressed as the sum of two consecutive positive integers. Try These 1. Express as sum of two consecutive integers: (i) 21² (ii) 13² (iii) 11² (iv) 19². Try These 2. Is the reverse true? Give example.

5. Product of two consecutive even or odd natural numbers. 11 × 13 = 143 = 12² - 1. Also 11 × 13 = (12 - 1)(12 + 1). Similarly, 13 × 15 = 14² - 1. 29 × 31 = 30² - 1. 44 × 46 = 45² - 1. In general, (a - 1)(a + 1) = a² - 1.

6. Some more patterns. 1² = 1. 11² = 121. 111² = 12321. 1111² = 1234321. 11111² = 123454321. 11111111² = 123456787654321. Another pattern: 7² = 49. 67² = 4489. 667² = 444889. 6667² = 44448889. 66667² = 4444488889. 666667² = 444444888889. Try These. Write the square using pattern: (i) 111111² (ii) 1111111². Try These. Find square using pattern: (i) 6666667² (ii) 66666667².

EXERCISE 5.1 1. Unit digit of the square of: (i) 81 is 1. (ii) 272 is 4. (iii) 799 is 1. (iv) 3853 is 9. (v) 1234 is 6. (vi) 26387 is 9. (vii) 52698 is 4. (viii) 99880 is 0. (ix) 12796 is 6. (x) 55555 is 5. 2. Not perfect squares. Give reason. (i) 1057 ends in 7. (ii) 23453 ends in 3. (iii) 7928 ends in 8. (iv) 222222 ends in 2. (v) 64000 ends with odd number of zeros. (vi) 89722 ends in 2. (vii) 222000 ends with odd number of zeros. (viii) 505050 ends with odd number of zeros. 3. Squares which would be odd. (i) 431 is odd, square is odd. (ii) 2826 is even, square even. (iii) 7779 is odd, square odd. (iv) 82004 is even, square even. So 431 and 7779 have odd squares. 4. Pattern: 100001² = 10000200001. 10000001² = 100000020000001. 5. Pattern: 1010101² = 1020304030201. 101010101² = 10203040504030201. 6. Pattern: 4² + 5² + 20² = 21². 5² + 6² + 30² = 31². 6² + 7² + 42² = 43². The third number is the product of the first and second numbers. The fourth number is the third number plus 1. 7. Without adding, find sum. (i) 1 + 3 + 5 + 7 + 9 = 5² = 25. (ii) 1 to 19 has 10 terms, so 10² = 100. (iii) 1 to 23 has 12 terms, so 12² = 144. 8. (i) Express 49 as sum of 7 odd numbers: 1 + 3 + 5 + 7 + 9 + 11 + 13. (ii) Express 121 as sum of 11 odd numbers: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. 9. Numbers between squares. (i) 12 and 13: 2 × 12 = 24. (ii) 25 and 26: 2 × 25 = 50. (iii) 99 and 100: 2 × 99 = 198.

Section 5.4 Finding the Square of a Number. To find 23² without multiplying 23 × 23: 23 = 20 + 3. So 23² = (20 + 3)² = 20(20 + 3) + 3(20 + 3) = 20² + 20 × 3 + 3 × 20 + 3² = 400 + 60 + 60 + 9 = 529. Example 1: (i) 39² = (30 + 9)² = 30² + 2 × 30 × 9 + 9² = 900 + 540 + 81 = 1521. (ii) 42² = (40 + 2)² = 40² + 2 × 40 × 2 + 2² = 1600 + 160 + 4 = 1764.

Section 5.4.1 Other patterns in squares. 25² = 625 = (2 × 3) hundreds + 25. 35² = 1225 = (3 × 4) hundreds + 25. 75² = 5625 = (7 × 8) hundreds + 25. 125² = 15625 = (12 × 13) hundreds + 25. For a number ending in 5, say a5, (a5)² = a(a + 1) hundred + 25. Try These. Find squares: (i) 15² (ii) 95² (iii) 105² (iv) 205².

Section 5.4.2 Pythagorean triplets. 3² + 4² = 9 + 16 = 25 = 5². Triplet: 3, 4, 5. 6, 8, 10 is also one. 5² + 12² = 169 = 13². Triplet: 5, 12, 13. For any natural number m > 1, 2m, m² - 1, and m² + 1 form a Pythagorean triplet. Example 2: Smallest member 8. Try m² - 1 = 8, so m = 3. Triplet: 6, 8, 10. 8 is not smallest. Try 2m = 8, so m = 4. m² - 1 = 15, m² + 1 = 17. Triplet: 8, 15, 17. Example 3: One member 12. Try m² - 1 = 12, m² = 13, not integer. Try m² + 1 = 12, m² = 11, not integer. Try 2m = 12, m = 6. m² - 1 = 35, m² + 1 = 37. Triplet: 12, 35, 37. Note: All triplets may not come from this form. 5, 12, 13 is another.

EXERCISE 5.2 1. Find square. (i) 32² = (30 + 2)² = 900 + 120 + 4 = 1024. (ii) 35² = 3 × 4 hundred + 25 = 1225. (iii) 86² = (80 + 6)² = 6400 + 960 + 36 = 7396. (iv) 93² = (90 + 3)² = 8100 + 540 + 9 = 8649. (v) 71² = (70 + 1)² = 4900 + 140 + 1 = 5041. (vi) 46² = (40 + 6)² = 1600 + 480 + 36 = 2116. 2. Write triplet with one member. (i) 6: 2m = 6, m = 3. 3² - 1 = 8, 3² + 1 = 10. Triplet 6, 8, 10. (ii) 14: 2m = 14, m = 7. 7² - 1 = 48, 7² + 1 = 50. Triplet 14, 48, 50. (iii) 16: 2m = 16, m = 8. 8² - 1 = 63, 8² + 1 = 65. Triplet 16, 63, 65. (iv) 18: 2m = 18, m = 9. 9² - 1 = 80, 9² + 1 = 82. Triplet 18, 80, 82.

Section 5.5 Square Roots. Situation a: Area of square is 144 cm². side² = 144. We need number whose square is 144. Situation b: Figure 5.1 shows a square ABCD with side 8 cm and diagonal AC. Using Pythagoras theorem, AB² + BC² = AC². So 8² + 8² = AC². 64 + 64 = 128 = AC². We need number whose square is 128. Situation c: Figure 5.2 shows a right triangle with hypotenuse 5 cm, one side 3 cm, and base x cm. 5² = x² + 3². 25 - 9 = x². 16 = x². We need number whose square is 16. Finding number with known square is finding square root.

Section 5.5.1 Finding square roots. Inverse of squaring is square root. 1² = 1, so √1 = 1. 2² = 4, √4 = 2. 3² = 9, √9 = 3. 9² = 81, and (-9)² = 81. So square roots of 81 are 9 and -9. Try These. (i) 11² = 121. What is √121? (ii) 14² = 196. What is √196? Think, Discuss and Write. (-1)² = 1. Is -1 a square root of 1? Yes. (-2)² = 4. Is -2 a square root of 4? Yes. (-9)² = 81. Is -9 a square root of 81? Yes. There are two integral square roots of a perfect square number. In this chapter, we take only the positive square root. Positive square root is denoted by √. For example, √4 = 2, √9 = 3. Table: 1² = 1, √1 = 1. 2² = 4, √4 = 2. 3² = 9, √9 = 3. 4² = 16, √16 = 4. 5² = 25, √25 = 5. 6² = 36, √36 = 6. 7² = 49, √49 = 7. 8² = 64, √64 = 8. 9² = 81, √81 = 9. 10² = 100, √100 = 10.

Section 5.5.2 Finding square root through repeated subtraction. Sum of first n odd natural numbers is n². Consider √81. 81 - 1 = 80. 80 - 3 = 77. 77 - 5 = 72. 72 - 7 = 65. 65 - 9 = 56. 56 - 11 = 45. 45 - 13 = 32. 32 - 15 = 17. 17 - 17 = 0. Obtained 0 at 9th step. Therefore √81 = 9. Try These. By repeated subtraction, find if perfect square and find root: (i) 121 (ii) 55 (iii) 36 (iv) 49 (v) 90.

Section 5.5.3 Finding square root through prime factorisation. 6 = 2 × 3, 36 = 2² × 3². 8 = 2³, 64 = 2⁶. 12 = 2² × 3, 144 = 2⁴ × 3². 15 = 3 × 5, 225 = 3² × 5². Each prime factor in the square occurs twice the number of times it occurs in the number. Find √324. 324 = 2² × 3⁴ = (2 × 3²)². So √324 = 2 × 3 × 3 = 18. For 256: 256 = 2⁸ = (2⁴)². √256 = 2⁴ = 16. Is 48 a perfect square? 48 = 2⁴ × 3. 3 is unpaired. Multiply by 3: 48 × 3 = 144 = 12². Divide by 3: 48 ÷ 3 = 16 = 4². Example 4: √6400. 6400 = 2⁸ × 5². √6400 = 2⁴ × 5 = 80. Example 5: Is 90 a perfect square? 90 = 2 × 3² × 5. 2 and 5 unpaired. Not a perfect square. Example 6: 2352 = 2⁴ × 3 × 7². 3 unpaired. Multiply by 3: 2352 × 3 = 7056. √7056 = 2² × 3 × 7 = 84. Example 7: 9408 = 2⁶ × 3 × 7². Divide by 3: 9408 ÷ 3 = 3136. √3136 = 2³ × 7 = 56. Example 8: Smallest square divisible by 6, 9, 15. LCM = 90 = 2 × 3² × 5. Multiply by 2 × 5 = 10. 90 × 10 = 900.

EXERCISE 5.3 1. Possible unit digits of square root: (i) 9801 ends in 1, so 1 or 9. (ii) 99856 ends in 6, so 4 or 6. (iii) 998001 ends in 1, so 1 or 9. (iv) 657666025 ends in 5, so 5. 2. Not perfect squares: (i) 153 ends in 3. (ii) 257 ends in 7. (iii) 408 ends in 8. (iv) 441 ends in 1, could be. 3. √100 by repeated subtraction: 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. 10 steps. √100 = 10. √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. 13 steps. √169 = 13. 4. Prime Factorisation: (i) √729 = √3⁶ = 3³ = 27. (ii) √400 = √2⁴ × 5² = 2² × 5 = 20. (iii) √1764 = √2² × 3² × 7² = 2 × 3 × 7 = 42. (iv) √4096 = √2¹² = 2⁶ = 64. (v) √7744 = √2⁶ × 11² = 2³ × 11 = 88. (vi) √9604 = √2² × 7⁴ = 2 × 7² = 98. (vii) √5929 = √7² × 11² = 7 × 11 = 77. (viii) √9216 = √2¹⁰ × 3² = 2⁵ × 3 = 32 × 3 = 96. (ix) √529 = √23² = 23. (x) √8100 = √2² × 3⁴ × 5² = 2 × 3² × 5 = 90. 5. Multiply to get perfect square: (i) 252 = 2² × 3² × 7. Multiply by 7. 1764. √1764 = 42. (ii) 180 = 2² × 3² × 5. Multiply by 5. 900. √900 = 30. (iii) 1008 = 2⁴ × 3² × 7. Multiply by 7. 7056. √7056 = 84. (iv) 2028 = 2² × 3 × 13². Multiply by 3. 6084. √6084 = 78. (v) 1458 = 2 × 3⁶. Multiply by 2. 2916. √2916 = 54. (vi) 768 = 2⁸ × 3. Multiply by 3. 2304. √2304 = 48. 6. Divide to get perfect square: (i) 252 = 2² × 3² × 7. Divide by 7. 36. √36 = 6. (ii) 2925 = 3² × 5² × 13. Divide by 13. 225. √225 = 15. (iii) 396 = 2² × 3² × 11. Divide by 11. 36. √36 = 6. (iv) 2645 = 5 × 23². Divide by 5. 529. √529 = 23. (v) 2800 = 2⁴ × 5² × 7. Divide by 7. 400. √400 = 20. (vi) 1620 = 2² × 3⁴ × 5. Divide by 5. 324. √324 = 18. 7. Donation ₹ 2401. x² = 2401. 2401 = 7⁴. √2401 = 7² = 49. 49 students. 8. 2025 plants. x² = 2025. 2025 = 3⁴ × 5². √2025 = 3² × 5 = 45. 45 rows, 45 plants. 9. Divisible by 4, 9, 10. LCM = 180 = 2² × 3² × 5. Multiply by 5. 900. 10. Divisible by 8, 15, 20. LCM = 120 = 2³ × 3 × 5. Multiply by 2 × 3 × 5 = 30. 120 × 30 = 3600.

Section 5.5.4 Division method. Table: 10² = 100 (smallest 3-digit), 31² = 961 (greatest 3-digit), 32² = 1024 (smallest 4-digit), 99² = 9801 (greatest 4-digit). If perfect square has n digits, root has n/2 digits if n even, or (n+1)/2 if n odd. Steps for √529: Place bars: 5̄ 29̄. Largest square ≤ 5 is 2² = 4. Remainder 1. Bring down 29 → 129. Double quotient 2 → 4. Guess digit 3. 43 × 3 = 129. Remainder 0. √529 = 23. Steps for √4096: Place bars: 40̄ 96̄. Largest square ≤ 40 is 6² = 36. Remainder 4. Bring down 96 → 496. Double quotient 6 → 12. Guess digit 4. 124 × 4 = 496. Remainder 0. √4096 = 64. Estimate digits for 14400: 1̄ 44̄ 00̄. 3 bars → 3 digits. Try These: (i) 25600 (5 digits) → 3 digits. (ii) 100000000 (9 digits) → 5 digits. (iii) 36864 (5 digits) → 3 digits. Example 9: (i) √729 = 27. (ii) √1296 = 36. Example 10: √5607. Quotient 74, remainder 131. 5607 - 131 = 5476. √5476 = 74. Example 11: √9999. Quotient 99, remainder 198. 9999 - 198 = 9801. √9801 = 99. Example 12: √1300. Quotient 36, remainder 4. Next square 37² = 1369. Add 1369 - 1300 = 69.

Section 5.6 Square Roots of Decimals. √17.64. Bars: 17̄ . 64̄. 4² = 16. Remainder 1. Bring down 64 → 164. Double 4 → 8. Decimal in quotient. 82 × 2 = 164. √17.64 = 4.2. Example 13: √12.25 = 3.5. Which way to move: 176.341. Bars on integral: 1̄ 76̄. Bars on decimal: 34̄ 10̄. Example 14: Area 2304 m². Side = √2304 = 48 m. Example 15: 2401 students. x² = 2401. x = √2401 = 49. 49 rows.

EXERCISE 5.4 1. Division method: (i) √2304 = 48. (ii) √4489 = 67. (iii) √3481 = 59. (iv) √529 = 23. (v) √3249 = 57. (vi) √1369 = 37. (vii) √5776 = 76. (viii) √7921 = 89. (ix) √576 = 24. (x) √1024 = 32. (xi) √3136 = 56. (xii) √900 = 30. 2. Digits in root: (i) 64 (2 digits) → 1. (ii) 144 (3 digits) → 2. (iii) 4489 (4 digits) → 2. (iv) 27225 (5 digits) → 3. (v) 390625 (6 digits) → 3. 3. Decimals: (i) √2.56 = 1.6. (ii) √7.29 = 2.7. (iii) √51.84 = 7.2. (iv) √42.25 = 6.5. (v) √31.36 = 5.6. 4. Subtract: (i) 402 → subtract 2, √400 = 20. (ii) 1989 → subtract 53, √1936 = 44. (iii) 3250 → subtract 1, √3249 = 57. (iv) 825 → subtract 41, √784 = 28. (v) 4000 → subtract 31, √3969 = 63. 5. Add: (i) 525 → add 4, √529 = 23. (ii) 1750 → add 14, √1764 = 42. (iii) 252 → add 4, √256 = 16. (iv) 1825 → add 24, √1849 = 43. (v) 6412 → add 149, √6561 = 81. 6. Area 441 m². Side = √441 = 21 m. 7. Right triangle ABC, ∠B = 90°. (a) AB = 6, BC = 8. AC² = 6² + 8² = 100. AC = 10 cm. (b) AC = 13, BC = 5. AB² = 13² - 5² = 144. AB = 12 cm. 8. 1000 plants. 31² = 961, 32² = 1024. Need 1024 - 1000 = 24 more. 9. 500 children. 22² = 484. Left out: 500 - 484 = 16.

WHAT HAVE WE DISCUSSED? 1. If a natural number m can be expressed as n², where n is also a natural number, then m is a square number. 2. All square numbers end with 0, 1, 4, 5, 6 or 9 at units place. 3. Square numbers can only have even number of zeros at the end. 4. Square root is the inverse operation of square. 5. There are two integral square roots of a perfect square number. Positive square root of a number is denoted by the symbol √. For example, 3² = 9 gives √9 = 3.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]