**Exercise 6.1**

Question 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) 20387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555

**Solution:**

The unit digit of square of a number having ‘a’ at its unit place ends with a×a.

(i) 81 - The given number has it's unit place is 1.

∴ Unit digit of the square of number 8(1)^{2} is equal to (1x1 = 1), as unit’s place is 1.

(ii) 272 - The given number has it's unit place is 2.

∴ Unit digit of the square of number 27(2)^{2} is equal to (2x2 = 4), as unit’s place is 4.

(iii) 799 - The given number has it's unit place is 9

∴ Unit digit of the square of number 79(9)^{2} is equal to (9x9= 81) , as unit’s place is 1.

(iv) 3853 - The given number has it's unit place is 3

∴ Unit digit of the square of number 385(3)^{2} is equal to (3x3=9) , as unit’s place is 9.

(v) 1234 - The given number has it's unit place is 4

∴ Unit digit of the square of number 123(4)^{2} is equal to (4x4=16), as unit’s place is 6.

(vi) 20387 - The given number has it's unit place is 7

∴ Unit digit of the square of number 2638(7 )^{2}is equal to(7x7=49), as unit’s place is 9.

(vii) 52698 - The given number has it's unit place is 8

∴ Unit digit of the square of number 5269(8)^{2} is equal to (8x8= 64), as unit’s place is 4.

(viii) 99880 - The given number has it's unit place is 0

∴ Unit digit of the square of number 9988(0)^{2} is equal to (0x0= 0), as unit’s place is 0.

(ix) 12796 - The given number has it's unit place is 6

∴ Unit digit of the square of number 1279(6)^{2} is equal to (6x6=36), as unit’s place is 6.

(x) 55555 - The given number has it's unit place is 5

∴ Unit digit of the square of number 5555(5)^{2} is equal to (5x5 = 25), as unit’s place is 5.

**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**

Solution:

We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.

(i) 1057 ends with 7 at unit place. So it is not a perfect square number.

(ii) 23453 ends with 3 at unit place. So it is not a perfect square number.

(iii) 7928 ends with 8 at unit place. So it is not a perfect square number.

(iv) 222222 ends with 2 at unit place. So it is not a perfect square number.

(v) 64000 ends with 3 zeros. So it cannot a perfect square number.

(vi) 89722 ends with 2 at unit place. So it is not a perfect square number.

(vii) 22000 ends with 3 zeros. So it can not be a perfect square number.

(viii) 505050 ends with 1 zero. So it is not a perfect square number.

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

**i. 431**

**ii. 2826**

**iii. 7779**

**iv. 82004**

Solution:

We know that the square of an odd number is odd and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

**4. Observe the following pattern and find the missing numbers. 11 ^{2} = 121**

**101 ^{2} = 10201**

**1001 ^{2} = 1002001**

**100001 ^{2} = 1 …….2………1**

**10000001 ^{2} = ……………………..**

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.

∴ 100001^{2} = 10000200001

10000001^{2} = 100000020000001

**5. Observe the following pattern and supply the missing numbers. 112 = 121**

**1012 = 10201**

**101012 = 102030201**

**10101012 = ………………………**

**…………2 = 10203040504030201**

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.

So, 10101012 =1020304030201

1010101012 =10203040505030201

**6. Using the given pattern, find the missing numbers. 1 ^{2} + 2^{2} + 2^{2} = 3^{2}**

**2 ^{2} + 3^{2} + 6^{2} = 7^{2}**

**3 ^{2} + 4^{2} + 12^{2} = 13^{2}**

**4 ^{2} + 5^{2} + _2 = 21^{2}**

**5 + _ ^{2} + 30^{2} = 31^{2}**

**6 + 7 + _ ^{2} = _ ^{2}**

Solution:

Given, 1^{2} + 2^{2} + 2^{2} = 3^{2}

i.e 1^{2} + 2^{2} + (1×2 )^{2} = ( 1^{2} + 2^{2} -1 × 2 )^{2}

2^{2} + 3^{2} + 6^{2} =7^{2}

∴ 2^{2} + 3^{2} + (2×3 )^{2} = (2^{2} + 3^{2} -2 × 3)^{2}

3^{2 }+ 4^{2} + 12^{2} = 13^{2}

∴ 3^{2} + 4^{2} + (3×4 )^{2} = (3^{2} + 4^{2} – 3 × 4)^{2}

4^{2} + 5^{2} + (4×5 )^{2} = (4^{2} + 5^{2} – 4 × 5)^{2}

∴ 4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + (5×6 )^{2} = (5^{2}+ 6^{2} – 5 × 6)^{2}

∴ 5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + (6×7 )^{2} = (6^{2} + 7^{2} – 6 × 7)^{2}

∴ 6^{2} + 7^{2} + 42^{2} = 43^{2}

**7. Without adding, find the sum.**

**i. 1 + 3 + 5 + 7 + 9**

Solution:

Sum of first five odd number = (5)^{2} = 25

**ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19**

Solution:

Sum of first ten odd number = (10)^{2} = 100

**iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23**

Solution:

Sum of first thirteen odd number = (12)^{2} = 144

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

**Solution:**

We know, sum of first n odd natural numbers is n^{2} . Since,49 = 7^{2}

∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

**(ii) Express 121 as the sum of 11 odd numbers. **Solution:

Since, 121 = 11^{2}

∴ 121 = sum of first 11 odd natural numbers = 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**

**ii. 25 and 26**

**iii. 99 and 100**

Solution:

Between n^{2} and (n+1)^{2}, there are 2n non–perfect square numbers.

i. 122 and 132 there are 2×12 = 24 natural numbers.

ii. 252 and 262 there are 2×25 = 50 natural numbers.

iii. 992 and 1002 there are 2×99 =198 natural numbers.

**Exercise 6.2**

**1. Find the square of the following numbers.**

**i. 32**

**ii. 35**

**iii. 86**

**iv. 93**

**v. 71**

**vi. 46**

Solution:

i. (32)^{2}

= (30 +2)^{2}

= (30)^{2} + (2)^{2} + 2×30×2 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 900 + 4 + 120

= 1024

ii. (35)^{2}

= (30+5 )^{2}

= (30)^{2} + (5)^{2} + 2×30×5 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 900 + 25 + 300

= 1225

iii. (86)^{2}

= (90 – 4)^{2}

= (90)^{2} + (4)^{2} – 2×90×4 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 8100 + 16 – 720

= 8116 – 720

= 7396

iv. (93)^{2}

= (90+3 )^{2}

= (90)^{2} + (3)^{2} + 2×90×3 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 8100 + 9 + 540

= 8649

v. (71)^{2}

= (70+1 )^{2}

= (70)^{2} + (1)^{2} +2×70×1 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 4900 + 1 + 140

= 5041

vi. (46)^{2}

= (50 -4 )^{2}

= (50)^{2} + (4)^{2} – 2×50×4 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 2500 + 16 – 400

= 2116

**2. Write a Pythagorean triplet whose one member is.**

**i. 6**

**ii. 14**

**iii. 16**

**iv. 18**

Solution:

For any natural number m, we know that 2m, m2–1, m2+1 is a Pythagorean triplet.

i. 2m = 6

⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

∴ (6, 8, 10) is a Pythagorean triplet.

ii. 2m = 14

⇒ m = 14/2 = 7

m2–1= 72–1 = 49–1 = 48

m2+1 = 72+1 = 49+1 = 50

∴ (14, 48, 50) is not a Pythagorean triplet.

iii. 2m = 16

⇒ m = 16/2 = 8

m2–1 = 82–1 = 64–1 = 63

m2+ 1 = 82+1 = 64+1 = 65

∴ (16, 63, 65) is a Pythagorean triplet.

iv. 2m = 18

⇒ m = 18/2 = 9

m2–1 = 92–1 = 81–1 = 80

m2+1 = 92+1 = 81+1 = 82

∴ (18, 80, 82) is a Pythagorean triplet.

**Exercise 6.3**

**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**

Solution:

i. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9^{2}=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unit’s digit of the square of a number having digit as unit’s

place 6 is 6 and also 4 is 6 [6^{2}=36 and 4^{2}=16, both the squares have unit digit 6].

∴ Unit’s digit of the square root of number 99856 is equal to 6.

iii. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9^{2}=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unit’s digit of the square of a number having digit as unit’s

place 5 is 5.

∴ Unit’s digit of the square root of number 657666025 is equal to 5.

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

**i. 153**

**ii. 257**

**iii. 408**

**iv. 441**

Solution:

We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.

i. 153⟹ Ends with 3.

∴, 153 is not a perfect square

ii. 257⟹ Ends with 7

∴, 257 is not a perfect square

iii. 408⟹ Ends with 8

∴, 408 is not a perfect square

iv. 441⟹ Ends with 1

∴, 441 is a perfect square.

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

Solution:

100

100 – 1 = 99

99 – 3 = 96

96 – 5 = 91

91 – 7 = 84

84 – 9 = 75

75 – 11 = 64

64 – 13 = 51

51 – 15 = 36

36 – 17 = 19

19 – 19 = 0

Here, we have performed subtraction ten times.

∴ √100 = 10

169

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

Here, we have performed subtraction thirteen times.

∴ √169 = 13

**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**

Solution:

(i) We have 729

Prime factors of 729

729 = 3 × 3 × 3 × 3 × 3 × 3 = 3^{2} × 3^{2} × 3^{2}

√729 = 3 × 3 × 3 = 27

(ii) We have 400

Prime factors of 400

400 = 2 × 2 × 2 × 2 × 5 × 5 = 2^{2} × 2^{2} × 5^{2}

√400 = 2 × 2 × 5 = 20

(iii) 1764

1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2^{2} × 3^{2} × 7^{2}

√1764 = 2 × 3 × 7 = 42

(iv) 4096

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2}

√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v) 7744

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

= 2^{2} × 2^{2} × 2^{2} × 11^{2}

√7744 = 2 × 2 × 2 × 11 = 88

(vi) Prime factorisation of 9604 is

9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2^{2} × 7^{2} × 7^{2}

√9604 = 2 × 7 × 7 = 98

(vii) Prime factorisation of 5929 is

5929 = 7 × 7 × 11 × 11 = 7^{2} × 11^{2}

√5929 = 7 × 11 = 77

(viii) Prime factorisation of 9216 is

9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3

= 2^{2} × 2^{2} × 2^{2} × 2^{2} × 2^{2} × 3^{2}

√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

(ix) Prime factorisation of 529 is

529 = 23 × 23 = 23^{2}

√529 = 23

(x) Prime factorisation of 8100 is

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2^{2} × 3^{2} × 3^{2} × 5^{2}

√8100 = 2 × 3 × 3 × 5 = 90

Question 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

Solution:

(i) Prime factorisation of 252 is

252 = 2 × 2 × 3 × 3 × 7

Here, the prime factorisation is not in pair. 7 has no pair.

Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.

The new square number is 252 × 7 = 1764

Square root of 1764 is

√1764 = 2 × 3 × 7 = 42

(ii) Prime factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

New square number = 180 × 5 = 900

The square root of 900 is

√900 = 2 × 3 × 5 = 30

Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

(iii) Prime factorisation of 1008 is

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, 7 has no pair.

New square number = 1008 × 7 = 7056

Thus, 7 is the required number.

Square root of 7056 is

√7056 = 2 × 2 × 3 × 7 = 84

(iv) Prime factorisation of 2028 is

2028 = 2 × 2 × 3 × 13 × 13

Here, 3 is not in pair.

Thus, 3 is the required smallest whole number.

New square number = 2028 × 3 = 6084

Square root of 6084 is

√6084 = 2 × 13 × 3 = 78

(v) Prime factorisation of 1458 is

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, 2 is not in pair.

Thus, 2 is the required smallest whole number.

New square number = 1458 × 2 = 2916

Square root of 1458 is

√2916 = 3 × 3 × 3 × 2 = 54

(vi) Prime factorisation of 768 is

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, 3 is not in pair.

Thus, 3 is the required whole number.

New square number = 768 × 3 = 2304

Square root of 2304 is

√2304 = 2 × 2 × 2 × 2 × 3 = 48

Question 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

Solution:

(i) Prime factorisation of 252 is

252 = 2 × 2 × 3 × 3 × 7

Here 7 has no pair.

7 is the smallest whole number by which 252 is divided to get a square number.

New square number = 252 ÷ 7 = 36

Thus, √36 = 6

(ii) Prime factorisation of 2925 is

2925 = 3 × 3 × 5 × 5 × 13

Here, 13 has no pair.

13 is the smallest whole number by which 2925 is divided to get a square number.

New square number = 2925 ÷ 13 = 225

Thus √225 = 15

(iii) Prime factorisation of 396 is

396 = 2 × 2 × 3 × 3 × 11

Here 11 is not in pair.

11 is the required smallest whole number by which 396 is divided to get a square number.

New square number = 396 ÷ 11 = 36

Thus √36 = 6

(iv) Prime factorisation of 2645 is

2645 = 5 × 23 × 23

Here, 5 is not in pair.

5 is the required smallest whole number.

By which 2645 is multiplied to get a square number

New square number = 2645 ÷ 5 = 529

Thus, √529 = 23

(v) Prime factorisation of 2800 is

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

Here, 7 is not in pair.

7 is the required smallest number.

By which 2800 is multiplied to get a square number.

New square number = 2800 ÷ 7 = 400

Thus √400 = 20

(vi) Prime factorisation of 1620 is

1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

Here, 5 is not in pair.

5 is the required smallest prime number.

By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324

Thus √324 = 18

Question 7.

The students of class VIII of a school donated ₹ 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.

Solution:

Total amount of money donated = ₹ 2401

Total number of students in the class = √2401

= 7×7×7×7

= (7×7)×(7×7)

= 49×49

= √(49×49)

= 49

∴ The number of students = 49

Question 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.

Solution:

Total number of rows = Total number of plants in each row = √2025

= 3×3×3×3×5×5

= (3×3)×(3×3)×(5×5)

= (3×3×5)×(3×3×5)

= 45×45

= √45×45

= 45

∴ The number of rows = 45 and the number of plants in each rows = 45.

Question 9.

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

Solution:

LCM of 4, 9, 10 = 180

The least number divisible by 4, 9 and 10 = 180

Now prime factorisation of 180 is

180 = 2 × 2 × 3 × 3 × 5

Here, 5 has no pair.

The required smallest square number = 180 × 5 = 900

Question 10.

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

Solution:

The smallest number divisible by 8, 15 and 20 is equal to their LCM.

LCM = 2 × 2 × 2 × 3 × 5 = 120

Here, 2, 3 and 5 have no pair.

The required smallest square number = 120 × 2 × 3 × 5 = 120 × 30 = 3600

**Exercise 6.4**

Question 1.

Find the square root of each of the following numbers by Long 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

Solution:

Question 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

Solution:

We know that if n is number of digits in a square number then

Number of digits in the square root = n/2 if n is even and n+1/2 if n is odd.

(i) 64

Here n = 2 (even)

Number of digits in √64 = 2/2 = 1

(ii) 144

Here n = 3 (odd)

Number of digits in square root = 3+1/2 = 2

(iii) 4489

Here n = 4 (even)

Number of digits in square root =4/2= 2

(iv) 27225

Here n = 5 (odd)

Number of digits in square root = 5+1/2 = 3

(iv) 390625

Here n = 6 (even)

Number of digits in square root = 6/2 = 3

Question 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

Solution:

Question 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

Solution:

(i)

Here remainder is 2

2 is the least required number to be subtracted from 402 to get a perfect square

New number = 402 – 2 = 400

Thus, √400 = 20

(ii)

Here remainder is 53

53 is the least required number to be subtracted from 1989.

New number = 1989 – 53 = 1936

Thus, √1936 = 44

(iii)

Here remainder is 1

1 is the least required number to be subtracted from 3250 to get a perfect square.

New number = 3250 – 1 = 3249

Thus, √3249 = 57

(iv)

Here, the remainder is 41

41 is the least required number which can be subtracted from 825 to get a perfect square.

New number = 825 – 41 = 784

Thus, √784 = 28

(v)

Here, the remainder is 31

31 is the least required number which should be subtracted from 4000 to get a perfect square.

New number = 4000 – 31 = 3969

Thus, √3969 = 63

Question 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

Solution:

(i)

Here remainder is 41

It represents that square of 22 is less than 525.

Next number is 23 an 23^{2} = 529

Hence, the number to be added = 529 – 525 = 4

New number = 529

Thus, √529 = 23

(ii)

Here the remainder is 69

It represents that square of 41 is less than in 1750.

The next number is 42 and 42^{2} = 1764

Hence, number to be added to 1750 = 1764 – 1750 = 14

Require perfect square = 1764

√1764 = 42

(iv)

Here the remainder is 27.

It represents that a square of 15 is less than 252.

The next number is 16 and 16^{2} = 256

Hence, number to be added to 252 = 256 – 252 = 4

New number = 252 + 4 = 256

Required perfect square = 256

and √256 = 16

(v)

Here, the remainder is 12.

It represents that a square of 80 is less than in 6412.

The next number is 81 and 81^{2} = 6561

Hence the number to be added = 6561 – 6412 = 149

The require perfect square is 6561 and √6561 = 81

Question 6.

Find the length of the side of a square whose area = 441 m^{2}

Solution:

Let the length of the side of the square be x m.

Area of the square = (side)^{2} = x^{2} m^{2}

x^{2} = 441 ⇒ x = √441 = 21

Thus, x = 21 m.

Hence the length of the side of square = 21 m.

Question 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

Solution:

(a) In right triangle ABC

AC^{2} = AB^{2} + BC^{2} [By Pythagoras Theorem]

⇒ AC^{2} = (6)^{2} + (8)^{2} = 36 + 64 = 100

⇒ AC = √100 = 10

Thus, AC = 10 cm.

(b) In right triangle ABC

AC^{2} = AB^{2} + BC^{2} [By Pythagoras Theorem]

⇒ (13)^{2} = AB^{2} + (5)^{2}

⇒ 169 = AB^{2} + 25

⇒ 169 – 25 = AB^{2}

⇒ 144 = AB^{2}

AB = √144 = 12 cm

Thus, AB = 12 cm.

Question 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 the same. Find the minimum number of plants he needs more for this.

Solution:

Let the number of rows be x.

And the number of columns also be x.

Total number of plants = x × x = x^{2}

x^{2} = 1000 ⇒ x = √1000

Here the remainder is 39

So the square of 31 is less than 1000.

Next number is 32 and 32^{2} = 1024

Hence the number to be added = 1024 – 1000 = 24

Thus the minimum number of plants required by him = 24.

Question 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 the number of columns. How many children would be left out in this arrangement?

Solution:

Let the number of children in a row be x. And also that of in a column be x.

Total number of students = x × x = x^{2}

x^{2} = 500 ⇒ x = √500

Here the remainder is 16

New Number 500 – 16 = 484

and, √484 = 22

Thus, 16 students will be left out in this arrangement.