8 Math Square And Square Roots

Finish Line & Beyond

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 0, 1, 4, 5, 6 or 9 at unit’s place. All numbers have numbers from 0 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: 0²=0 1²=1 2²=4 3²=9 4²=16 5²=25 6²=36 7²=49 8²=64 9²=81

0 1 4 9 6 5 6 9 4 1

is is is is is is is is is is

the the the the the the the the the the

number number number number number number number number number number

at at at at at at at at at at

unit’s unit’s unit’s unit’s unit’s unit’s unit’s unit’s unit’s unit’s

place place place place place place place place place 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 Answer: 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 Answer: 9²=81 So, 799 will have 1 at unit’s place (iv) 3853 Answer: 3²=9 So, 3853² will have 9 at unit’s place.

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Finish Line & Beyond (v) 1234 Answer: 4²=16 So, 1234² will have 6 at unit’s place (vi) 26387 Answer: 7²=49 So, 26387² will have 9 at unit’s place (vii) 52698 Answer: 8²=64 So, 52698² will have 4 at unit’s place (viii) 99880 Answer: 0²=0 So, 99880² will have 0 at unit’s place (ix) 12796 Answer: 6²=36 So, 12796² will have 6 at unit’s place (x) 55555 Answer: 5²=25 So, 55555² will have 5 at unit’s place 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 Answer: (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 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) 82004 Answer: (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 = 121 101² = 10201

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Finish Line & Beyond 1001² = 1002001 100001² = 1 0000 2 0000 1 10000001² = 100000020000001 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 1. 5. Observe the following pattern and supply the missing numbers. 11² = 1 2 1 101² = 1 0 2 0 1 10101² = 102030201 1010101² = 1020304030201 101010101² =102030405040201 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 + 20²= 21² 5² + 6²+ 30² = 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) 1 + 3 + 5 + 7 + 9 Answer: 1 + 3 = 2² = 4

1 + 3 + 5 = 3² = 9 1 + 3 + 5 + 7 = 4² = 16 1 + 3 + 5 + 7 + 9 = 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) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19 Answer: There are 10 odd numbers in the given equation So, sum =10²=100 (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 Answer: Sum = 12²=144

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Finish Line & Beyond 8. (i) Express 49 as the sum of 7 odd numbers. As you know 49=7² So, 7² can be expressed as follows: 1+3+5+7+9+11+13 (ii) Express 121 as the sum of 11 odd numbers. 11²=1+3+5+7+9+11+13+15+17+19+21 9. How many numbers lie between squares of the following numbers? (i) 12 and 13 Answer: 12²=144 13²=169 Now, 169-144=25 So, there are 25-1=24 numbers lying between 12² and 13² (ii) 25 and 26 Answer: 25²=625 26²=676 Now, 676-625=51 So, there are 51-1=50 numbers lying between 25² and 26² (iii) 99 and 100 Answer: 99²=9801 100²=10000 Now, 10000-9801=199 So, there are 199-1=198 numbers lying between 99² and 100²

Exercise 2 1. Find the square of the following numbers. (i) 32 Answer: 32²=32 × 32= 1024 But above method can be tough to calculate. It is easier to calculate such values by using algebraic identities. So, 32²=(30+2) ² Using (a+b) ² = a²+b²+2ab We get (30+2) ²= 30²+2²+2 × 30 × 2 =900+4+120=1024

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Finish Line & Beyond (ii) 35 Answer: (35)²=(30+5) ² =30²+5²+2 × 30 × 5 =900+25+300=1225 (iii) 86 Answer: 86²=(80+6) ² =80²+6²+2 × 80 × 6 =6400+36+960=7396 (iv) 93 Answer=93²=(90+3² =90²+3²+2 × 90 × 3 =8100+9+540 =8649 (v) 71 Answer: 71²=(70+1) ² =70²+1²+2 × 70 × 1 =4900+1+140 =5040 (vi) 46 Answer: 46²=(40+6) ² =40²+6²+2 × 40 × 6 =1600+36+480=2116 2. Write a Pythagorean triplet whose one member is. (i) 6 Answer: 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=10 ⇒ m²-1=3²-1=8 Test: 6²+8²=36+64=100=10² Hence, the triplet is 6, 8, and 10

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Finish Line & Beyond Note: Most of the Pythagorean triplets are in the ratio of 3:4:5 (ii) 14 Answer: Let us assume, 2m=14, then m=7 So, m²+1=7²+1=50 And, m²-1=7²-1=48 Test: 14²+48²=196+1304=2500=50² Hence, the triplet is 14, 48, and 50 (iii) 16 Answer: Let us assume 2m=16, then m=8 So, m²+1=8²+1=65 And, m²-1=8²-1=63 Test: 16²+63²=256+3969=4225=65² Hence, the triplet is 16, 63, and 65 (iv) 18 Answer: Let us assume 2m=18, then m=9 So, m²+1=9²+1=82 m²-1=9²-1=80 test: 18²+80²=6724=82²

Exercise 3 1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers? (i) 9801 Answer: Since 1² and 9² give 1 at unit’s place, so these are the possible values of unit digit of the square root. (ii) 99856 Answer: 4²=16 and 6²=36, hence, 4 and 6 are possible (iii) 998001 Answer: Same as question 1—(i) (iv) 657666025 Answer: 5²=25, hence 5 is possible.

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Finish Line & Beyond 2. Without doing any calculation, find the numbers which are surely not perfect squares. (i) 153 (ii) 257 (iii) 408 (iv) 441 Answer: 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 0, 1, 4, 5, 6, 9 3. Find the square roots of 100 and 169 by the method of repeated subtraction. Answer: Repeated subtraction: 1. 100-1=99 2. 99-3=96 3. 96-5=91 4. 91-7=84 5. 84-9=75 6. 75-11=64 7. 64-13=51 8. 51-15=36 9. 36-17=19 10. 19-19=0 We get 0 at 10th step So,

100 =10

1. 169-1=168 2. 168-3=165 3. 165-5=160 4. 160-7=153 5. 153-9=144 6. 144-11=133 7. 133-13=120 8. 120-15=105 9. 105-17=88 10. 88-19=69 11. 69-21=48 12. 48-23=25 13. 25-25=0 We get 0 at 13th step So,

169 = 13

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Finish Line & Beyond 4. Find the square roots of the following numbers by the Prime Factorisation Method. (i) 729

3 729 3 243 Answer:

3 81 3 27

39 3 ⇒ 729= 3² × 3² × 3² ⇒ 729 = 3 × 3 × 3 = 27 (ii) 400

2 400 2 200 Answer:

2 100 2 50 5 25 5

⇒ 400 = 2² × 2² × 5² ⇒ 400 = 2 × 2 × 5 = 20 (iii) 1764 Answer: 1764= 2 × 882 = 2 × 2 × 441 = 2 × 2 × 3 × 147 = 2 × 2 × 3 × 3 × 49 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²

⇒

1764 = 2 × 3 × 7 = 42

(iv) 4096 Answer: 4096 = 2 × 2048

= 2 × 2 × 1024 = 2 × 2 × 2 × 512 = 2 × 2 × 2 × 2 × 256 = 2 × 2 × 2 × 2 × 2 × 128 = 2 × 2 × 2 × 2 × 2 × 2 × 64 = 2× 2× 2× 2× 2× 2× 8× 8 = 2² × 2² × 2² × 8² ⇒ 4096 = 2 × 2 × 2 × 8 = 64

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Finish Line & Beyond (v) 7744 Answer: 7744= 2 × 3872

= 2 × 2 × 1936 = 2 × 2 × 2 × 968 = 2 × 2 × 2 × 2 × 484 = 2 × 2 × 2 × 2 × 2 × 242 = 2 × 2 × 2 × 2 × 2 × 2 × 121 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 = 8² × 11² ⇒ 7744 = 88 (vi) 9604 Answer: 9604 = 4 × 2401

= 4 × 7 × 343 = 4 × 7 × 7 × 49 = 4× 7× 7× 7× 7 = 2² × 7² × 7² ⇒ 9604 = 98

(vii) 5929 Answer: 5929 = 11 × 539

= 11 × 11 × 49 = 11 × 11 × 7 × 7 = 11² × 7² ⇒ 5929 = 77

(viii) 9216

9216 = 9 × 1024 = 9 × 4 × 256 = 9 × 4 × 4 × 64 = 9× 4× 4× 8× 8 = 3² × 4² × 8² ⇒ 9216 = 96

Answer:

(ix) 529

529 = 23 × 23 = 23² ⇒ 529 = 23

Answer:

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Finish Line & Beyond (x) 8100 Answer: 8100 = 9 × 900

= 9 × 9 × 10 × 10 = 9² × 10² ⇒ 8100 = 90

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) 252

252 = 2 × 126 = 2 × 2 × 63 = 2 × 2 × 3× 3× 7

Answer:

Here, 2 and 3 are in pairs but 7 needs a pair, so 252 will become a perfect square when multiplied by 7. (ii) 180

180 = 2 × 2 × 45 = 2 × 2 × 3× 3× 5

Answer:

180 needs to be multiplied by 5 to become a perfect square. (iii) 1008

1008 = 2 × 2 × 252 = 2 × 2 × 2 × 2 × 63 = 2 × 2 × 2 × 2 × 3× 3× 7

Answer:

1008 needs to be multiplied by 7 to become a perfect square (iv) 2028 Answer: 2028 = 4 × 507

= 4 × 3 × 169 = 2 × 2 × 3 × 13 × 13

2028 needs to be multiplied by 3 to become a perfect square. (v) 1458

1458 = 2 × 729 = 2 × 3 × 3 × 81 = 2 × 3× 3× 3× 3× 3× 3

Answer:

1458 needs to be multiplied by 2 to become a perfect square.

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Finish Line & Beyond (vi) 768 Answer: 768 = 2 × 2 × 192

= 2 × 2 × 2 × 2 × 48 = 2 × 2 × 2 × 2 × 2 × 2 × 12 = 2× 2× 2× 2× 2× 2× 2× 2× 3

768 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) 252 Answer: 252 = 2 × 2 × 63

= 2 × 2 × 3× 3× 7

252 needs to be divided by 7 to become a perfect square. (ii) 2925 Answer: 2925 = 5 × 5 × 117

= 5 × 5 × 3 × 3 × 13

2925 needs to be divided by 13 to become a perfect square (iii) 396 Answer: 396 = 2 × 2 × 99

= 2 × 2 × 3 × 3 × 11

396 needs to be divided by 11 to become a perfect square (iv) 2645 Answer: 2645 = 5 × 529 5 × 23 × 23 2645 needs to be divided by 5 to become a perfect square. (v) 2800 Answer: 2800 = 2 × 2 × 7 × 10 × 10 2800 needs to be divided by to become a perfect square. (vi) 1620 Answer: 1620 = 2 × 2 × 405

= 2 × 2 × 3 × 3 × 45 = 2 × 2 × 3× 3× 3× 3× 5

1620 needs to be divided by 5 to become a perfect square

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Finish Line & Beyond 7. The students of Class VIII of a school donated Rs 2401 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. Answer: We need to calculate the square root of 2401 to get the solution

2401 = 7 × 7 × 7 × 7 ⇒ 2401 = 7 × 7 = 49 There are 49 students, each contributing 49 rupees 8. 2025 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.

2025 = 5 × 5 × 3 × 3 × 3 × 3 2025 = 5 × 3 × 3 = 45

Answer:

⇒

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 10. Answer: Let us find LCM of 4, 9 and 10

4 = 2× 2 9 = 3× 3 10 = 5 × 2

So, LCM = 2² × 3² × 5 = 180 Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square. The Required number = 180 × 5 = 900 10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20. Answer:

8 = 2× 2× 2 15 = 3 × 5 20 = 2 × 2 × 5 So, LCM = 2 × 2 × 2 × 5 × 3 = 120

As 3 and 5 are not in pair in LCM’s factor so we need to multiply 120 by 5 and three to make it a perfect square. Required Number= 180 × 3 × 5 = 2700

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Finish Line & Beyond Exercise 4 1. Find the square root of each of the following numbers by Division method. (i) 2304 Answer: 48 4 4

2304

88 8

16 704 704 0

2304 = 48 (ii) 4489 Answer: 67 6 6

4489

127 7

36 889 889 0

4489 = 67 (iii) 3481 Answer: 59 5 5

3481

109 9

25 981 981 0

3481 = 59

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Finish Line & Beyond (iv) 529 Answer: 23 2 2 43 3

5 29 4 129 129 0

529 = 23 (v) 3249 Answer: 57 5 5

3249

107 7

25 749 749 0

3249 = 57 (vi) 1369 Answer: 37 3 1369 3 9 67 7

469 469 0

1369 = 37

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Finish Line & Beyond (vii) 5776 Answer: 76 7 7

5776

146 6

49 876 876 0

5776 = 76 (viii) 7921 Answer: 89 8 8

7921

169 9

64 1521 1521 0

7921 = 89 (ix) 576 Answer: 24 2 2

576

44 4

4 176 176 0

576 = 24

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Finish Line & Beyond

(x) 1024 Answer: 32 3 3

1024

62 2

9 124 124 0

1024 = 32 (xi) 3136 Answer: 56 5 5

3136

106 6

25 636 636 0

3136 = 56 (xii) 900 Answer: 30 3 3

900

60 0

9 000 000 0

900 = 30

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Finish Line & Beyond

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) 27225 (v) 390625 Answer: If there are even number of digits in square then number of digits in square root =

n 2

If there are odd number of digits in square then number of digits in square root=

n+ 1 2

(i) 1, (ii) 2, (iii) 2, (iv) 3, (v) 3 3. Find the square root of the following decimal numbers. (i) 2.56 Answer:

1 1 26 6

1.6 2.56 1 156 156 0

2.56 = 1.6 (ii) 7.29 Answer:

2 2 47 7

2.7 7.29 4 329 329 0

7.29 = 2.7

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Finish Line & Beyond

(iii) 51.84 Answer:

7 7 142 2

7.2 51.84 49 284 284 0

51.84 = 7.2 (iv) 42.25 Answer:

6 6 125 5

6.5 42.25 36 625 625 0

42.25 = 65 (v) 31.36 Answer:

5 5 106 6

5.6 31.36 25 636 636 0

31.36 = 5.6

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Finish Line & Beyond

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) 402 Answer: 2 2 2 4

402 4 002

It is clear that if 2 is subtracted then we will get 400, which is a perfect square. (ii) 1989 Answer:

4 4 8

1989 16 389

Here, 84X4=336 which is less than 389 And, 85X5=425, which is more than 389 Hence the required difference =389-336=53 1989-53=1936 is a perfect square. (iii) 3250 Answer:

5 5 10

3250 25 750

Here, 107X7=749 is less than 750 108X8=864 is more than 750 Hence, the required difference = 750-749=1 3250-1=3249 is a perfect square.

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Finish Line & Beyond

(iv) 825 Answer:

2 2 4

825 4 425

Here, 48X8=384 is less than 425 49X9=441 is more than 425 Hence, the required difference= 425-384=41 825-41=784 is a perfect square. (v) 4000 Answer:

6 6 12

4000 36 400

Here, 123X3=369 is less than 400 124X4=496 is more than 400 Hence, the required difference = 400-369=31 4000-31=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) 525 Answer:

2 2 4

525 4 125

Here, 43X3=129 is more than 125 42X2=84 is less than 125 Hence, required addition= 129-125=4 525+4=529 is a perfect square.

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Finish Line & Beyond

(ii) 1750 Answer:

4 1750 4 16 16 150 Here, 161X1=161 is 11 more than 150 So, 1750+11=1761 is a perfect square (iii) 252 Answer:

1 252 1 1 2 152 Here, 25X5=125 is less than 152 26X6=156 is more than 152 Required difference= 156-152=4 So, 252+4=256 is a perfect square (iv) 1825 Answer: ` 4 4 8

1825 16 225

Here, 82X2=164 is less than 225 83X3=249 is more than 225 Required difference= 249-225=24 So, 1825+24=1849 is a perfect square (v) 6412 Answer:

8 8 16

6412 64 12

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Finish Line & Beyond Here, we need 161X1=161 Required difference=161-12=149 So, 6412+149=6561 is a perfect square 6. Find the length of the side of a square whose area is 441 m². Answer: Area of Square = Side²

⇒ Side = Area 441 = 3 × 3 × 7 × 7 ⇒ 441 = 3 × 7 = 21 7. In a right triangle ABC, ∠ B = 90° . (a) If AB = 6 cm, BC = 8 cm, find AC Answer= AC²=AB²+BC² =6²+8²=36+64=100 AC=

100 = 10

(b) If AC = 13 cm, BC = 5 cm, find AB Answer: AB²=AC²-BC² =13²-5²=169-25=144 AB = 144 = 12 8. A gardener has 1000 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. Answer:

3 3 6

1000 9 100

Here, 61X1=61 is less than 100 62X2=124 is more than 100 Hence, the required difference= 100-61=39 Min. number of plants required= 1000-39=961

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Finish Line & Beyond

9. There are 500 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. Answer:

2 2 4

500 4 100

Here, 42X2=84 is less than 100 43X3=129 is more than 100 Hence, the required difference = 100-84=16 So, 16 children will be left out in the arrangement.

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