1.2.2 Perfect squares – Chapter 2 – Class VIII

Observe that, 1 = 1, 4 = 2 x 2, 9 = 3 x 3, 100 = 10 x 10. If a is an integer and b = a x a, we say b is a perfect square. Hence 1, 4, 9, 16, 25 are all perfect squares. Since 0 = 0 x 0, we see that 0 is a perfect square. If a is an integer, we denote a x a = a2. We read it as square of a or simply a square. Thus 36 = 62 and 81 = 92. Thus a perfect square is of the m2, where m is an integer.

For example, 4 = 2 x 2 and 4 = (-2) x (-2); in the second representation, we again have equal integers, but negative this time.

Thus a perfect square is either equal to 0 or must be a positive integer. It can be a negative integer.

Look at the following table

a1238-7-1220-15
a21496449144400225

We see that squares of 2, 8, -12, 20 are even numbers and the squares of 1, 3, -7, -15 are odd numbers.

Statement 1: The Square of an even integer is even and the square of an odd integer is odd.

Consider first 10 perfect squares,

122232425262728292102
149162536496481100

If we observe that the units place in these squares are 1, 4, 9, 6, 5, 6, 9, 4, 1 and 0 in that order. Thus only the digits which can occupy digit’s place in perfect squares are 1, 4, 5, 6 and 9.

Statement 2: A perfect square always ends in one of the digits 0, 1, 4, 5, 6 and 9. If the last digit of a number is 2, 3, 7 or 8, it cannot be a perfect square.


Perfect squares – Exercise 1.2.2

  1. Express the following statements mathematically:

i) square of 4 is 16

Solution:

42 = 16


ii) square of 8 is 64

Solution:

82 = 64


iii) square of 15 is 225

Solution:

152 = 225


  1. Identify the perfect squares among the following numbers:

1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000

Solution:

Perfect squares among the above are:

1, 36, 49, 81, 169, 625, 125, 900, 100


  1. Make a list of all perfect squares from 1 to 500

Solution:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484


  1. Write the 3 digit numbers ending with 0, 1, 4, 5, 6 , 9 one for each digit but one of them is a perfect square.

Solution:

We can take 200, 201, 204, 205, 206 and 209. None of these is perfect square lies between 142 = 196 and 152 = 225.

We can also take 300, 301, 304, 305, 306 and 309. None of these is a perfect square lies between 172 = 289 and 182 = 324.


  1. Find the numbers from 100 to 400 that end with 0, 1, 4, 5, 6 or 9, which are perfect squares.

Solution:

102 = 100, 112 = 121, 122 = 144, 132 = 169, 142 = 196, 152 = 225, 162 = 256, 172 = 289, 182 = 324, 192 = 361, 202 = 400


1.2.3 Some facts related to perfect squares – Perfect squares

There are some nice properties about perfect squares. Let us study them here:

  1. Look at following table:
a41020251003001000
a21610040062510000900001000000
The number of zeros at the end of a20220446

We observe that, the number of zeros at end of a square is always an even number.

STATEMENT 3: if a number has k zeros at the end, then its square ends in 2k zeros

Thus, if a number has odd number of zeros, it cannot be perfect square.

  1. Look at the adjoining table:
aa2The remainder of a2when divided by 3The remainder of a2when divided by 4
1111
2410
3901
52511
86410
1112111
-63600

We can see that, the remainder of a perfect square is either 1 or 0. we can also note that remainder of a perfect square when divided by 4 are either 0 or 1.

Since, when a number is divided by 3, the reminder used to be 0, 1 or 2. Similarly, when a number is divided by 4 the remainder used to be 0, 1, 2 or 3.

 

STATEMENT 4: The reminder of a perfect square when divided by 3 is either 0 or 1, but never be 2. The reminder of a perfect square when divided by 4 is either 0 or 1, but never be 2 and 3.

STATEMENT 5: When the product of 4 consecutive integers is added to 1, the resulting number is perfect square.

Ex:

(1 x 2 x 3 x 4) + 1 = 24 + 1 = 25 = 52

(5 x 6 x 7 x 8) + 1 = 1680 + 1 = 412

STATEMENT 6: The sum of the first n odd natural numbers is equal to n2, for every natural number n.

Ex:

1 = 1 = 1; 1 + 3 = 4 = 2; 1 + 3 + 5 = 9 = 3; 1 + 3 + 5 + 7 = 16 = 42  ; 1 + 3 + 5 + 7 + 9 = 25 = 52

STATEMENT 7: consider the number N = 1000….01, where zeros appear k times. (For example, for k = 6, you get N = 10000001; there are 6 zeros in the middle.) Then N2 = 1000…02000…01, where the number of zeros on both the sides of 2 is k.

Ex:

112 = 121

1012 = 10201

10012 = 1002001

100012 = 100020001 and so on.

STATEMENT 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Ex:

Perfect Squares

The dots are now arranged in shapes. Now count the number of dots in each triangle. (Single dot is considered as the generate triangle.) They are 1, 3, 6, 10, 15, 21, 28, 36 and so on. These are called triangular numbers.

For nth triangular number, we form a triangle of dots with n- rows and each row contains as many points as index of that row.  If you want find the 8th triangular number, the number of points in the 8th triangle is

1 + 2 + 3 + 4+ 5 + 6 + 7 + 8 = 36

The first few triangular numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91.

Take any two consecutive triangular number and find their sum. For example : 10 + 15 = 25 = 52; 21 + 28 = 49 = 72. Which holds the statement 8.


Perfect Squares – Exercise 1.2.3

  1. Find the sum 1 + 3 + 5 + ……. +51 (the sum of all odd numbers from 1 to 51) without actually adding them.

Solution:

[We have 5 odd numbers from 1 to 10, therefore there are 5×5 = 25 odd numbers are there from 1 to 50, we know 51 is also an odd number…. So, there are 25+1 = 26 ]

There are 26 odd numbers are there from 1 to 51.

Thus, 1 + 3 + 5 + …….. + 51 = 262 = 676.


  1. Express 144 as a sum of 12 odd numbers.

Solution:

We know that, the sum of the first n odd natural numbers is equal to n2, for every natural number n.

Therefore, the sum of first 12 odd natural numbers is equal to 122 = 144.

Thus,

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 22 = 122 = 144.


  1. Find the 14th and 15th triangular numbers, and find their sum. Verify the statement 8 for this sum.

Solution:

We know, statement 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Sum of 14th triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105

Sum of 15th triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120

Now, let us verify, statement 8: The sum of nth and (n+1)th triangular number is (n+1)2.

i.e., Sum of 14th triangular number(n) + sum of 15th triangular number(n+1) = 105 + 120

= 225 = 152 = (n+1)2, which holds the statement 8.


  1. What are the remainders of a perfect square when divided by 5?

Solution:

aa2The remainders of a2 when divided by 5
111
244
5250
6361
-7494
111211

The remainders of perfect squares when divided by 5 are 0, 1 and 4.


1.2.4 Methods of squaring a number – Perfect Squares

Many times it is easy to find the square of a number without actually multiplying the number to itself.

Consider 42 = 40 + 2

Thus,

422 = (40+2)2

= 402 + 2(40)(2) + 22

Now it is easy to recognise 402 = 1600; 2x40x2 = 160; 22 = 4.

Therefore, 422 = 1600 + 160 + 4 = 1764.

 

Example: Find 892

Solution: 892 = (80 + 9)2

= 802 + 2 x 80 x 9 + 92

= 6400 + 1440 + 81

= 7921


Perfect Squares –  Exercise 1.2.4

  1. Find the squares of:

i) 31

Solution:

31 = 30 + 1

312 = (30 + 1)2

= 302 + 2 x 30 x 1 +1

= 900 + 60 + 1

= 961


ii) 72

Solution:

72 = 70 + 2

722 = (70 + 2)2

= 702 + 2 x 70 x 2 + 22

= 4900 + 280 + 4

= 5184


iii) 37

Solution:

37 = 30 + 7

372 = (30 + 7)2

= 302 + 2 x 30 x 7 + 72

= 900 + 420 + 49

= 1369


iii) 166

Solution:

166 = 160 + 6

1662 = (160 + 6)2

= 1602 + 2 x 160 x 6 + 62

= 25600 + 1920 + 36

= 27556


2. Find the squares of:

i) 85

Solution:

852 = (80 + 5)2

= 802 + 2 x 80 x 5 + 52

= 1600 + 800 + 25

=2425


ii) 115

Solution:

1152 = (100 + 15)2

= 1002 + 2 x 100 x 15 + 152

= 10000 + 3000 + 225

= 13225


iii) 165

1652 = (160 + 5)2

= 1602 + 2 x 160 x 5 + 52

= 25600 + 1600 + 25

= 27225


  1. Find the squares of 1468 by writing this as 1465+3

14682 = (1465 + 3)2

= (1465)2 + 2 x 1465 x 3 + 32

= 2146225 + 8790 + 9

= 2155024


1.2.5 Square roots – Perfect Squares

Consider the following perfect squares:

1 = 12, 4 = 22, 9 = 32, 16 = 42, 49 = 72, 196 = 142.

In each case the number is obtained by the product of two equal numbers. Here we say 1 is the square root of 1; 2 is the square root of 4; Hence 3 is the square root of 9 and so on.

 

Suppose N is a natural number such that N= M2. The number M is called a square root of N.

 

We have seen earlier m2 = m x m = (-m) x (-m). Thus m2 has 2 roots m and –m. For example, 16 = 42 = (-4)2, thus both 4 and -4 are the roots of 16. Mathematically both 4 and -4 are accepted as the square root of 16. Therefore, whenever the word square root is used, it is always meant to be the positive square root. The square root on n is denoted by √N.


Squares, Square roots, cube and cube roots


 

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