8 Math Square and 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 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 0 is the number at unit’s place1²=1 1 is the number at unit’s place2²=4 4 is the number at unit’s place3²=9 9 is the number at unit’s place4²=16 6 is the number at unit’s place5²=25 5 is the number at unit’s place6²=36 6 is the number at unit’s place7²=49 9 is the number at unit’s place8²=64 4 is the number at unit’s place9²=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

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=4So, 272² will have 4 at unit’s place

(iii) 799

Answer: 9²=81So, 799 will have 1 at unit’s place

(iv) 3853

Answer: 3²=9So, 3853² will have 9 at unit’s place.

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(v) 1234

Answer: 4²=16So, 1234² will have 6 at unit’s place

(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

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 = 121101² = 10201




1001² = 1002001100001² = 1 0000 2 0000 110000001² = 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 1101² = 1 0 2 0 110101² = 1020302011010101² = 1020304030201101010101² =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 = 3²2² + 3² + 6² = 7²3² + 4² + 12² = 13²4² + 5² + 20²= 21²5² + 6²+ 30² = 31²6² + 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: 4²231 ==+9²3531 ==++16²47531 ==+++25²597531 ==++++

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 equationSo, sum =10²=100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Answer: Sum = 12²=144




8. (i) Express 49 as the sum of 7 odd numbers.

As you know 49=7²So, 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=14413²=169

Now, 169-144=25So, there are 25-1=24 numbers lying between 12² and 13²

(ii) 25 and 26

Answer: 25²=62526²=676Now, 676-625=51So, there are 51-1=50 numbers lying between 25² and 26²

(iii) 99 and 100

Answer: 99²=9801100²=10000Now, 10000-9801=199So, 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) ²Using (a+b) ² = a²+b²+2abWe get (30+2) ²= 30²+2²+2× 30× 2=900+4+120=1024




(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=6Then m=3⇒ m²+1=3²+1=10⇒ m²-1=3²-1=8

Test: 6²+8²=36+64=100=10²Hence, the triplet is 6, 8, and 10




Note: Most of the Pythagorean triplets are in the ratio of 3:4:5

(ii) 14

Answer: Let us assume, 2m=14, then m=7So, m²+1=7²+1=50And, 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=8So, m²+1=8²+1=65And, 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=9So, m²+1=9²+1=82m²-1=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.




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=992. 99-3=963. 96-5=914. 91-7=845. 84-9=756. 75-11=647. 64-13=518. 51-15=369. 36-17=1910. 19-19=0We get 0 at 10th step

So, 100 =10

1. 169-1=1682. 168-3=1653. 165-5=1604. 160-7=1535. 153-9=1446. 144-11=1337. 133-13=1208. 120-15=1059. 105-17=8810. 88-19=6911. 69-21=4812. 48-23=2513. 25-25=0

We get 0 at 13th step

So, 13169 =




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

Answer:

393

273

813

2433

7293

⇒ 729= 3²3²²3 ××27333729 =××=⇒

(ii) 400

Answer:

5

255

502

1002

2002

4002

5²2²²2400 ××=⇒20522400 =××=⇒

(iii) 1764 Answer: 1764= 147322441228822 ×××=××=×

773322493322 ×××××=××××= = 7²3²²2 ××427321764 =××=⇒

(iv) 4096

Answer: 204824096 ×=102422 ××=512222 ×××=2562222 ××××=12822222 ×××××=64222222 ××××××=88222222 ×××××××=

8²2²2²²2 ×××=6482224096 =×××=⇒




(v) 7744

Answer: 7744= 38722 ×193622 ××=968222 ×××=4842222 ××××=

= 24222222 ×××××121222222 ××××××=

1111222222 ×××××××=11²²8 ×=

887744 =⇒

(vi) 9604

Answer: 240149604 ×=34374 ××=49774 ×××=77774 ××××=

7²7²²2 ××=989604 =⇒

(vii) 5929

Answer: 539115929 ×=491111 ××=771111 ×××=

7²²11 ×=775929 =⇒

(viii) 9216

Answer: 102499216 ×=25649 ××=64449 ×××=88449 ××××=

8²4²²3 ××=969216 =⇒

(ix) 529

Answer: 2323529 ×=²23=

23529 =⇒




(x) 8100

Answer: 90098100 ×=101099 ×××=

10²²9 ×=908100 =⇒

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

Answer: 1262252 ×=6322 ××=

73322 ××××=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

Answer: 4522180 ××=53322 ××××=

180 needs to be multiplied by 5 to become a perfect square.

(iii) 1008

Answer: 252221008 ××=632222 ××××=

7332222 ××××××=1008 needs to be multiplied by 7 to become a perfect square

(iv) 2028

Answer: 50742028 ×=16934 ××=

1313322 ××××=2028 needs to be multiplied by 3 to become a perfect square.

(v) 1458

Answer: 72921458 ×=81332 ×××=

3333332 ××××××=1458 needs to be multiplied by 2 to become a perfect square.




(vi) 768

Answer: 19222768 ××=482222 ××××=

12222222 ××××××=322222222 ××××××××=

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: 6322252 ××=73322 ××××=

252 needs to be divided by 7 to become a perfect square.

(ii) 2925

Answer: 117552925 ××=133355 ××××=

2925 needs to be divided by 13 to become a perfect square

(iii) 396

Answer: 9922396 ××=113322 ××××=

396 needs to be divided by 11 to become a perfect square

(iv) 2645

Answer: 52952645 ×= 23235 ××2645 needs to be divided by 5 to become a perfect square.

(v) 2800

Answer: 10107222800 ××××=2800 needs to be divided by to become a perfect square.

(vi) 1620

Answer: 405221620 ××=453322 ××××=

5333322 ××××××=1620 needs to be divided by 5 to become a perfect square




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

77772401 ×××=49772401 =×=⇒

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.

Answer: 3333552025 ×××××=453352025 =××=⇒

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 10224 ×=339 ×=2510 ×=

So, LCM = 18053²²2 =××Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square.The Required number = 9005180 =×

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Answer: 2228 ××=

5315 ×=52220 ××=

So, LCM = 12035222 =××××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= 270053180 =××




Exercise 4

1. Find the square root of each of the following numbers by Division method.(i) 2304 Answer:

4844

042316

88 8

704 704 0

482304 =

(ii) 4489

Answer:

6766

894436

127 7

889 889 0

674489 =

(iii) 3481

Answer:

5955

8134 25

109 9

981 981

0

593481 =




(iv) 529

Answer:

2322

5294

43 3

129129 0

23529 =

(v) 3249

Answer:

5755

4932 25

107 7

749 749 0

573249 =

(vi) 1369

Answer: 37

33

6913 9

67 7

469 469 0

371369 =




(vii) 5776

Answer:

7677

765749

146 6

876 876 0

765776 =

(viii) 7921

Answer:

8988

217964

169 9

15211521 0

897921 =

(ix) 576

Answer:

2422

7654

44 4

176176 0

24576 =




(x) 1024

Answer:

3233

2410 9

62 2

124 124 0

321024 =

(xi) 3136

Answer:

5655

3631 25

106 6

636 636 0

563136 =

(xii) 900

Answer:

3033

0099

60 0

000000 0

30900 =




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 =2n

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

21+n

(i) 1, (ii) 2, (iii) 2, (iv) 3, (v) 3

3. Find the square root of the following decimal numbers.(i) 2.56

Answer:

1.611

2.561

26 6

156156 0

6.156.2 =

(ii) 7.29

Answer:

2.722

7.294

47 7

329329 0

7.229.7 =




(iii) 51.84

Answer:

7.277

51.8449

142 2

284 284 0

2.784.51 =

(iv) 42.25

Answer:

6.566

42.2536

125 5

625 625 0

6525.42 =

(v) 31.36

Answer:

5.655

31.3625

106 6

636 636 0

6.536.31 =




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:

222

0244

4 002

It is clear that if 2 is subtracted then we will get 400, which is a perfect square.

(ii) 1989

Answer:

44

198916

8 389

Here, 84X4=336 which is less than 389And, 85X5=425, which is more than 389

Hence the required difference =389-336=531989-53=1936 is a perfect square.

(iii) 3250

Answer:

55

325025

10 750

Here, 107X7=749 is less than 750 108X8=864 is more than 750Hence, the required difference = 750-749=13250-1=3249 is a perfect square.




(iv) 825

Answer:

22

8254

4 425

Here, 48X8=384 is less than 425 49X9=441 is more than 425Hence, the required difference= 425-384=41825-41=784 is a perfect square.

(v) 4000

Answer:

66

400036

12 400

Here, 123X3=369 is less than 400 124X4=496 is more than 400Hence, the required difference = 400-369=314000-31=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:

22

5254

4 125

Here, 43X3=129 is more than 125 42X2=84 is less than 125Hence, required addition= 129-125=4525+4=529 is a perfect square.




(ii) 1750

Answer:

44

175016

16 150 Here, 161X1=161 is 11 more than 150So, 1750+11=1761 is a perfect square

(iii) 252

Answer:

11

2521

2 152Here, 25X5=125 is less than 152 26X6=156 is more than 152Required difference= 156-152=4So, 252+4=256 is a perfect square

(iv) 1825

Answer:

`44

182516

8 225

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

(v) 6412

Answer:

88

641264

16 12




Here, we need 161X1=161Required difference=161-12=149So, 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=7733441 ×××=2173441 =×=⇒

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= 10100 =

(b) If AC = 13 cm, BC = 5 cm, find AB

Answer: AB²=AC²-BC²=13²-5²=169-25=144AB = 12144 =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:

33

10009

6 100

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




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:

22

5004

4 100

Here, 42X2=84 is less than 100 43X3=129 is more than 100Hence, the required difference = 100-84=16

So, 16 children will be left out in the arrangement.



8 Math Square and Square Roots

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