Square & Square Roots

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Square & Square Roots 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, 1, 4, 5, 6 or 9 at unit s place.

1 Square & Square Roots 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, 1, 4, 5, 6 or 9 at unit s place. All numbers have numbers from to 9 at their unit place and the number at the unit place of square of that number behaves as per the number at unit place in the original number. Following is the illustration of this property: ²= is the number at unit s place 1²=1 1 is the number at unit s place 2²=4 4 is the number at unit s place 3²=9 9 is the number at unit s place 4²=16 6 is the number at unit s place 5²=25 5 is the number at unit s place 6²=36 6 is the number at unit s place 7²=49 9 is the number at unit s place 8²=64 4 is the number at unit s place 9²=81 1 is the number at unit s 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 Exercise 1 1. What will be the unit digit of the squares of the following numbers? (i) 81 As 1² ends up having 1 as the digit at unit s place so 81² will have 1 at unit s place. (ii) 272 Asnwer: 2²=4 So, 272² will have 4 at unit s place (iii) 799 9²=81 So, 799 will have 1 at unit s place (iv) ²=9 So, 3853² will have 9 at unit s place.

2 (v) ²=16 So, 1234² will have 6 at unit s place (vi) ²=49 So, 26387² will have 9 at unit s place (vii) ²=64 So, 52698² will have 4 at unit s place (viii) 9988 ²= So, 9988² will have at unit s place (ix) ²=36 So, 12796² will have 6 at unit s place (x) ²=25 So, 55555² will have 5 at unit s place 2. The following numbers are obviously not perfect squares. Give reason. (i) 157 (ii) (iii) 7928 (iv) (v) 64 (vi) (vii) 222 (viii) 555 (i), (ii), (iii), (iv), (vi) don t have any of the, 1, 4, 5, 6, and 9 at unit s place, so they are not perfect squares. (v), (vii) and (viii) don t have even number of zeroes at the end so they are not perfect squares. 3. The squares of which of the following would be odd numbers? (i) 431 (ii) 2826 (iii) 7779 (iv) 824 (i) and (iii) will have odd numbers as their square, because an odd number multiplied by another odd number always results in an odd number. 4. Observe the following pattern and find the missing digits. 11² = ² = 121

3 11² = ² = ² = 121 Start with 1 followed as many zeroes as there are between the first and the last one, followed by two again followed by as many zeroes and end with Observe the following pattern and supply the missing numbers. 11² = ² = ² = ² = ² = Start with 1 followed by a zero and go up to as many number as there are number of 1s given, follow the same pattern in reverse order. 6. Using the given pattern, find the missing numbers. 1² + 2² + 2² = 3² 2² + 3² + 6² = 7² 3² + 4² + 12² = 13² 4² + 5² + 2²= 21² 5² + 6²+ 3² = 31² 6² + 7² + 42² = 43² If the square of a number is added with square of its prime factors we get square of a number which is 1 more than the original number. 7. Without adding, find the sum. (i) = 2² = = 3² = = 4² = = 5² = 25 In other words this is a way of finding the sum of n odd numbers starting from 1. Sum of n odd numbers starting from 1 = n² (ii) I There are 1 odd numbers in the given equation So, sum =1²=1 (iii) Sum = 12²=144

4 8. (i) Express 49 as the sum of 7 odd numbers. As you know 49=7² So, 7² can be expressed as follows: (ii) Express 121 as the sum of 11 odd numbers. 11²= How many numbers lie between squares of the following numbers? (i) 12 and 13 12²=144 13²=169 Now, =25 So, there are 25-1=24 numbers lying between 12² and 13² (ii) 25 and 26 25²=625 26²=676 Now, =51 So, there are 51-1=5 numbers lying between 25² and 26² (iii) 99 and 1 99²=981 1²=1 Now, 1-981=199 So, there are 199-1=198 numbers lying between 99² and 1² Exercise 2 1. Find the square of the following numbers. (i) 32 32²=32 32= 124 But above method can be tough to calculate. It is easier to calculate such values by using algebraic identities. So, 32²=(3+2) ² Using (a+b) ² = a²+b²+2ab We get (3+2) ²= 3²+2² =9+4+12=124

5 (ii) 35 (35)²=(3+5) ² =3²+5² =9+25+3=1225 (iii) 86 86²=(8+6) ² =8²+6² = =7396 (iv) 93 Answer=93²=(9+3² =9²+3² = =8649 (v) 71 71²=(7+1) ² =7²+1² = =54 (vi) 46 46²=(4+6) ² =4²+6² = = Write a Pythagorean triplet whose one member is. (i) 6 As we know 2m, m²+1 and m²-1 form a Pythagorean triplet for any number, m>1. Let us assume 2m=6 Then m=3 m²+1=3²+1=1 m²-1=3²-1=8 Test: 6²+8²=36+64=1=1² Hence, the triplet is 6, 8, and 1

6 Note: Most of the Pythagorean triplets are in the ratio of 3:4:5 (ii) 14 Let us assume, 2m=14, then m=7 So, m²+1=7²+1=5 And, m²-1=7²-1=48 Test: 14²+48²= =25=5² Hence, the triplet is 14, 48, and 5 (iii) 16 Let us assume 2m=16, then m=8 So, m²+1=8²+1=65 And, m²-1=8²-1=63 Test: 16²+63²= =4225=65² Hence, the triplet is 16, 63, and 65 (iv) 18 Let us assume 2m=18, then m=9 So, m²+1=9²+1=82 m²-1=9²-1=8 test: 18²+8²=6724=82² Exercise 3 Finish Line & Beyond 1. What could be the possible one s digits of the square root of each of the following numbers? (i) 981 Since 1² and 9² give 1 at unit s place, so these are the possible values of unit digit of the square root. (ii) ²=16 and 6²=36, hence, 4 and 6 are possible (iii) 9981 Same as question 1 (i) (iv) ²=25, hence 5 is possible.

7 2. Without doing any calculation, find the numbers which are surely not perfect squares. (i) 153 (ii) 257 (iii) 48 (iv) 441 Option 1 can be a perfect square, others can t be perfect squares because the unit digit of a perfect square can be only from, 1, 4, 5, 6, 9 3. Find the square roots of 1 and 169 by the method of repeated subtraction. Repeated subtraction: = = = = = = = = = = We get at 1 th step So, 1 = = = = = = = = = = = = = = We get at 13 th step So, 169 = 13

8 4. Find the square roots of the following numbers by the Prime Factorisation Method. (i) = 3 ² 3² 3² 729 = = 27 (ii) = 2² 2² 5² 4 = = 2 (iii) = = = = = = 2 ² 3² 7² 1764 = = 42 (iv) = = = = = = = = 2 ² 2² 2² 8² 496 = = 64

9 (v) = = = = = = = = 8 ² 11² 7744 = 88 (vi) = = = = = 2 ² 7² 7² 964 = 98 (vii) = = = = 11 ² 7² 5929 = 77 (viii) = = = = = 3 ² 4² 8² 9216 = 96 (ix) = = 23² 529 = 23

10 (x) = 9 9 = = 9 ² 1² 81 = 9 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) = = = Here, 2 and 3 are in pairs but 7 needs a pair, so 252 will become a perfect square when multiplied by 7. (ii) = = needs to be multiplied by 5 to become a perfect square. (iii) = = = needs to be multiplied by 7 to become a perfect square (iv) = 4 57 = = needs to be multiplied by 3 to become a perfect square. (v) = = = needs to be multiplied by 2 to become a perfect square.

11 (vi) = = = = needs to be multiplied by 3 to become a perfect square. 6. 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) = = needs to be divided by 7 to become a perfect square. (ii) = = needs to be divided by 13 to become a perfect square (iii) = = needs to be divided by 11 to become a perfect square (iv) = needs to be divided by 5 to become a perfect square. (v) = needs to be divided by to become a perfect square. (vi) = = = needs to be divided by 5 to become a perfect square

12 7. The students of Class VIII of a school donated Rs 241 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. We need to calculate the square root of 241 to get the solution 241 = = 7 7 = 49 There are 49 students, each contributing 49 rupees 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. 225 = = = 45 There are 45 rows with 45 plants in each of them. 9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 1. Let us find LCM of 4, 9 and 1 4 = = = 5 2 So, LCM = 2 ² 3² 5 = 18 Now the LCM gives us a clue that if 18 is multiplied by 5 then it will become a perfect square. The Required number = 18 5 = 9 1. Find the smallest square number that is divisible by each of the numbers 8, 15 and 2. 8 = = = So, LCM = = 12 As 3 and 5 are not in pair in LCM s factor so we need to multiply 12 by 5 and three to make it a perfect square. Required Number= = 27

13 Exercise 4 1. Find the square root of each of the following numbers by Division method. (i) = 48 (ii) = 67 (iii) = 59

14 (iv) = 23 (v) = 57 (vi) = 37

15 (vii) = 76 (viii) = 89 (ix) = 24

16 (x) = 32 (xi) = 56 (xii) = 3

17 2. Find the number of digits in the square root of each of the following numbers (without any calculation). (i) 64 (ii) 144 (iii) 4489 (iv) (v) If there are even number of digits in square then number of digits in square root = 2 n If there are odd number of digits in square then number of digits in square root= n (i) 1, (ii) 2, (iii) 2, (iv) 3, (v) 3 3. Find the square root of the following decimal numbers. (i) = (ii) = 2.7

18 (iii) = (iv) = (v) = 5.6

19 4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) It is clear that if 2 is subtracted then we will get 4, which is a perfect square. (ii) Here, 84X4=336 which is less than 389 And, 85X5=425, which is more than 389 Hence the required difference = = =1936 is a perfect square. (iii) Here, 17X7=749 is less than 75 18X8=864 is more than 75 Hence, the required difference = = =3249 is a perfect square.

20 (iv) Here, 48X8=384 is less than X9=441 is more than 425 Hence, the required difference= = =784 is a perfect square. (v) Here, 123X3=369 is less than 4 124X4=496 is more than 4 Hence, the required difference = 4-369= =3969 is a perfect square. 5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) Here, 43X3=129 is more than X2=84 is less than 125 Hence, required addition= = =529 is a perfect square.

21 (ii) Here, 161X1=161 is 11 more than 15 So, =1761 is a perfect square (iii) Here, 25X5=125 is less than X6=156 is more than 152 Required difference= =4 So, 252+4=256 is a perfect square (iv) 1825 ` Here, 82X2=164 is less than X3=249 is more than 225 Required difference= =24 So, =1849 is a perfect square (v)

22 Here, we need 161X1=161 Required difference=161-12=149 So, =6561 is a perfect square 6. Find the length of the side of a square whose area is 441 m². Area of Square = Side² Side = Area 441 = = 3 7 = In a right triangle ABC, B = 9. (a) If AB = 6 cm, BC = 8 cm, find AC Answer= AC²=AB²+BC² =6²+8²=36+64=1 AC= 1 = 1 (b) If AC = 13 cm, BC = 5 cm, find AB AB²=AC²-BC² =13²-5²=169-25=144 AB = 144 = A gardener has 1 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this Here, 61X1=61 is less than 1 62X2=124 is more than 1 Hence, the required difference= 1-61=39 Min. number of plants required= 1-39=961

23 9. There are 5 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement Here, 42X2=84 is less than 1 43X3=129 is more than 1 Hence, the required difference = 1-84=16 So, 16 children will be left out in the arrangement.

Question: 1 - What will be the unit digit of the squares of the following numbers?

Square And Square Roots Question: 1 - What will be the unit digit of the squares of the following numbers? (i) 81 Answer: 1 Explanation: Since, 1 2 ends up having 1 as the digit at unit s place so 81 2