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6 SQUARES AND SQUARE ROOTS

Exercise 6.1

Q.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

Ans. (i) The unit digit of 812 = 1 [12 = 1] (ii) The unit digit of 2722 = 4 [22 = 4] (iii) The unit digit of 7992 = 1 [92 = 81] (iv) The unit digit of 38532 = 9 [32 = 9] (v) The unit digit of 12342 = 6 [42 = 16] (vi) The unit digit of 263872 = 9 [72 = 49] (vii) The unit digit of 526982 = 4 [82 = 64] (viii) The unit digit of 998802 = 0 [02 = 0] (ix) The unit digit of 12796 = 6 [62 = 36] (x) The unit digit of 55555 = 5 [52 = 25]

Q.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

Ans. (i) 1057 – it is not a perfect square as it ends with 7. (ii) 23453 – it is not a perfect square as it ends with 3. (iii) 7928 – it is not a perfect square as it ends with 8.

(iv) 222222 – it is not a perfect square as it ends with 2.


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(v) 64000 – it is not a perfect square as it does not have even number of zeroes at the end.

(vi) 89722 – it is not a perfect square as it ends with 2. (vii) 222000 – it is not a perfect square as it ends with an

odd number of zeroes. (viii) 505050 – it is not a perfect square as it ends with an

odd number of zeroes.

Q.3. The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Ans. (i), (iii) Since, 431 and 7779 are odd numbers, so their squares will be odd numbers.

Q.4. Observe the following pattern and find the missing digits.

112 = 121 1012 = 10201 10012 = 1002001 1000012 = 1 ……. 2 …….. 1 100000012 = …………………

Ans. Using the given pattern, we have 1000012 = 10000200001 100000012 = 100000020000001

Q.5. Observe the following pattern and supply the missing numbers.

112 = 121 1012 = 10201 101012 = 102030201 10101012 = ……………….. ……….. 2 = 10203040504030201


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Ans. Using the given pattern, we have 10101012 = 1020304030201

1010101012 = 10203040504030201

Q.6. Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + ___2 = 212

52 + ___2 + 302 = 312

62 + 72 + ___2 = ___2

Ans. 12 + 22 + 22 = 32 22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

Q.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

Ans. (i) 1 + 3 + 5 + 7 + 9 Sum of first five odd numbers = (5)2 = 25 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 Sum of first ten odd numbers = (10)2 = 100 (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 Sum of first twelve odd numbers = (12)2 = 144

Q.8. (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers.


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Ans. (i) 49 = (7)2 = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 (ii) 121 = (11)2 = Sum of first 11 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q.9. How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Ans. (i) There are 2n numbers between the squares of the numbers n and (n + 1)

Here n = 12 and n + 1 = 13 ∴ 2 × 12 i.e., 24 numbers lie between 12 and 13. (ii) Here n = 25 and n + 1 = 26 ∴ Numbers between the squares of 25 and 26 are 2 × 25

i.e. 50 (iii) Here n = 99 and n + 1 = 100 ∴ Numbers between the squares of 99 and 100 are 2 × 99

i.e. 198

Exercise 6.2

Q.1. Find the squares of the following numbers. (i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Ans. (i) 322 = (30 + 2)2 = (30 + 2) (30 + 2) = 30(30 + 2) + 2 (30 + 2) = 900 + 60 + 60 + 4 = 1024


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(ii) 352 = (5 + 30)2 = (5 + 30) (5 + 30) = 5(5 + 30) + 30 (5 + 30) = 25 + 150 + 150 + 900 = 25 + 300 + 900 = 1225 (iii) 862 = (80 + 6)2 = (80 + 6) (80 + 6) = 80(80 + 6) + 6 (80 + 6) = 6400 + 480 + 480 + 36 = 7396 (iv) 932 = (90 + 3)2 = (90 + 3) (90 + 3) = 90(90 + 3) + 3 (90 + 3) = 8100 + 270 + 270 + 9 = 8649 (v) 712 = (70 + 1) (70 + 1) = 70(70 + 1) + 1 (70 + 1) = 4900 + 70 + 70 + 1 = 5041 (vi) 462 = (40 + 6)2 = (40 + 6) (40 + 6) = 40(40 + 6) + 6 (40 + 6) = 1600 + 240 + 240 + 36 = 2116

Q.2. Write a Pythagorean triplet whose one member is. (i) 6 (ii) 14 (iii) 16 (iv) 18

Ans. For any natural number m > 1; (2m)2, (m2 – 1)2 and (m2 + 1)2 forms a Pythagorean triplet.

(i) Let 2m = 6 ------------- (i)


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⇒ m = 3 (integer) Also, m2 – 1 = 32 – 1 = 9 – 1 = 8 -------------- (ii) And, m2 + 1 = 32 + 1 = 9 + 1 = 10 ------------- (iii) From eqn. (i), (ii) and (iii), we get the required triplet as 6, 8, 10. (ii) Let 2m = 14 ------------- (i)

⇒ m = 142

= 7 (integer)

∴ m2 – 1 = (7)2 – 1 = 49 – 1 = 48 ------------ (ii) And, m2 + 1 = (7)2 + 1 = 49 + 1 = 50 ------------ (iii) From eqn. (i), (ii) and (iii), we get the required triplet as 14, 48, 50. (iii) Let 2m = 16 -------------- (i) ⇒ m = 8 (integer) ∴ m2 – 1 = (8)2 – 1 = 63 ------------- (ii) And m2 + 1 = (8)2 + 1 = 65 -------------- (iii) From eqn. (i), (ii) and (iii), we get the required triplet as 16, 63, 65. (iv) Let 2m = 18 ------------ (i) ⇒ m = 9 (integer) ∴ m2 – 1 = (9)2 – 1 = 81 – 1 = 80 ------------- (ii) And m2 + 1 = (9)2 + 1 = 81 + 1 = 82 ------------- (iii) From eqn. (i), (ii) and (iii), we get the required triplet, as 18, 80, 82.


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Exercise 6.3

Q.1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Ans. (i) As we know, 12 = 1 × 1 = 1; One’s digit is 1. And, 92 = 9 × 9 = 81; One’s digit is 1. Thus, the one’s digit of the square root of 9801 can

be 1 or 9. (ii) As we know, 42 = 4 × 4 = 16; One’s digit is 6 And, 62 = 6 × 6 = 36; One’s digit is 6 Thus, the one’s digit of the square root of 99856 can be

4 or 6. (iii) As discussed in part (i), the one’s digit of the square

root of 998001 can be 1 or 9. (iv) As we know, 52 = 5 × 5 = 25 One’s digit is 5 Thus, the one’s digit of the square root of 657666025

can be 5.

Q.2. Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Ans. (i) 153 is not a perfect square because it ends with 3. (ii) 257 is not a perfect square because it ends with 7. (iii) 408 is not a perfect square because it ends with 8.


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Q.3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Ans. Method of repeated subtraction

For 169 For 100 Step (i) 169 – 1 = 168 Step (i) 100 – 1 = 99 Step (ii) 168 – 3 = 165 Step (ii) 99 – 3 = 96 Step (iii) 165 – 5 = 160 Step (iii) 96 – 5 = 91 Step (iv) 160 – 7 = 153 Step (iv) 91 – 7 = 84 Step (v) 153 – 9 = 144 – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – Step (x) 9 – 19 = 0 Step (xiii) 25 – 25 = 0 From 100 and 169 we have subtracted successive odd

numbers starting from 1 and obtained 0 at 10th and 13th steps respectively.

∴ 169 = 13

and 100 = 10

Q.4. Find the square roots of the following numbers by the Prime Factorisation Method

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

Ans. (i) 729 = 3 × 3 × 3 × 3 × 3 × 3

∴ 729 = 3 × 3 × 3 = 27


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(ii) 400 = 2 × 2 × 2 × 2 × 5 × 5

∴ 400 = 2 × 2 × 5 = 20

(iii) 1764 = 2 × 2 × 3 × 3 × 7 × 7

∴ 1764 = 2 × 3 × 7 = 42

(iv) 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64


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(v) 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

∴ 7744 = 2 × 2 × 2 × 11 = 8 × 11 = 88

(vi) 9604 = 2 × 2 × 7 × 7 × 7 × 7

∴ 9604 = 2 × 7 × 7 = 98

(vii) 5929 = 7 × 7 × 11 × 11

∴ 5929 = 11 × 7 = 77


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(viii) 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

∴ 9216 = 2 × 2 × 2 × 2 × 2 × 3 = 32 × 3 = 96

(ix) 529 = 23 × 23

∴ 529 = 23

(x) 8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

∴ 8100 = 2 × 3 × 3 × 5 = 90


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Q.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 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768

Ans. (i) 252 = 2 × 2 × 3 × 3 × 7 Here, 7 is unpaired So, we will have to multiply 252 by 7 to make it a perfect square. ∴ 252 × 7 = 2 × 2 × 3 × 3 × 7 × 7

⇒ 1764 = 2 × 3 × 7 = 42

(ii) 180 = 2 × 2 ×6 3 × 3 × 5 Here, 5 is unpaired So, we will have to multiply 180 by 5 to make it a perfect square. ∴ 180 × 5 = 2 × 2 × 3 × 3 × 5 × 5 ⇒ 900 = 2 × 3 × 5 = 30

(iii) 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 Here, 7 is unpaired So, we will have to multiply 1008 by 7 to make it a perfect square. ∴ 1008 × 7 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

⇒ 7056 = 2 × 2 × 3 × 7 = 84


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(iv) 2028 = 2 × 2 × 3 × 13 × 13 Here, 3 is unpaired So, we will have to multiply 2028 by 3 to make it a perfect square. ∴ 2028 × 3 = 2 × 2 × 3 × 3 × 13 × 13

⇒ 6084 = 2 × 3 × 13 = 78

(v) 1458 = 3 × 3 × 3 × 3 × 3 × 3 × 2 Here, 2 is unpaired So, we will have to multiply 1458 by 2 to make it a perfect square. ∴ 1458 × 2 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

⇒ 2916 = 2 × 3 × 3 × 3 = 2 × 27

= 54

(vi) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 Here, 3 is unpaired So, we will have to multiply 768 by 3 to make it a perfect square. ∴ 768 × 3 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

⇒ 2304 = 2 × 2 × 2 × 2 × 3 = 48


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Q.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 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620

Ans. (i) 252 = 2 × 2 × 3 × 3 × 7 Here, 7 is unpaired. If we divide 252 by 7,

we get a perfect square. 252 ÷ 7 = 36 = 2 × 2 × 3 × 3 Which is a perfect square. Therefore, the required smallest number is 7.

Also, 36 = 2 × 3 = 6

(ii) 2925 = 3 × 3 × 5 × 5 × 13 Here, 13 is unpaired. If we divide

2925 by 13, we get a perfect square. 2925 ÷ 13 = 225 = 3 × 3 × 5 × 5 Which is a perfect square. Therefore, the required smallest number is 13.

Also, 225 = 3 × 5 = 15

(iii) 396 = 2 × 2 × 3 × 3 × 11 Here, 11 is unpaired. If we divide

396 by 11, we get a perfect square. 396 ÷ 11 = 36 = 2 × 2 × 3 × 3


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Which is a perfect square. Therefore, the required smallest number is 11.

Also, 36 = 2 × 3 = 6

(iv) 2645 = 5 × 23 × 23 Here, 5 is unpaired. If we divide 2645 by 5,

we get a perfect square. 2645 ÷ 5 = 529 = 23 × 23 Which is a perfect square. Therefore, the required smallest number is 5.

Also, 529 = 23

(v) 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7 Here, 7 is unpaired. If we divide 2800 by 7,

we get a perfect square. 2800 ÷ 7 = 400 = 2 × 2 × 2 × 2 × 5 × 5 Which is a perfect square. Therefore, the required smallest number is 7.

Also, 400 = 2 × 2 × 5 = 20

(vi) 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5 Here, 5 is unpaired. If we divide 1620 by 5,

we get a perfect square. 1620 ÷ 5 = 324 = 2 × 2 × 3 × 3 × 3 × 3 Which is a perfect square. Therefore, the required smallest number is 5.

Also, 324 = 2 × 3 × 3 = 18


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Q.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.

Ans. Let the number of students in the class VIII = x Then, the rupees donated by each student = Rs x According to question, x × x = 2401

⇒ x = 2401

= 7 × 7 × 7 × 7 ∴ x = 7 × 7 = 49 Hence, the number of students in the class is 49.

Q.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.

Ans. Let the number of plants in each row = x Then, the number of rows = x According to question, x × x = 2025

⇒ x = 2025

⇒ x = 3 × 3 × 3 × 3 × 5 × 5 = 3 × 3 × 5 ∴ x = 45 Hence, the number of rows and the numbers of plants in

each row is 45.


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Q.9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Ans. L.C.M. of 4, 9 and 10 = 2 × 2 × 3 × 3 × 5 = 180 Here 5 is unpaired. If we multiply 180 by 5, then we will get a perfect square Since, 180 × 5 = 2 × 2 × 3 × 3 × 5 × 5 Or, 900 = 2 × 2 × 3 × 3 × 5 × 5

⇒ 900 = 2 × 3 × 5 = 30 Hence, required number is 900

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

Ans. L.C.M. of 8, 15 and 20 = 2 × 2 × 2 × 3 × 5 = 120 Here, 2, 3 and 5 are unpaired. If we

multiply 120 by 2, 3 and 5 we will get a perfect square.

Since, 120 × 2 × 3 × 5 = 2 × 2 × 3 × 3 × 5 × 5

∴ 3600 = 2 × 2 × 3 × 5 = 60 Hence, required number is 3600.

Exercise 6.4

Q.1. Find the square root of each of the following numbers by Division Method.

(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900


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Ans. (i)

Therefore, 2304 = 48

(ii)


(iii)


(iv)

Therefore, 529 = 23

(v)



19

(vi)


(vii)


(viii)


(ix)

Therefore, 576 = 24

(x)



20

(xi)


(xii)

Therefore, 900 = 30

Q.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

Ans. We know that if a perfect square is of n-digits, then its

square root will have 2n digits, if n is even or + 1

2n⎛

⎜⎝ ⎠

⎞⎟ , if n

is odd: (i) Here, n = 2 (even) So, number of digits in the square root of 64

= 2n = 2

2 = 1

(ii) Here n = 3 (odd) So, number of digits in the square root of 144

= + 12

n = 3 + 12

= 42

= 2


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(iii) Here n = 4 (even) So, number of digits in the square root of 4489

= 2n = 4

2 = 2

(iv) Here n = 5 (odd) So, number of digits in the square root of 27225

= + 12

n = 5 + 12

= 3

(v) Here n = 6 (even) So, number of digits in the square root of 390625

= 2n = 6

2 = 3

Q.3. Find the square root of the following decimal numbers. (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36

Ans. (i) Therefore, 2.56 = 1.6

(ii)

Therefore, 7.29 = 2.7


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(iii)


(iv)


(v)


Q.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 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000

Ans. (i) Here, the remainder is 2. It shows that 202 is less than 402 by 2. This means if we subtract the remainder from the given number, we will get a perfect square.

Therefore, the required perfect square is 402 – 2 = 400.

and, 400 = 20


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(ii) Here, the remainder is 53. It shows that 442 is less than 1989 by 53. So, the number to be subtracted from 1989 to make it a

perfect square is 53. Therefore, required perfect square = 1989 – 53 = 1936

And, 1936 = 44

(iii) Here, the remainder is 1. It shows that 572 is less than 3250 by 1 So, the number to be subtracted from

3250 to make it a perfect square is 1. Therefore, required perfect square = 3250 – 1 = 3249 and, 3249 = 57

(iv) Here, the remainder is 41. It shows that 282 is less than 825 by 41. So, the number to be subtracted from


(v) Here, the remainder is 31. It shows that 632 is less than 4000 by 31. So, the number to be subtracted from



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Q.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 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412

Ans. (i) Here, the remainder is 41. It shows that 222 is less

than 525 by 41. ⇒ 222 < 525 So, next perfect square number is 232 = 529 Hence, the number to be added = 232 – 525 = 529 – 525 = 4 Square root of 529 = 529 = 23

(ii) Here, the remainder is 69. It shows that 412 is less than 1750 by 69.

⇒ 412 < 1750

So, next perfect square number is 422 = 1764 Hence, the number to be added = 422 – 1750 = 1764 – 1750 = 14

Square root of 1764 = 1764

= 42


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(iii) 252 : Here, we get remainder as 27. It shows that 152 is less than 252 by 27. ⇒ 152 < 252 So, next perfect square number

is 162 = 256 Hence, the number to be added

= 162 – 252 = 256 – 252 = 4 Square root of 256 = 256 = 16

(iv) 1825 : Here, we get remainder as 61. It shows that 422 is less than 1825 by 61.

⇒ 422 < 1825 So, next perfect square number is

432 = 1849. Hence, the number to be added

= 432 – 1825 = 1849 – 1825 = 24 Square root of 1849 = 1849 = 43

(v) 6412 : Here, we get remainder as 12. It shows that 802 is less than 6412 by 12. ⇒ 802 < 6412 So, next perfect square number is

812 = 6561 Hence, the number to be added is 812 – 6412 = 6561 – 6412 = 149 Square root of 6561 = 6561 = 81


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Q.6. Find the length of the side of a square whose area is 441 m2.

Ans. Let side of square = x metre We know that area of square = side2

⇒ x2 = 441

⇒ x = 441 = 21 m Hence, length of the side of the square is 21 m.

Q.7. In a right triangle ABC, ∠B = 90°. (a) If AB = 6 cm, BC = 8 cm, find AC (b) If AC = 13 cm, BC = 5 cm, find AB

Ans. (a) We know that AC2 = AB2 + BC2 (By Pythagoras Theorem) = (6)2 + (8)2 = 36 + 64 ⇒ AC2 = 100 Or, AC = 10 cm

(b) We have AC2 = AB2 + BC2 (By Pythagoras Theorem) ⇒ (13)2 = (AB)2 + (5)2 ⇒ 169 = (AB)2 + 25 ⇒ AB2 = 169 – 25 = 144 Or, AB = 12 cm

Q.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.


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Ans. Here, we get the remainder as 39.

It shows that 312 < 1000 So, next perfect square is 322 = 1024 So, the number to be added

= 322 – 1000 = 1024 – 1000 = 24 So, gardener needs 24 more plants.

Q.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.

Ans. Here, we get the remainder as 16. It show that 222 < 500 So, number to be subtracted from

500 to make it a perfect square = 16 Hence, 16 children would be left out in this arrangement.

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