ppt

1. B.J.P.S Samiti’s
M.V.HERWADKAR ENGLISH MEDIUM HIGH SCHOOL
Square and Square Roots
Program:
Semester:
Course: NAME OF THE COURSE
Sandhya S.Kulkarni 1

2. Squares and Square Roots
In general, if a natural number m can be
expressed as n2, where n is also a natural
number, then m is a square number
Example: 1X1=1, 2X2 = 4, 3X3 = 9, 4X4 = 16
5X5 = 25, 6X6 = 36
The numbers 1, 4, 9, 16 ... are square numbers. These
numbers are also called perfect squares
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 2

3. Properties of Square numbers
• A perfect square number ends in 0, 1, 4, 5, 6 or 9 at
unit’s place. Example:9 ,16, 25, 36, 49
• If a number has 1 or 9 in the units place, then it’s
square ends in 1. Example: 92 =81, 112 =121
• when a square number ends in 6, the number whose
square it is, will have either 4 or 6 in unit’s place
Example: 42 =16, 62=36, 142=196, 162=256
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 3

4. Properties of square numbers (contd)
•There are 2n non perfect square numbers
between the squares of the numbers n and (n + 1)
Example: For n=5 and n+1= 6,between n2 =25 and
(n+1)2 =36 there are 2n=2X5=10 non perfect square
numbers i.e.26, 27, 28, 29, 30, 31, 32, 33, 34, 35
•Sum of first ‘ n’ odd natural numbers is n2
Example: Sum of the first 4 odd natural numbers
1+3+5+7=(4)2 = 16
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 4

5. Properties of square numbers (contd)
•Square of even number is always even
number
Example: (2)2=4, (4)2=16, (10)2=100
•Square of odd number is always odd number
Example: ( 3)2 =9,(5 )2 =25, (7 )2 =49
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 5

6. Properties of square numbers (contd)
•Square numbers can have only even number of
zeros at the end
Example:100, 10000. 1000 is not a square number
•For any odd or even integer ‘a’, (a-1)(a+1)=a2 -1
Example:For a=4, (4-1)(4+1)= 42 -1=15
Example:For a=5, (5-1)(5+1)= 52 -1= 24
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 6

7. Properties of square numbers (contd)
•Square root is the inverse operation of square
Example:When 22 =4, sqrt(4)=sqrt(2X2)=2
•The square of any odd number as the sum of
two consecutive positive integers For example:
•32 = 9 = 4 + 5
•52 = 25 = 12 + 13
•72 = 49 = 24 + 25
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 7

8. Properties of square numbers (contd)
• If a,b,c are three numbers such that a2 = b2+c2 then a,
b,c are called Pythagorean triplet
Example: 52=32+42 hence 5, 3, 4 are Pythagorean
triplet
Example: 102=82+62 hence 10, 8, 6 are Pythagorean
triplet
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 8

9. Finding the square of a number
Finding the Square of a Number without
multiplying it with itself
Example:
392 = (30 + 9)2 = 30(30 + 9) + 9(30 + 9)
= 302 + 30 × 9 + 9 × 30 + 92
= 900 + 270 + 270 + 81 = 1521
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 9

10. Finding the square of a number
•Considering a number with unit digit 5 such as a5
(a5)2 = a(a + 1) hundred + 52
Example: (25)2 = (a5)2 here a = 2 and hence
(25)2 = 2( 2 + 1)hundred + 25 = 600 + 25 =625
Example: (35)2 = (a5)2 here a = 3 and hence
(35)2 = 3( 3 + 1)hundred + 25 = 700 + 25 =625
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 10

11. Square Roots
• Finding square roots
• By observation
Knowing 22= 4, √4 = 2
Knowing 32= 9, √9 = 3
Knowing 62= 36, √36 = 6
Knowing 92= 81, √81 = 9
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 11

12. Finding square root
• Finding square root through repeated subtraction
• Knowing that the sum of the first n odd natural numbers is n2 and
every square number can be expressed as a sum of successive odd
natural numbers starting from 1.
Consider 81. Then,
(i) 81 – 1 = 80 (ii) 80 – 3 = 77 (iii) 77 – 5 = 72 (iv) 72 – 7 = 65
(v) 65 – 9 = 56 (vi) 56 – 11 = 45 (vii) 45 – 13 = 32 (viii) 32 – 15 = 17
(ix) 17 – 17 = 0 At 9th step ‘0’ occurs and hence √81 = 9
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 12

13. Finding square root
• Finding square root through prime factorisation
Example
• 36 = 2 × 2 × 3 × 3 Hence √36 = 2X3 = 6
• 64 = 2 × 2 × 2 × 2 × 2 × 2 Hence √64 = 2X2X2 = 8
• 144 = 2 × 2 × 2 × 2 × 3 × 3 Hence √144 = 2X2X3 = 12
• 225 = 3 × 3 × 5 × 5 Hence √225 = 3X5 = 15
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 13

14. Finding square root
• Factorisation to find the factors by dividing the given
number by prime numbers like 2, 3 , 5, 7, 11, 13 etc.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 14

15. Finding square root
Factorisation to find the factors by dividing the given
number by prime numbers like 2, 3 , 5, 7, 11, 13 etc.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 15
2 324 162
2 162 81
3 81 27
3 27 9
3 9 3
3 3 1

16. Finding square root
• Considering a number with unit digit 5 such as
a5
(a5)2 = a(a + 1) hundred + 52
Example: (25)2 = (a5)2 here a = 2 and hence
(25)2 = 2( 2 + 1)hundred + 25 = 600 + 25 =625
Example: (35)2 = (a5)2 here a = 3 and hence
(35)2 = 3( 3 + 1)hundred + 25 = 700 + 25 =625
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 16

17. • Consider a number with unit digit 5, i.e., a5
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 17

18. • Consider a number with unit digit 5, i.e., a5
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 18