Squares and Square Roots Class VIII PDF

# A Textbook of Mathematics for Class VIII ## Squares and Square Roots ### 6.1 Introduction You know that the area of a square = side × side (where 'side' means the length of a side'). Study the following table. | Side of a square (in cm) | Area of the square (in cm²) | |---|---| | 1 | 1 × 1 = 1...

# A Textbook of Mathematics for Class VIII ## Squares and Square Roots ### 6.1 Introduction You know that the area of a square = side × side (where 'side' means the length of a side'). Study the following table. | Side of a square (in cm) | Area of the square (in cm²) | |---|---| | 1 | 1 × 1 = 1 = 1² | | 2 | 2 × 2 = 4 = 2² | | 3 | 3 × 3 = 9 = 3² | | 5 | 5 × 5 = 25 = 5² | | 8 | 8 × 8 = 64 = 8² | | a | a × a = a² | What is special about the numbers 4, 9, 25, 64 and other such numbers? Since, 4 can be expressed as 2 × 2 = 2², 9 can be expressed as 3 × 3 = 3², all such numbers can be expressed as the product of the number with itself. Such numbers like 1, 4, 9, 16, 25,... are known as **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 that 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 following numbers and their squares. | Number | Square | |---|---| | 1 | 1 × 1 = 1 | | 2 | 2 × 2 = 4 | | 3 | 3 × 3 = 9 | | 4 | 4 × 4 = 16 | | 5 | 5 × 5 = 25 | | 6 | | | 7 | | | 8 | | | 9 | | | 10 | | Can you complete it? From the above table, can we enlist the square numbers between 1 and 100? Are there any natural square numbers upto 100 left out? You will find that the rest of the numbers are not square numbers. The numbers 1, 4, 9, 16 ...... are square numbers. These numbers are also called **perfect squares**. ### Try These 1. Find the perfect square numbers between (i) 30 and 40 (ii) 50 and 60 ### 6.2 Properties of Square Numbers Following table shows the squares of numbers from 1 to 20. | Number | Square | Number | Square | |---|---|---|---| | 1 | 1 | 11 | 121 | | 2 | 4 | 12 | 144 | | 3 | 9 | 13 | 169 | | 4 | 16 | 14 | 196 | | 5 | 25 | 15 | 225 | | 6 | 36 | 16 | 256 | | 7 | 49 | 17 | 289 | | 8 | 64 | 18 | 324 | | 9 | 81 | 19 | 361 | | 10 | 100 | 20 | 400 | Study the square numbers in the above table. What are the ending digits (that is, digits in the one's place) of the square numbers? All these numbers end with 0, 1, 4, 5, 6 or 9 at units place. None of these end with 2, 3, 7 or 8 at unit's place. Can we say that if a number ends in 0, 1, 4, 5, 6 or 9, then it must be a square number? Think about it. ### Try These Can we say whether the following numbers are perfect squares? How do we know? 1. (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 1069 (vi) 2061 ## A Textbook of Mathematics for Class VIII 2. Write five numbers which you can decide by looking at their one's digit that they are not square numbers. Write five numbers which you cannot decide just by looking at their unit's digit (or one's place) whether they are square numbers or not. Study the following table of some numbers and their squares and observe the one's place in both. ### Table 1 | Number | Square | Number | Square | Number | Square | |---|---|---|---|---|---| | 1 | 1 | 11 | 121 | 21 | 441 | | 2 | 4 | 12 | 144 | 22 | 484 | | 3 | 9 | 13 | 169 | 23 | 529 | | 4 | 16 | 14 | 196 | 24 | 576 | | 5 | 25 | 15 | 225 | 25 | 625 | | 6 | 36 | 16 | 256 | 30 | 900 | | 7 | 49 | 17 | 289 | 35 | 1225 | | 8 | 64 | 18 | 324 | 40 | 1600 | | 9 | 81 | 19 | 361 | 45 | 2025 | | 10 | 100 | 20 | 400 | 50 | 2500 | The following square numbers end with digit 1. | Square | Numbers | |---|---| | 1 | 1 | | 81 | 9 | | 121 | 11 | | 161 | 19 | | 441 | 21 | ### Try These Which of 123², 77², 82², 161², 109² would end with digit 1? Write the next two square numbers which end in 1 and their corresponding numbers. You will see that if a number has 1 or 9 in the unit's place, then it's square ends in 1. Let us consider square numbers ending in 6. | Square | Numbers | |---|---| | 16 | 4 | | 36 | 6 | | 196 | 14 | | 256 | 16 | We can see that when a square number ends in 6, the number whose square it is, will have either 4 or 6 in unit's place. Can you find more such rules by observing the numbers and their squares (Table 1)? ### Try These What will be the "one's digit" in the square of the following numbers? (i) 1234 (ii) 26387 (iii) 52698 (iv) 99880 (v) 21222 (vi) 9106 Consider the following numbers and their squares. We have one zero - 10² = 100 - 20² = 400 - 80² = 6400 But we have two zeros We have two zeros - 100² = 10000 - 200² = 40000 - 700² = 490000 - 900² = 810000 But we have four zeros If a number contains 3 zeros at the end, how many zeros will its square have? What do you notice about the number of zeros at the end of the number and the number of zeros at the end of its square? Can we say that square numbers can only have even number of zeros at the end? See Table 1 with numbers and their squares. What can you say about the squares of even numbers and squares of odd numbers? ### Try These 1. The square of which the following numbers would be an odd number/an even number? Why? - (i) 727 - (ii) 158 - (iii) 269 - (iv) 1980 2. What will be the number of zeros in the square of the following numbers? - (i) 60 - (ii) 400 ## A Textbook of Mathematics for Class VIII ### 6.3 Some More Interesting Patterns #### Adding triangular numbers Do you remember triangular numbers (numbers whose dot patterns can be arranged as triangles)? * 1 * ** * *** * **** * ***** If we combine two consecutive triangular numbers, we get a square number like * ** * ** * ** * ** * ** * *** * *** * *** 1 + 3 = 4 = 2² * *** * *** * *** * *** * *** * **** * **** * **** * **** 3 + 6= 9 = 3² * **** * **** * **** * **** * **** * ***** * ***** * ***** * ***** * ***** 6 + 10= 16 = 4² #### Numbers between square numbers Let us now see if we can find some interesting pattern between two consecutive square numbers. 6 non square numbers between the two square numbers 9 (=3²) and 16 (= 4²) 8 non square numbers between the two square numbers 16 (= 4²) and 25 (= 5²) | 1(=1²) | 2,3,4(-2²) | 5,6,7,8,9 (3²) | 10, 11, 12, 13, 14, 15, 16 (=4²) | 17, 18, 19,20, 21, 22, 23, 24, 25 (=5²) | |---|---|---|---|---| | Two non square numbers between the two square numbers | 1(=1²) and 4 (= 2²) | 4 non square numbers between the two square numbers | 4 (= 2²)and9 (= 3²) | | Between 1² (= 1) and 2² (= 4) there are two (2 × 1) non square numbers 2, 3. Between 2² (= 4) and 3² (= 9) there are four (i.e., 2 × 2) non square numbers 5, 6, 7, 8. Now, 3² = 9, 4² = 16 Therefore, 4² - 3² = 16 - 9 = 7 Between 9(=3²) and 16(=4²) the numbers are 10, 11, 12, 13, 14, 15 that is six non square numbers which is less than the difference of two squares. We have 4² = 16 and 5² = 25 Therefore, 5² - 4² = 9 Between 16 (= 4²) and 25 (= 5²) the numbers are 17, 18, ...... 24 that is, eight non square numbers which is 1 less than the difference of two squares. Consider 7² and 6². Can you say how many numbers are there between 7² and 6²? If we think of any natural n and (n + 1), then, (n + 1)² - n² = (n² + 2n+1)-n² = 2n+1 We find that between n²and (n + 1)² there are 2n numbers which is 1 less than the difference of two squares. Thus, in general we can say that there are 2n non perfect numbers between the squares of the numbers n and (n + 1). Check for n = 5, n = 6 etc., and verify. ## A Textbook of Mathematics for Class VIII ### Try These 1. How many natural numbers lie between 9² and 10² ? Between 11² and 12²? 2. How many non square numbers lie between the following pairs of numbers (i) 100² and 101² (ii) 90² and 91² (iii) 1000² and 1001² 3. Adding odd numbers Consider the following - 1 [one odd number] - 1 + 3 [sum of first two odd numbers] - 1 +3 + 5 [sum of first three odd numbers] - 1 +3 + 5 + 7 [...] - 1 +3+5+7+9 [...] - 1 +3+5+7+ 9 +11 [...] - = 1 = 1² - = 4 = 2² - = 9 = 3² - = 16 = 4² - = 25 = 5² - = 36=6² So we can say that the sum of first n odd natural numbers is n². Looking at it in a different way, we can say: 'If the number is a square number, it has to be the sum of successive odd numbers starting from 1. Consider those numbers which are not perfect squares, say 2,3,5,6 Can you express these numbers as a sum of successive odd natural numbers beginning from 1? You will find that these numbers cannot be expressed in this form. Consider the number 25. Successively subtract 1, 3, 5, 7, 9, ... from it - (i) 25-1 = 24 - (ii) 24-3 = 21 - (iii) 21-5 = 16 - (iv) 16-7=9 - (v) 9-9=0 This means, 25 = 1+3+5+7 + 9. Also, 25 is a perfect square. Now consider another number 38, and again do as above. - (i) 38-1 = 37 - (ii) 37-3=34 - (iii) 34-5=29 - (iv) 29-7=22 - (v) 22-9 = 13 - (vi) 13-11 = 2 - (vii) 2-13=-11 This shows that we are not able to express 38 as the sum of consecutive odd numbers starting with 1. Also, 38 is not a perfect square. So we can also say that if a natural number cannot be expressed as a sum of successive odd natural numbers starting with 1, then it is not a perfect square. We can use this result to find whether a number is a perfect square or not. ### 6.4 A sum of consecutive natural numbers Consider the following | First number | Second number | |---|---| | 3² - 1 | 3² + 1 | | | 2 | - 3² = 9 = 4 + 5 - 5² = 25 = 12 + 13 - 7² = 49 = 24 + 25 - 9² = 81 = 40 + 41 - 11² = 121 = 60 + 61 - 15² = 225 = 112 + 113 ### Try These 1. Express the following as the sum of two consecutive integers. - (i) 21² - (ii) 13² - (iii) 11² - (iv) 19² 2. Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is perfect square of a number? Give example to support your answer. ### 6.5 Product of two consecutive even or odd natural numbers 11 x 13 = 143 = 122-1 Also, 11 × 13 = (12-1) × (12 + 1) Therefore, 11 × 13 = (12-1) × (12 + 1) = 12² - 1 Similarly, 13 × 15 = (14 – 1) × (14 + 1) = 14² – 1 29 × 31 = (30-1) × (30 + 1) = 30² - 1 44 × 46 = (45-1) × (45 + 1) = 45² - 1 So in general we can say that (a + 1) × (a-1) = a²-1 ### 6.6 Some more patterns in square numbers Observe the squares of numbers; 1, 11, 111 ... etc. they give a beautiful pattern: - 1² = 1 - 11² = 121 - 111² = 12321 - 1111² = 1234321 - 11111² = 123454321 - 111111² = 12345654321 - 1111111² = 1234567654321 Another interesting pattern. - 7² = 49 - 67² = 4489 - 667² = 444889 - 6667² = 44448889 - 66667² = 4444488889 - 666667² = 444444888889 ### Try These Write the square, making use of the above pattern. - (i) 111111² - (ii) 1111111² ### Try These Write the square, making use of the above pattern. - (i) 6666667² - (ii) 66666667² ## A Textbook of Mathematics for Class VIII ### 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 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 3. The squares of which of the following would be odd numbers? - (i) 431 - (ii) 2826 - (iii) 7779 - (iv) 82004 4. Observe the following pattern and find the missing digits. - 11² = 121 - 101² = 10201 - 1001²= 1002001 - 100001² = 1......2......1 - 10000001² = ....... 5. Observe the following pattern and supply the missing numbers. - 11² = 121 - 101² = 10201 - 10101² = 102030201 - 1010101² = 10203040504030201 6. Using the given pattern, find the missing numbers - 1² + 2² + 2² = 3² - 2² + - 3² + 4² + 12² = 13² - 4² + 5² - 5² + 6² + 30² = 31² - 6² + 7² + 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 8. (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers. 9. How many numbers lie between squares of the following numbers? - (i) 12 and 13 - (ii) 25 and 26 - (iii) 99 and 100 ## A Textbook of Mathematics for Class VIII ### 6.4 Finding the Square of a Number Squares of small numbers like 3, 4, 5, 6, 7, etc. are easy to find. But can we find the squares of 23 so quickly? The answer is not so easy and we may need to multiply 23 by 23. There is a way to find this without having to multiply 23 x 23. We know 23 = 20 + 3 Therefore, 23² = (20+3)² = 20(20+3) + 3(20+3) = 20² +20 ×3+3×20+3² = 400+60+60+9= 529 **Example 1**: Find the square of the following numbers without actual multiplication. - (i) 39 - (ii) 42 **Solution**: - (i) 39² = (30+9)² = 30(30+9) + 9(30+9) = 30² +30×9 + 9×30+9² = 900+270+270+81 = 1521 - (ii) 42² = (40+2)² = 40(40+2) + 2(40 + 2) = 40² +40 × 2+2×40+2² = 1600+80+80+4= 1764 ### 6.4.1 Other patterns in squares Consider the following pattern: - 25² = 625 = (2 x 3) hundreds + 25 - 35² = 1225 = (3 × 4) hundreds + 25 - 75² = 5625 = (7 × 8) hundreds + 25. - 125² = 15625 = (12×13) hundreds + 25 Now can you find the square of 95? ### Try These Find the squares of the following numbers containing 5 in unit's place. - (i) 15 - (ii) 95 - (iii) 105 - (iv) 205 Consider a number with unit digit 5, i.e., a5 (a5)² = (10a + 5)² = 10a (10a +5) + 5(10a + 5) = 100a² + 50a + 50a +25 = 100a (a + 1) + 25 = a (a + 1) hundred + 25 ### 6.4.2 Pythagorean triplets Consider the following - 3² + 4² = 9 + 16 = 25 = 5² This collection of numbers 3, 4, and 5 is known as **Pythagorean triplet**. 6, 8, 10 is also a Pythagorean triplet, since - 6² + 8² = 36 + 64 = 100 = 10² Again, observe that - 5² + 12² = 25 + 144 = 169 = 13². The numbers 5, 12, 13 form another triplet. Can you find more such triplets? For any natural number m > 1, we have (2m)² + (m² - 1)² = (m² + 1)². So, 2m, m² - 1, m² + 1 forms a Pythagorean triplet. Try to find some more Pythagorean triplets using this form. **Example 2**: Write a Pythagorean triplet whose smallest member is 8. **Solution**: We can get Pythagorean triplets by using general form 2m, m² - 1, m² + 1. Let us first take m²-1=8 which gives m²=8+1=9 Therefore, m = 3 2m = 6 and m²+1 = 9 + 1 = 10 The triplet is thus 6, 8, 10. But 8 is not the smallest member of this. So, let us try 2m = 8 m = 4 m² + 1 = 16 + 1 = 17 m² - 1 = 16 - 1 = 15 The triplet is 8, 15, 17 with 8 as the smallest member. **Example 3**: Find a Pythagorean triplet in which one member is 12. **Solution**: If we take m² - 1 = 12 Then, m² = 12 + 1 = 13 So the value of m² will not be an integer. Then we try to take m² + 1 = 12. Again m² = 11 will not give an integer value for m. So, let us take 2m = 12 then m=6 Thus, m²-1 = 36-1= 35 and m²+1 = 36+1 = 37 Therefore, the required triplet is 12, 35, 37. Note: All Pythagorean triplets may not be obtained using this form. For example another triplet 5, 12, 13 also has 12 as a member. ## A Textbook of Mathematics for Class VIII ### Exercise 6.2 1. Find the Square of the following numbers. - (i) 32 - (ii) 35 - (iii) 14 - (iv) 93 - (v) 71 - (vi) 46 - (iii) 86 2. Write a Pythagorean triplet whose one member is. - (i) 6 - (ii) 16 (iii) 18 ## A Textbook of Mathematics for Class VIII ### 6.5 Square roots Study the following situations (a) Area of a square is 144 cm². What could be the side of the square? We know that the area of a square = side². If we assume the length of the side to be 'a', then 144 = a². To find the length of side it is necessary to find a number whose square is 144. (b) What is the length of a diagonal of a square of side 8 cm (Fig 6.1)? Can we use Pythagoras theorem to solve this? We have, AB² + BC² = AC². i.e., 8² + 8² = AC² or 64 + 64 = AC² or 128 = AC² Again to get AC we need to think of a number whose square is 128. (c) In a right triangle the length of the hypotenuse and a side are respectively 5 cm and 3 cm (Fig 6.2). Can you find the third side? Let x cm be the length of the third side. Pythagoras theorem 52 = x² + 32 25 - 9 = x² 16 = x² Again, to find x we need a number whose square is 16. ### 6.5.1 Finding Square Roots The inverse (opposite) operation of addition is subtraction and the inverse operation of multiplication is division. Similarly, finding the square root is the inverse operation of squaring. We have, - 1² = 1, therefore square root of 1 is 1 - 2² = 4, therefore square root of 4 is 2 - 3² = 9, therefore square root of 9 is 3 Since - 9² = 81, - (9)² = 81 of 81 are 9 and - 9. We say that square roots of 81 are 9 and - 9. ### Try These - (i) 11² = 121. What is the square root of 121? - (ii) 14² = 196. What is the square root of 196? ## A Textbook of Mathematics for Class VIII ### Think, Discuss and Write (-1)²= 1. Is-1, a square root of 1? (-2)² = 4. Is -2, a square root of 4? (-9)² = 81. Is - 9 a square root of 81?. From the above, you may say that there are two integral square roots of a perfect square number. In this chapter, we shall take up only positive square root of a natural number. Positive square root of a number is denoted by the symbol√. For example: √4 = 2 (not-2); √9 = 3 (not-3) etc. | Statement | Inference | Statement | Inference | |---|---|---|---| | 1² = 1 | √1=1 | 6² = 36 | √36 = 6 | | 2² = 4 | √4=2 | 7² = 49 | √49 = 7 | | 3² = 9 | √9=3' | 8² = 64 | √64= 8 | | 4² = 16 | √16=4 | 9² = 81 | √81=9 | | 5² = 25 | √25=5 | 10² = 100 | √100=10 | ### 6.5.2 Finding square root through repeated subtraction Do you remember that the sum of the first n odd natural numbers is n2? That is, every square number can be expressed as a sum of successive odd natural numbers starting from 1. Consider√81. Then, - (i) 81-1 = 80 - (ii) 80-3=77 - (iii) 77-5= 72 - (vi) 72-7= 65 - (vii) 45-13 = 32 - (v) 65-9= 56 - (vi) 56-11 = 45 - (viii) 32-15 = 17 - (ix) 17-17=0 From 81 we have subtracted successive odd numbers starting from 1 and obtained 0 at 9th step. Therefore √81=9. Can you find the square root of 729 using this method? Yes, but it will be time consuming. Let us try to find it in a simpler way. ### Try These By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root. - (i) 121 - (ii) 55 - (iii) 36 - (iv) 49 - (v) 90 ### 6.5.3 Finding square root through prime factorisation Consider the prime factorisation of the following numbers and their squares. | Prime factorisation of a Number | Prime factorisation of its Square | |---|---| | 6=2×3 | 36 = 2×2×3×3 | | 8=2×2×2 | 64 = 2 ×2×2×2×2×2 | | 12 =2×2×3 | 144 = 2×2×2×2×3×3 | | 15=3×5 | 225 = 3×3×5×5 | How many times does 2 occur in the prime factorisation of 6? Once. How many times does 2 occur in the prime factorisation of 36? Twice. Similarly, observe the occurrence of 3 in 6 and 36 of 2 in 8 and 64 etc. You will find that each prime factor in the prime factorisation of the square of a number, occurs twice the number of times it occurs in the prime factorisation of the number itself. Let us use this to find the square root of a given square number, say 324. We know that the prime factorisation of 324 is 324=2×2×3×3×3×3 | 2 | 324 | | |---|---|---| | 2 | 162 | the | | 3 | 81 | say | | 3 | 27 | | | 3 | 9 | | | 3 | 3 | | By pairing the prime factors we get 324 = 2×2×3×3×3×3=22×32×32 = (2×3×3)²