CHAPTER- 6 SQUARES AND SQUARE ROOTS

Explanation:

The unit digit of 81 is ‘1’. Unit digit of ‘1×1’ is 1. So,

Unit digit in square of 81 is 1.

Explanation:

 The unit digit of 272 is ‘2’. Unit digit of ‘2×2’ is 4. So,

Unit digit in square of 272 is 4.

Explanation:

The unit digit of 799 is ‘9’. Unit digit of ‘9×9’ is 1. So,

Unit digit in square of 799 is 1.

Explanation:

The unit digit of 3853 is ‘3’. Unit digit of ‘3×3’ is 9. So,

Unit digit in square of 3853 is 9.

Explanation:

The unit digit of 1234 is ‘4’. Unit digit of ‘4×4’ is 6. So,

Unit digit in square of 1234 is 6.

Explanation:

The unit digit of 26387 is ‘7’. Unit digit of ‘7×7’ is 9. So,

Unit digit in square of 26387 is 9.

Explanation:

The unit digit of 52698 is ‘8’. Unit digit of ‘8×8’ is 4. So,

Unit digit in square of 52698 is 4.

Explanation:

 The unit digit of 99880 is ‘0’. Unit digit of ‘0×0’ is 0. So,

Unit digit in square of 99880 is 0.

Explanation:

The unit digit of 12796 is ‘6’. Unit digit of ‘6×6’ is 6. So,

Unit digit in square of 12796 is 6.

Explanation:

The unit digit of 55555 is ‘5’. Unit digit of ‘5×5’ is 5. So,

Unit digit in square of 55555 is 5.

Explanation:

If the unit digit of any natural number is 2, 3, 7 or 8 then it is not a perfect square.

If the unit digit of any number is 0, 1, 4, 5, 6, 9 then it may or may not be a perfect square. To be a perfect square the number has to be the sum of successive odd numbers starting from 1.

i) 1057 ⇨ unit digit is 7

ii) 23453 ⇨ unit digit is 3

iii) 7928 ⇨ unit digit is 8

iv) 222222 ⇨ unit digit is 2

v) 64000 ⇨ It is not a sum of successive odd numbers.

vi) 89722 ⇨ unit digit is 2

vii) 222000 ⟹ It is not a sum of successive odd numbers.

viii) 505050 ⟹ It is not a sum of successive odd numbers.

Explanation:

If the unit digit of any number is odd then its square is also an odd number. Similar is the case for even numbers.

i) 431 is an odd number so it’s square is also odd number.

ii) 2826 is an even number so it’s square is also even number.

iii) 7779 is an odd number so it’s square is also odd number.

iv) 82004 is an even number so it’s square is also even number.

Explanation:

After observing the patterns we conclude that the result of square of these number have odd number of digits in it and it contains digit 2 in the middle. Also, there are zero between left most 1 and middle digit 2 and right most 1 and middle digit 2. The number of zero between these are equal to the number of zeros in the original number. Hence, the total number of zeros in the result is 2 times the number of zeros in the original number.

So,

⇨ 1000012 will be equal to 10000200001

⇨ 100000012 will be equal to 100000020000001

Explanation:

After observing the patterns we conclude that the total number of digit in the result is an odd number with digit x/2 (x is total number of digit in the original number) in the middle. The total number of zeros in the result is 2 times the number of zeros in the original number. These zeros are placed after every digit and the result starts and ends with digit 1 and increases after every zero. The result is symmetric about the middle digit. 

So, 

⇨ 10101012 will be equal to 1020304030201

⇨ 1010101012 will be equal to 10203040505030201

Explanation:

If we remove the squares from all the term on the both sides, then we can observe that the third term in LHS is equal to the multiplication of first and second term of LHS. The term in RHS is 1 greater than the third term of LHS.  

⇨ 12 + 22 + 22

⇨ 1 + 2 + (1×2) 

⇨ 22 + 32 + 62 = 72

⇨ 2 + 3 + (2×3) = 2×3 + 1

⇨ 32 + 42 + 122 = 132

⇨ 3 + 4 + (3×4) = 3×4 + 1

⇨ 42 + 52 + _2 = 212  

⇨ 4 + 5 + 4×5 = 4×5 + 1

⇨ 42 + 52 + 202 = 212

⇨ 52 + _2 + (5×6)2 = 312

⇨ 5 + 6 + 5×6 = 5×6 + 1

⇨ 52 + 62 + 302 = 312

⇨ 62 + 72 + _2 = _2

⇨ 6 + 7 + 6×7 = 6×7 + 1

⇨ 62 + 72 + 422 = 432

Explanation:

i) There are 5 odd numbers in the term. So,

The sum of first 5 odd numbers will be 52 which is equal to 25. So,

⇨ 1 + 3 + 5 + 7 + 9 = 25

ii) There are 10 odd numbers in the term. So,

The sum of first 10 odd numbers will be 102 which is equal to 100. So,

⇨ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

iii) There are 12 odd numbers in the term. So,

The sum of first 12 odd numbers will be 122 which is equal to 144. So,

⇨ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 144

Explanation:

If a number is a perfect square of any number (let x) then the perfect square can be represented as sum of first x odd natural numbers.

i) 49 = 72. Hence, 49 can be represented as sum of first 7 odd natural numbers. So,

⇨ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 

ii) 121 = 112. Hence, 121 can be represented as sum of first 11 odd natural numbers. So,

⇨ 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Explanation:

Total Number of natural numbers present between n2 and (n+1)2 is-

((n+1)2 – n2) – 1 = n2 + 1 + 2×n – n2 – 1 = 2n

i) So, natural numbers between 122 and 132 is = 2×12 = 24

ii) So, natural numbers between 252 and 262 is = 2×25 = 50

iii) So, natural numbers between 992 and 1002 is = 2×99 = 198

Explanation:

(32)2

= (30 + 2)2

= 30(30+2) + 2(30+2)

= 30×30 + 30×2 + 2×30 + 2×2

= 900 + 60 + 60 + 4 

= 1024

Explanation:

 (35)2

= (30+5)2

= 30(30+5) + 5(30+5)

= 30×30 + 30×5 + 5×30 + 5×5

= 900 + 150 + 150 + 25 

= 1225

Explanation:

 (86)2

= (90 – 4)2

= 90(90– 4) – 4(90– 4)

= 90×90 – 90×4 – 4×90 + 4×4

= 8100 – 360 – 360 + 16 

= 7396

Explanation:

 (93)2

= (90+3)2

= 90(90+3) + 3(90+3)

= 90×90 + 90×3 + 3×90 + 3×3

= 8100 + 270 + 270 + 9 

= 8649

Explanation:

 (71)2

= (70+1)2

= 70(70+1) + 1(70+1)

= 70×70 + 70×1 + 1×70 + 1×1

= 4900 + 70 + 70 + 1 

= 5041

Explanation:

 (46)2

= (50 – 4)2

= 50(50– 4) – 4(50– 4)

= 50×50 – 50×4 – 4×50 + 4×4

= 2500 – 200 – 200 + 16 

= 2116

Explanation:

We know that, for a natural number x its Pythagorean triplets are 2x, x2–1, x2+1.

2x = 6

⇨ x = 6/2 

⇨ x = 3

x2 –1 = 32 – 1 = 9 – 1 = 8

x2+1 = 32 + 1 = 9 + 1 = 10

Hence, 6,8,10 is a Pythagorean Triplet.

Explanation:

2x = 14

⇨ x = 14/2 

⇨ x = 7

x2 – 1 = 72 – 1 = 49 – 1 = 48

x2 + 1 = 72 + 1 = 49 + 1 = 50

Hence, 14, 48, 50 is a Pythagorean Triplet.

Explanation:

 2x = 16

⇨ x = 16/2 

⇨ x = 8

x2 – 1 = 82 – 1 = 64 – 1 = 63

x2 + 1 = 82 + 1 = 64 + 1 = 65

Hence, 16, 63, 65 is a Pythagorean Triplet.

Explanation:

2x = 18

⇨ x = 18/2 

⇨ x = 9

x2 – 1 = 92 – 1 = 81 – 1 = 80

x2 + 1 = 92 + 1 = 81 + 1 = 82

Hence, 18, 80, 82 is a Pythagorean Triplet.

Explanation:

Unit digit in square of any natural number with different unit digit are given below-

1 ⇨ 1, 2 ⇨ 4, 3 ⇨ 9, 4 ⇨ 6, 5 ⇨ 5, 6 ⇨ 6, 7 ⇨ 9, 8 ⇨ 4, 9 ⇨ 1

i) Unit digit of 9801 is 1. So, it’s square root may have unit digit of 1 or 9.

ii) Unit digit of 99856 is 6. So, it’s square root may have unit digit of 4 or 6.

iii) Unit digit of 998001 is 1. So, it’s square root may have unit digit of 1 or 9.

iv) Unit digit of 657666025 is 5. So, it’s square root must have unit digit of 5.

Explanation:

If the unit digit of any natural number is 2, 3, 7 or 8 then it is not a perfect square.

If the unit digit of any number is 0, 1, 4, 5, 6, 9 then it may or may not be a perfect square. To be a perfect square the number has to be the sum of successive odd numbers starting from 1.

i) 153 ⇨ Unit digit is 3.

So, 153 is not a perfect square

ii) 257 ⇨ Unit digit is 7.

So, 257 is not a perfect square

iii) 408 ⇨ Unit digit is 8.

So, 408 is not a perfect square

iv) 441 ⇨ Unit digit is 1.

441 can also be represented as sum of first 21 odd numbers.

So, 441 is a perfect square.

Explanation:

In this method we subtract all the odd numbers starting from 1 with the given number. If the given number becomes 0 at any step then that number of step will be the square root of the given number.

⇨ for 100

100 – 1 = 99 ⇨ subtraction no. 1

99 – 3 = 96 ⇨ subtraction no. 2

96 – 5 = 91 ⇨ subtraction no. 3

91 – 7 = 84 ⇨ subtraction no. 4

84 – 9 = 75 ⇨ subtraction no. 5

75 – 11 = 64 ⇨ subtraction no. 6

64 – 13 = 51 ⇨ subtraction no. 7

51 – 15 = 36 ⇨ subtraction no. 8

36 – 17 = 19 ⇨ subtraction no. 9

19 – 19 = 0 ⇨ subtraction no. 10

The given number becomes 0 at the step no. 10. Hence, square root of 100 is 10.

⇨ for 169

169 – 1 = 168 ⇨ subtraction no. 1

168 – 3 = 165 ⇨ subtraction no. 2

165 – 5 = 160 ⇨ subtraction no. 3

160 – 7 = 153 ⇨ subtraction no. 4

153 – 9 = 144 ⇨ subtraction no. 5

144 – 11 = 133 ⇨ subtraction no. 6

133 – 13 = 120 ⇨ subtraction no. 7

120 – 15 = 105 ⇨ subtraction no. 8

105 – 17 = 88 ⇨ subtraction no. 9

88 – 19 = 69 ⇨ subtraction no. 10

69 – 21 = 48 ⇨ subtraction no. 11

48 – 23 = 25 ⇨ subtraction no. 12

25 – 25 = 0 ⇨ subtraction no. 13

The given number becomes 0 at the step no. 13. Hence, square root of 169 is 13.

Explanation:

⇨ 729 = 3×3×3×3×3×3 (pairing the factors)

⇨ 729 = (3×3×3)×(3×3×3) (form one pair)

⇨ 729 = (3×3×3)2 (put square root on LHS and RHS)

⇨ √729 = 3×3×3 

⇨ √729 = 27

Explanation:

400

⇨ 400 = 5×5×2×2×2×2×1 (prime factors)

⇨ 400 = (5×5)×(2×2)×(2×2) (pairing the factors)

⇨ 400 = (5×2×2)×(5×2×2) (forming a single pair)

⇨ 400 = (5×2×2)2 (put square root on LHS and RHS)

⇨ √400 = 5×2×2 

⇨ √400 = 20

Explanation:

1764

⇨ 1764 = 7×7×3×3×2×2 (prime factors)

⇨ 1764 = (7×3×2)×(7×3×2) (form one pair)

⇨ 1764 = (7×3×2)2 (put square root on LHS and RHS)

⇨ √1764 = 7 ×3×2 

⇨ √1764 = 42

Explanation:

 4096

⇨ 4096 = 2×2×2×2×2×2×2×2×2×2×2×2 (prime factors)

⇨ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2) (from one pair)

⇨ 4096 = (2×2×2×2×2×2)2 (put square root on LHS and RHS)

⇨ √4096 = 2×2×2×2×2×2 

⇨ √4096 = 64

Explanation:

7744

⇨ 7744 = 11×11×2×2×2×2×2×2 (prime factors)

⇨ 7744 = (11×2×2×2)×(11×2×2×2) (form one pair)

⇨ 7744 = (11×2×2×2)2 (put square root on LHS and RHS)

⇨ √7744 = 2×2×2×11 

⇨ √7744 = 88

Explanation:

9604

⇨ 9604 = 7 × 7 × 7 × 7 × 2 × 2 (prime factors)

⇨ 9604 = (7 × 7 ×2) × (7 × 7 ×2) (form one pair)

⇨ 9604 = (2×7×7)2 (put square root on LHS and RHS)

⇨ √9604 = 2×7×7

⇨ √9604 = 98

Explanation:

 5929

⇨ 5929 = 11×11×7×7 (prime factors)

⇨ 5929 = (7×11)×(7×11) (form single pair)

⇨ 5929 = (7×11)2 (put square root on LHS and RHS)

⇨ √5929 = 7×11 

⇨ √5929 = 77

Explanation:

9216

⇨ 9216 = 3×3×2×2×2×2×2×2×2×2×2×2 (prime factors)

⇨ 9216 = (3×3)×(2×2)×(2×2)×(2×2)×(2×2)×(2×2)

⇨ 9216 = (3 × 2 × 2 × 2 × 2 × 2) × (3 × 2 × 2 × 2 × 2 × 2) (form one pair)

⇨ 9216 = (96 × 96) 

⇨ 9216 = (96)2 (put square root on LHS and RHS)

⇨ √9216 = 96

Explanation:

⇨ 529 = 23×23 (prime factors)

⇨ 529 = (23)2 (put square root on LHS and RHS)

⇨ √529 = 23

Explanation:

8100

⇨ 8100 = 5×5×3×3×3×3×2×2×1 (prime factors)

⇨ 8100 = (5×3×3×2)×(5×3×3×2) (form one pair)

⇨ 8100 = 90×90 

⇨ 8100 = (90)2 (put square root on LHS and RHS)

⇨ √8100 = 90

Explanation:

i) 252 

⇨ 252 = 7×3×3×2×2 (prime factors)

⇨ 252 = (3×3)×(2×2)×7

7 remains unpaired. So, smallest number to be multiplied is 7.

Number becomes = 252×7 = 1764

⇨ 1764 = (7×7)×(3×3)×(2×2)

⇨ 1764 = (7×3×2)2 (put square root on LHS and RHS)

⇨ √1764 = 7×3×2 = 42

ii) 180

⇨ 180 = 3×3×2×2×5 (prime factors)

⇨ 180 = (3×3)×(2×2)×5

5 remains unpaired. So, smallest number to be multiplied is 5.

Number becomes = 180×5 = 900

⇨ 900 = (5×5)×(3×3)×(2×2)

⇨ 900 = (5×3×2)2 (put square root on LHS and RHS)

⇨ √900 = 5×3×2 = 30

iii) 1008

⇨ 1008 = 3×3×2×2×2×2×7 (prime factors)

⇨ 1008 = (3×3)×(2×2)×(2×2)×7

7 remains unpaired. So, smallest number to be multiplied is 7.

Number becomes = 1008×7 = 7056

⇨ 7056 = (7×7)×(2×2)×(2×2)×(3×3)

⇨ 7056 = (7×3×2×2)2 (put square root on LHS and RHS)

⇨ √7056 = 3×2×2×7 = 84

iv) 2028

⇨ 2028 = 13×13×3×2×2 (prime factors)

⇨ 2028 = (13×13)×(2×2)×3

3 remains unpaired. So, smallest number to be multiplied is 3.

Number becomes = 2028×3 = 6084

⇨ 6084 = (13×13)×(3×3)×(2×2)

⇨ 6084 = (13×3×2)2 (put square root on LHS and RHS)

⇨ √6084 = 13×3×2 = 78

v) 1458

⇨ 1458 = 3×3×3×3×3×3×2

⇨ 1458 = (3×3)×(3×3)×(3×3)×2

2 remains unpaired. So, smallest number to be multiplied is 2.

Number becomes = 1458 × 2 = 2916

⇨ 2916 = (2×2)×(3×3)×(3×3)×(3×3)

⇨ 2916 = (2×3×3×3)2 (put square root on LHS and RHS)

⇨ √2916 = 2×3×3×3 = 54

vi) 768

⇨ 768 = 3×2×2×2×2×2×2×2×2

⇨ 768 = (2×2)×(2×2)×(2×2)×(2×2)×3

3 remains unpaired. So, smallest number to be multiplied is 3.

Number becomes = 768×3 = 2304

⇨ 2304 = (3×3)×(2×2)×(2×2)×(2×2)×(2×2)

⇨ 2304 = (3×2×2×2×2)2 (put square root on LHS and RHS)

⇨ √2304 = 3×2×2×2×2 = 48

Explanation:

i) 252

⇨ 252 = 2×2×3×3×7 (prime factors)

⇨ 252 = (3×3)×(2×2)×7

7 remains unpaired. So, smallest number to be divided is 7.

Number becomes = 252 ÷ 7 = 36

⇨ 36 = (3×3)×(2×2)

⇨ 36 = (3×2)2 (put square root on LHS and RHS)

⇨ √36 = 3×2 = 6

ii) 2925

⇨ 2925 = 5×5×3×3×13

⇨ 2925 = (5×5)×(3×3)×13

13 remains unpaired. So, smallest number to be divided is 13.

Number becomes = 2925 ÷ 13 = 225

⇨ 225 = (5×5)×(3×3)

⇨ 225 = (3×5)2 (put square root on LHS and RHS)

⇨ √36 = 3×5 = 15

iii) 396

⇨ 396 = 3×3×2×2×11

⇨ 396 = (3×3)×(2×2)×11

13 remains unpaired. So, smallest number to be divided is 13.

Number becomes = 396 ÷ 11 = 36

⇨ 36 = (3×3)×(2×2)

⇨ 36 = (3×2)2 (put square root on LHS and RHS)

⇨ √36 = 3×2 = 6

iv) 2645

⇨ 2645 = 23×23×5

⇨ 2645 = (23×23)×5

13 remains unpaired. So, smallest number to be divided is 13.

Number becomes = 2645 ÷ 5 = 529

⇨ 529 = 23×23

⇨ 529 = (23)2 (put square root on LHS and RHS)

⇨ √529 = 23

v) 2800

⇨ 2800 = 5×5×2×2×2×2×7 (prime factors)

⇨ 2800 = (5×5)×(2×2)×(2×2)×7

13 remains unpaired. So, smallest number to be divided is 13.

Number becomes = 2800 ÷ 7 = 400

⇨ 400 = (5×5)×(2×2)×(2×2)

⇨ 400 = (2×2×5)2 (put square root on LHS and RHS)

⇨ √400 = 2×2×5

⇨ √400 = 20

vi) 1620

⇨ 1620 = 3×3×3×3×2×2×5

⇨ 1620 = (3×3)×(3×3)×(2×2)×5

13 remains unpaired. So, smallest number to be divided is 13.

Number becomes = 1620 ÷ 5 = 324

⇨ 324 = (3×3)×(3×3)×(2×2)

⇨ 324 = (2×3×3)2 (put square root on LHS and RHS)

⇨ √324 = (2×3×3)

⇨ √324 = 18

Explanation:

Assume the number of students in class is y.

Rs. y was donated by each student.

Total amount given by class = y×y = y2 

It is given, y2 = Rs. 2401

⇨ y2 = 7×7×7×7 (prime factors)

⇨ y2 = (7×7)×(7×7)

⇨ y2 = (49×49) (take square root)

⇨ y = √(49×49)

⇨ y = 49

So, number of students in class VIII is = 49

Explanation:

Assume number of rows is, y.

So, each row contains y number of plants.

Total number of plants in the garden = y × y = y2

It has been given that,

⇨ y2 = 2025

⇨ y2 = 3×3×3×3×5×5 (prime factors)

⇨ y2 = (3×3)×(5×5)×(3×3)

⇨ y2 = (3×5×3)×(3×5×3)

⇨ y2 = 45×45 (put square root on LHS and RHS)

⇨ y = √45×45

⇨ y = 45

So, number of rows is = 45

Number of plants each row contains is = 45.

Explanation:

First let us find smallest number divisible by 4, 9 and 10. So,

L.C.M of 4, 9 and 10 is (2×2×9×5) = 180.

Now,

⇨ 180 = 2×2×9×5 (prime factors)

⇨ 180 = (3×3)×2×2×5

⇨ 180 = (3×3)×(2×2)×5

5 remains unpaired. So, let us multiply the number 5.

Number becomes = 180×5 = 900

So, the smallest perfect square divisible by 4, 9 and 10 = 900

Explanation:

First let us find smallest number divisible by 4, 9 and 10. So,

L.C.M of 8, 15 and 20 is (2×2×5×2×3) = 120.

Now,

⇨ 120 = 2×2×3×5×2 (prime factors)

⇨ 120 = (2×2)×3×5×2

3, 5 and 2 remains unpaired.

So, multiply 120 with (3×5×2) = 30 to get a perfect square.

New number = 120×30 = 3600

So, the smallest perfect number divisible by 8, 15 and 20 = 3600

Explanation:

2304

Explanation:

4489

Explanation:

3481

Explanation:

529

Explanation:

 3249

Explanation:

1369

Explanation:

5776

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Place a bar over every pair of digits starting from the one’s digit. The number of bars on a number represent the number of digits in its square root.

i) 64 ⇨ 64
There is only one bar so number of digits in square root is 1.

ii) 144 ⇨ 1 44

There are 2 bars so number of digits is 2.

iii) 4489 ⇨ 44 89

There are 2 bars so number of digits is 2.

iv) 27225 ⇨ 2 72 25

There are 3 bars so number of digits is 3.

v) 390625 ⇨ 39 06 25

There are 3 bars so number of digits is 3.

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

i) 402

The remainder is 2. So, subtract 2 from the number.

New number = 400

ii) 1989

The remainder is 53. So, subtract 53 from the number.

New number = 1936

iii) 3250

The remainder is 1. So, subtract 1 from the number.

New number = 3249

iv) 825

The remainder is 41. So, subtract 41 from the number.

New number = 784

v) 4000

The remainder is 31.So, subtract 31 from the number.

New number = 3969

Explanation:

525

Here, Quotient = 22, Remainder = 41

Number to be added 

= 2×Quotient – Remainder + 1

= 2×22 – 41 + 1

= 4

So, Perfect square number = 525 + 4 = 529

Explanation:

1750

Here, Quotient = 41, Remainder = 69

Number to be added 

= 2×Quotient – Remainder + 1

= 2×41 – 69 + 1

= 14

So, Perfect square number = 1750 + 14 = 1764

Explanation:

 252

Here, Quotient = 15, Remainder = 27

Number to be added 

= 2×Quotient – Remainder + 1

= 2×15 – 27 + 1

= 4

So, Perfect square number = 252 + 4 = 256

Explanation:

1825

Here, Quotient = 42, Remainder = 61

Number to be added 

= 2×Quotient – Remainder + 1

= 2×42 – 61 + 1

= 24

So, Perfect square number = 1825 + 24 = 1849

Explanation:

6412

Here, Quotient = 80, Remainder = 12

Number to be added 

= 2×Quotient – Remainder + 1

= 2×80 – 12 + 1

= 149

So, Perfect square number = 6412 + 149 = 6561

Explanation:

Let us assume square’s side is = x

It is given that it has area of = 441 m2

⇨ x2 = 441 m2 

⇨ x = √441 m

⇨ x = 21

Length of the side is 21m.

Explanation:

(a)

It is given that, BC = 8 cm, AB = 6 cm

Using Pythagoras theorem-

⇨ AC2 = AB2 + BC2

⇨ AC2 = 62 + 82

⇨ AC2 = 36 + 64

⇨ AC2 = 100

⇨ AC = √100 = 10

⇨ AC = 10 cm.

(b)

It is given that, BC = 5 cm, AC = 13 cm

Using Pythagoras theorem-

⇨ AC2 = AB2 + BC2

⇨ 132 = AB2 + 52

⇨ 169 = AB2 + 25

⇨ AB2 = 169 – 25

⇨ AB = √144 = 12

⇨ AB = 12 cm

Explanation:

Let us assume the number of columns and rows = a

Total number of spots for planting in the area = a × a = a2.

It has been given that, a2 = 1000

⇨ a = √1000

Here, Quotient = 31, Remainder = 39

Number to be added 

= 2×Quotient – Remainder + 1

= 2×31 – 39 + 1

= 24

So, number of plants that are = 24.

Explanation:

Let us assume the number of columns and rows = a

Total number of spots for planting in the area = a × a = a2.

It has been given that, a2 = 500

⇨ a = √500

Here remainder is 16, so no of students left out = 16.