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CONTENTS OF THIS PAGE:
What is square of a number?
How to find the square root of a number by prime factorisation method?
How to find the square root of a number by long division method?
Algebraic Identities (1 to 4)
Cylinder: (video)
How to find the Curved Surface Area of a cylinder (CSA)?
How to find the Total Surface Area of a cylinder (TSA)?
How to find the Volume of a cylinder?
How to find the radius of a cylinder given its height and CSA?
How to find the radius and height of a cylinder given CSA and ratio of radius and height?
Cone: (video)
How to find the Curved Surface Area of a cone (CSA)?
How to find the Total Surface Area of a cone (TSA)?
How to find the Volume of a cone?
How to find the radius of a cone given its height and slant height?
How to find the slant height of a cone given its radius and height?
How to find the height of a cone given its volume and radius?
SQUARE OF A NUMBER:
Square of a number is the result obtained when the number is multiplied by itself.
Square of 2 is 2 x 2 =4.
Square of -6 is -6 x-6 = 36
This means square of a number is always positive.
PERFECT SQUARES:
Perfect square is the product of two equal integers.
Squares end with the following numbers 1, 4, 5, 6, 9 but not all numbers ending with such numbers are perfect squares.
Perfect squares end with even number of zeros as seen in 100, 400, 900, 10000, 490000 and so on but not all numbers ending with even number of zeros are perfect squares.
This is just to indicate that there is a possibility of such numbers to be perfect squares.
It can be definitely said that numbers ending with 2, 3, 7, 8 and also an odd number of zeros are not perfect squares.
SQUARES OF NUMBERS 1 TO 10:
12 = 1 x 1 = 1
22 = 2 x 2 = 4
32 = 3 x 3 = 9
42 = 4 x 4 = 16
52 = 5 x 5 = 25
62 = 6 x 6 = 36
72 = 7 x 7 = 49
82 = 8 x 8 = 64
92 = 9 x 9 = 81
102 = 10 x 10 = 100
SQUARE ROOT OF A NUMBER by PRIME FACTORISATION METHOD:
We use only prime numbers to divide the given number and continue the division till we get all the prime factors of the number. We express the given number as the product of its prime factors and to find the square root, we take one factor out of a pair of equal factors. The product of such factors will be the square root of the given number. It will be easy to do prime factorisation using divisibility rules. To know more about divisibility rules kindly click DIVISIBILITY RULES .
To know the divisibility rules for 7, 8, 9, 10 and 11, kindly click Divisibility rules for 7,8,9,10 and 11
Let us see some examples for finding the square root of a number using prime factorisation method:
SQUARE ROOT OF A NUMBER:
LONG DIVISION METHOD:
We can find the square root of a number using long division method. Let us see a detailed explanation of this method here.
Move towards the left of the given number and put a bar for each pair. If the given number has an odd number of digits, put a bar for the left out single digit also. As regards to the divisor and quotient, same number has to be written in both the places. Choose a number such that its square is lesser than the number in the first bar. Along with the remainder got from the first step, write the next pair from the dividend. For every step, we start with a new divisor which is got by multiplying the quotient by 2. Following this, write the same number both in the divisor and quotient and repeat the usual division. Repeat this process for every new step till a remainder zero or a number lesser than the divisor is got. Finally, when the division process is completely over, whatever quotient we have get is the SQUARE ROOT of the given number. If the remainder is zero, then we call the given number a PERFECT SQUARE.
LONG DIVISION METHOD TO FIND THE SQUARE ROOT OF A NUMBER
ALGEBRAIC IDENTITIES
Algebraic identities are proved results that hold good for any value of the variable. The following examples show that the identities can be applied to any value of the variable a and b given below. Identities are mainly used for simplifying algebraic expressions and also to find the value of the expressions at specific values of the variables.
Some more identities have been discussed in the next section.
IDENTITY 3
IDENTITY 4
EVALUATE USING ALGEBRAIC IDENTITIES:
232 = (20 + 3)2 Using (a + b)2 = a2 + 2ab + b2
= (20)2 + 2(20)(3) + (3)2
= 400 + 120 + 9
= 529
992 = (100 – 1)2 using (a-b)2=a2-2ab+b2
= (100)2 – 2 (100) (1) + (1)2
= 10000 – 200 + 1
= 9800 + 1
= 9801
842 – 162 = (84+16) (84-16) using a2-b2= (a + b) (a - b)
= (100) (68)
= 6800
103 x 104 = (100 + 3) (100 + 4) using (x + a)(x + b) = x2+(a + b)x + ab
= (100)2 +(3+4)100 + (3)(4)
= 10000 + 700 + 12
= 10712
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