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3 Squares and Square Roots introduction In earlier classes you have learnt that area of a square = side × side i.e. area of a square = (side)2 (i.e. square of side). If the side of a square is given x units, then Area of square = (x × x) sq. units = x2 sq. units So, square of a number is the product of the number by itself. For example: 52 = 5 × 5, 72 = 7 × 7 In this chapter, we will study about square numbers or perfect squares, properties and patterns of square numbers, Pythagorean triplets, square roots and various methods to find square roots of natural numbers, square roots of decimals and fractions. Square nuMberS or Perfect SquareS Consider the following table: Table 1 Natural number Square 2 1 1 =1×1=1 2 22 = 2 × 2 = 4 3 32 = 3 × 3 = 9 4 42 = 4 × 4 = 16 5 52 = 5 × 5 = 25 6 62 = 6 × 6 = 36 7 72 = 7 × 7 = 49 8 82 = 8 × 8 = 64 9 92 = 9 × 9 = 81 10 102 = 10 × 10 = 100 You can see that 1, 4, 9, 16, 25, ….. are the natural numbers which are the squares of natural numbers. Such numbers are called square numbers or perfect squares. Squares of natural numbers are called square numbers or perfect squares. 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 or perfect square. Squares and Square Roots 43 From table 1, you can write all the square numbers between 1 and 100. You will find that rest of the natural numbers are not perfect squares. Hence, we can say that All the natural numbers are not perfect squares. How do we know, whether a given natural number is a perfect square? Observe the following: (i)Let us consider a square number 16. It can be expressed as 16 = 2 # 2 × 2 # 2 S Prime factorization 2 2 2 2 S Observation: 16 can be expressed as the product of pairs of equal prime factors. (ii)Let us consider another square number 144. It can be expressed as 144 = 2 # 2 × 2 # 2 × 3 # 3 S S Prime factorization 2 144 2 72 2 36 2 18 3 9 3 3 1 S Observation: 144 can also be expressed as the product of pairs of equal prime factors. (iii)Now consider a non-square number 48. It can be expressed as 48 = 2 # 2 × 2 # 2 × 3 S S Observation: 48 cannot be expressed as the product of pairs of equal prime factors. (3 cannot be paired) 16 8 4 2 1 Prime factorization 2 2 2 2 3 Thus, we conclude that 48 24 12 6 3 1 A perfect square can always be expressed as the product of pairs of equal prime factors. Example 1. Is 324 a perfect square? Solution. Given number is 324. It can be expressed as 324 = 2 # 2 × 3 # 3 × 3 # 3 S S S Since 324 can be expressed as the product of pairs of equal prime factors. Hence, 324 is a perfect square. Prime factorization 2 324 2 162 3 81 3 27 3 9 3 3 1 Learning Mathematics–VIII 44 Example 2. Is 1152 a perfect square? Prime factorization Solution. Given number is 1152. 2 1152 2 576 2 288 2 144 2 72 2 36 2 18 3 9 3 3 1 It can be expressed as 1152 = 2 × 2 # 2 × 2 # 2 × 2 # 2 × 3 # 3 S S S S Since one 2 is left unpaired, ∴ 1152 cannot be expressed as the product of pairs of equal prime factors. Hence, 1152 is not a perfect square. Example 3. Show that 676 is a perfect square. Find the number whose square is 676. Solution. Given number is 676. It can be expressed as 676 = 2 # 2 × 13 # 13 S S Since 676 can be expressed as the product of pairs of equal prime factors. Hence, 676 is a perfect square. Also 676 = (2)2 × (13)2 = (2 × 13)2 = (26)2 Hence, 26 is the number whose square is 676. Prime factorization 2 676 2 338 13 169 13 13 1 Example 4. Find the smallest natural number by which 720 should be multiplied to make it a perfect square. Prime factorization Solution. Given number is 720. It can be expressed as S S S 720 = 2 # 2 × 2 # 2 × 3 # 3 × 5 Since 5 is left unpaired, so to make 720 a perfect square 5 should be paired. So, the given number should be multiplied by 5. Hence, the smallest natural number by which 720 be multiplied to make it a perfect square is 5. 2 720 2 360 2 180 2 90 3 45 3 15 5 5 1 Example 5. Find the smallest natural number by which 2527 should be divided to make it a perfect square. Prime factorization Solution. Given number is 2527. It can be expressed as S 2527 = 7 × 19 # 19 7 2527 19 361 19 19 1 Since 7 is left unpaired, so to make 2527 a perfect square it should be divided by 7. Hence, the smallest natural number by which 2527 be divided to make it a perfect square is 7. Squares and Square Roots 45 Exercise 3.1 1. Which of the following natural numbers are perfect squares? Give reasons in support of your answer. (i) 729 (ii) 5488 (iii) 1024 (iv) 243 2. Show that each of the following numbers is a perfect square. Also find the number whose square is the given number. (i) 1296 (ii) 2304 (iii) 3025 3. Find the smallest natural number by which 1008 should be multiplied to make it a perfect square. 4. Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number. 5. Find the area of a square lawn, whose one side is 16 m long. Properties of square numbers Property 1 Consider the following table: Table 2 Natural number 1 2 3 4 5 6 7 8 9 10 Square 1 4 9 16 25 36 49 64 81 100 Natural number 11 12 13 14 15 16 17 18 19 20 Square 121 144 169 196 225 256 289 324 361 400 Natural number 21 22 23 24 25 26 27 28 29 30 Square 441 484 529 576 625 676 729 784 841 900 From the above table, we can observe that unit digits (i.e. digit in one’s place) of square numbers are 0, 1, 4, 5, 6 or 9. 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? Answer is no, since the numbers 30, 21, 34, 45, 76 or 89 are not square numbers. Hence, we can conclude that: A square number always ends with 0, 1, 4, 5, 6 or 9 at unit’s place but converse is not true. Property. A number having 2, 3, 7 or 8 at its unit place is never a square number. Example 1. Can we say whether the following numbers are perfect squares? Give reasons. (i)1057 (ii) 23453 (iii) 7928 (iv) 22222 Solution. (i)1057 is not a perfect square, since it ends with 7 at unit’s place. (ii)23453 is not a perfect square, since it ends with 3 at unit’s place.