Associative Property of Matrix

Introduction 
Associative property is property of some binary operation ( operation that involves two operands )

ASSOCIATIVITY means within an expression containing two or more occurences in a row of the same associative operator, the order in which the operation are performed does not matter as long as the same sequence of the operands is not changed.

In associative law if we change or re-arrange the parenthesis the result will not change.

Both commutative law and associative law are different. In commutative law order in which operands appears can be changed which is not allowed in case of associative law.

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Associative Property of Matrix Addition
Associative property of  matrix addition states that

Let  A, B and C be n * n matrices then,

( A + B ) + C = A + ( B + C )

statement :   ( A + B ) + C = A + ( B + C )

let us prove the above statement using example

let  A =           `[[a,b],[c,d]]`                                   

let B =   `[[a,b],[c,d]]`


and  C be

`[[a,b],[c,d]]`

To prove :  ( A + B ) + C = A + ( B + C )

so let us first see  ( A + B ) + C

A + B = `[[a,b],[c,d]]+[[a,b],[c,d]]`

( A + B ) + C = ` [[2a,2b],[2c,2d]]+[[a,b],[c,d]] = [[3a,3b],[3c,3d]]` 


next find  A + ( B + C )

( B + C ) = `[[2a,2b],[2c,2d]]`

A + (B + C) =`[[a,b],[c,d]] + [[2a,2b],[2c,2d]] =[[3a,3b],[3c,3d]]`

solving  ( A + B ) + C =  A  + ( B + C ) we came to know that in associative property of multiplication even if we change or rearrange the parenthesis the value of the matrix does not change. It remains same.

Associative Property of  Multiplication

Let  A , B  , C  be n* n matrices . Then ( A B ) C = A ( B C )

A = `[[a,b],[c,d]]` B = `[[a,b],[c,d]]` C = `[[a,b],[c,d]]`

first let us find  (AB )C

(AB) = `[[a,b],[c,d]]``[[a,b],[c,d]]`

` =[[a2+bc,ab+bd],[ac+bd,bc+d2]]`

(AB )C = `[[a2+bc,ab+bd],[ac+bd,bc+d2]]` * `[[a,b],[c,d]]`

= `[[a3+bc2+a2b+bcd,a2b+bcd+ab2+bd2],[a2c+bcd+abc+d2c,ab+bd2+b2c+d3]]`    

Next fine  A ( B C )

(BC) = `[[a,b],[c,d]] *[[a,b],[c,d]]`

= `[[a2+bc,ab+bd],[ac+bd,bc+d2]]`

A(BC) = `[[a2+bc,ab+bd],[ac+bd,bc+d2]]` * `[[a,b],[c,d]]`

= `[[a3+bc2+a2b+bcd,a2b+bcd+ab2+bd2],[a2c+bcd+abc+d2c,ab+bd2+b2c+d3]]`

Solving  ( A B ) C and  A ( B C ) we can observe that even if we rearrange the parenthesis the value does not change. This shows the associative property of multiplication.
.

Solving Trinomial Inequalities

Introduction :
An algebraic expression is a combination of constants and variables.  An algebraic expression containing three terms is known as trinomial.

Inequality is a relationship between two expressions that are not equal, often written in th form of an equation but with symbols '<' or '>' meaning 'is greater than ' or  'is less than'.

Trinominal inequality is an algebraic expression where the two expressions are not equal and the symbols used is '<' or ' >'.

Steps for solving trinominal inequalities:

Bring all the terms on one side of the inequality so that we get a trinomial.
Find the factors of the trinominal by using algebraic identities or by splitting middle method.
Apply the inequality to all the factors of the trinomial so obtained.

Examples of Trinominal Inequality - I

1) Solve x^2 +1>2x

Solution:  x^2 + 1 > 2x

subtract 2x on both sides

x^2 - 2x + 1 > 2x - 2x

x^2 - 2x + 1  > 0

we know, (a-b) 2 = a^2 - 2ab + b^2

plug in a = x and b = 1, we get,

x^2 - 2x + 1 = ( x -1)2

Thus,  x^2 - 2x + 1> 0

(x-1) 2 > 0

x - 1 > 0

Add 1 on both sides

x -1 + 1> 0 +1

x > 1

2)Solve the inequality  x^2 + 5x < 6

Solution:  x^2 + 5x < 6

Subtract 6 on both sides

x 2 + 5x - 6 < 6 -6

x 2 + 5x - 6 < 0

Now, we will find the factors of the trinomial,

x 2 + 5x - 6

Clearly the last term of the trinomial is not a perfect square.

We apply splitting the middle term method,

Coefficient of first term = 1,    Last term = 6

Least common factor of first term and last term = 6

Factors of 6 are,

2 x 3 = 6     and    6 x 1 = 6

We take the factors,

6 x 1 = 6   because  6-1 = 5 = middle term

Therefore,   x 2 + 5x - 6  =   x 2 + 6x -x -6

=  x(x+6) -1(x+6)

= (x-1)(x+6)

Therefore,     x^2 +5x -6 < 0

(x-1)(x+6) <0

(x-1) <0                 or               (x+5) < 0

x -1+1< 0+1          or               x +5 -5 < 0 -5

x < 1                        or              x < (-5)

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Solving Word Problems of Trinomial Inequalities
Solving word problems of trinomial inequalities means we first convert the word problem to mathematical form and then follow the steps to solve the inequality.

1) Square of a number is greater than the sum 4 and thrice the number. Find the number.

Solution:  Let the number be x

Square of a number = x^2

Sum of 4 and thrice the number = 4 + 3x

Therefore the inequality formed is,

x^2 > 4 + 3x

subtract  (4+3x) on both sides

x^2 -(4 + 3x) > 4 + 3x - ( 4+3x)

x^2 -4 -3x > 0

x^2 -3x -4 > 0

Now consider the trinomial

x^2 -3x - 4

First term and last term of the trinomial are perfect squares

Here, we will solve the trinominal by splitting the middle term because

middle term `!=` 2 * first term * last term.

Now,  coefficient of first term = 1

Coefficient of first term * last term = 1*4 = 4

Factors of 4 are,

2 x 2 = 4           and              4 x 1 = 4

we will take, 4 x 1 = 4     because -4 +1 = -3 = middle term

Therefore,  x^2 - 3x - 4 = x^2 -4x + x - 4

=  x(x-4)+1(x-4)

=  (x-4)(x+1)

Therefore,    x^2 -3x -4 > 0

(x-4)(x+1) >0

(x-4) > 0                   or          (x+1) > 0

x-4+4 > 0 + 4          or         x +1-1> 0-1

x > 4                           or         x > -1

The number is either 4 or (-1)

Semi Circle Formula

Introduction 

Draw a circle on a piece of paper. Plot the center of the circle. Draw a line through the center of the circle and cut it along the line. We get two plane figures. Each plane figure is bounded by a  long curve and a straight line. We could observe that this plane figure is exactly half of the circle. Such plane figure is called as semi circle.

Since plane figure is two dimensional, semi circle possess area and perimeter. We know that angle at the center of the circle is 360 degrees. Angle at the center of the semi circle is half of 360 degrees = 180 degrees.

Choose any point on the circumference of the semi circle and join them to the ends of the diameter of the semi circle. This is a triangle. Angle formed at that point of the triangle is right angle.

Formulas on Semi Circle:

A semicircle has only one diameter.

semi circle

Area of the semicircle is half of the area of the circle.

Area of semicircle = Area of circle
2

= `(pir^2)/2` sq. units

Circumference of the semi circle = πr units.
If we know the length of the straight line (diameter, d)  in the semicircle, then we could find the radius by dividing it with 2.

Circumference of the semicircle if diameter is known = `(pid)/2` units

Perimeter of the semi circle = Circumference of the circle + diameter
= πr + d

= πr + 2r units

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Problems on Semicircle Using Formulas:

Ex 1: Find the area, circumference of the semi circle whose diameter is 14 cm.

Sol:

Step 1: Write the given details.

Diameter = 14 cm

Step 2: Find the radius

radius = 14/2 = 7 cm

Step 3: Find the area of the semicircle using the formula

Area of the semi circle = `(pir^2)/2` sq. units

= `22/7 xx 7^2/2`

= 77 cm2

Step 4: Find the circumference of the semi circle

circumference of the semi circle = πr units

= `22/7 xx 7 = 22 cm`

Ex 2: The government has planned to construct a botanical garden in semicircular shape with perimeter of 36 hectare. Calculate the area  of the garden.

Sol:

Step 1: Write the given details

Perimeter = 36 cm

Step 2: To find the area, we need radius. So, write the perimeter formula and equate it to 36 and find the radius

πr + 2r= 36

3.14r + 2r = 36

r(3.14 + 2) = 36

5.14r = 36

r = 7.003 = 7 hectare

Step 3: Calculate the area

Area of the semicircle = `(pir^2)/2`

`= 22/7 xx 7^2/2 = 77 ` sq. hectare

Triangular Pyramid How many Faces

Introduction

Triangular pyramid how many faces a pyramid is a solid three dimensional shape formed by  connecting the base which can be any polygon and the apex . Pyramids have a base which is a polygon with ' n ' sides then the number of faces it has will be equal to ' n '. If the number of faces is ' n ' then the number of vertices will be ' n + 1 '  and the number of edges are  ' 2n ' . Pyramids are specified according to their base shape , if it is unspecified then it is normally assumed the pyramid has a square base .

Triangular pyramid means it has a triangle base , when a pyramid is made up of four equilateral triangles then it is called a tetrahedron . A tetrahedron has four equilateral triangular faces (any one of the face can be considered as a base) , four vertices's and six edges . In a tetrahedron all angles and all edges are equal .

Triangular Pyramid How many Faces : Surface Area and Volume.

Triangular pyramid how many faces? A triangular pyramid has 3 faces  and one base , we can say it has all together 4 faces as we see a Tetrahedron has 4 faces each one is an equilateral triangle so any face may be considered as the base .

Surface Area

The surface area of a triangular pyramid is given by...

S.A = Area of base + 3 ( Area of a face ) = 1/2 s a + 3 ( 1/2 s l )

S.A  =  1/2 x s a + 3/2 s l  sq.units.   where ' a ' is the height of base triangle , ' s ' is the side length of the base triangle which is again the base of the face triangle , ' l ' is the slant height or the height of the face triangle .

For a tetrahedron surface area  S.A  =  4 ( 1/2 s a )  =  2 s a  ,  where ' s ' is the base of any face and ' a ' is the height or slant height .

Volume

Volume of a triangular pyramid is given by...

V  =  1/3 times Area of base triangle times height of the pyramid .

V =  1/3 x ( 1/2 s a ) h   =   1/6 s a h    cubic units

Triangular Pyramid How many Faces:example Problems

Ex 1:  Find the surface area and volume of the given triangular pyramid .

Sol :   Surface area  S.A  =  area of base + area of three faces            ( s = 6 cm , a = 4 cm , l = 8 cm  , h = 5 cm )

area of base  = 1/2 sa  =  1/2 x 6 x 4  = 12 cm2,  area of three faces  = 3/2 (sl) =   3/2 ( 6 x 8 )  =  3 ( 24 )  =  72 cm2

Therefore S.A  =  12 cm   +  72 cm      =   842 cm

Volume  V  =  1/6 sah    =   1/6 x 6 x 4 x 5  =  20 cm3

Ex 2 : The surface area of a tetrahedron is 112 m2 , find the area of the base.

Sol: A tetrahedron has four congruent faces , the area of  all four faces are equal and any face can be considered as base.

So , The area of the base in the given tetrahedron  =  112/4  =  28 m2   .

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Length of Lines

Introduction:

The Length of lines is the main topic in Geometry. The line length between two points, it is used in all geometric concepts. Let P1 (x1, y1) and P2 (x2, y2) be two distinct points in the Cartesian plane and denote the line length between P1 and P2 by d (P1, P2) or by P1P2. The length formula is used to find the length between two points.  With the help of geometric properties and the length formula we have to identify the given points make which shape.


Brief Description of Line Length Formula

Consider the line segment P1 P2. Here two cases arise.

Case (I):

The segment P1 P2 is parallel to the x-axis Then y1 = y2. Draw P1L and P2M, perpendicular to the x-axis. Then d (P1, P2) is equal to the length between A and B. But A is (x1, 0) and B is (x2, 0). So the length AB =|x1 -X2|. Then d (P1, P2) =|x1 -X2|.

Case (II):

The segment P1 P2 is parallel to the y-axis. Then x1 = x2 .Draw P1A and P2B, perpendicular to the y-axis. Then d (P1, P2) is equal to the line length between A and B. But A is (0, y1) and B is (0, y2). So the length AB =| y1- y2|.then d (P1, P2) = | y1- y2|

`sqrt(((x2-x1)^2 + (y2-y1)^2))`

This is called the line length formula which gives the line length d between the two given points (x1, y1) and (x2, y2). We observe that d (P1, P2) = d (P2, P1). The formula has been derived for two points which are not on a horizontal line or vertical line.

Examples Problems

Given points are P (5,4) and Q (8, 6). Find the length of PQ using line length formula.

Solution:

Let d be the line length between P and Q. Here we have to find the length between two points using line length formula.

Then d (P, Q) =`sqrt(((x2-x1)^2 + (y2-y1)^2))`

= `sqrt(((8-5)^2 + (6-4)^2))`

Simplifying this we get the length is v13.

Example 2:

Given points are W(2, -3), X(6, 5), Y(-2, 1) and Z(-6, -7), Show that this points make a square.

Solution: One way of showing that WXYZ is a rhombus is to show that all its sides are of equal length. One way is showing that a rhombus is not a square is to show that the diagonals are of unequal length. Here we find

WX= `sqrt(((6-2)^2 + (3+1)^2))`

Simplifying this we get v80

XY = `sqrt(((-2-6)^2 + (1-5)^2))`

Simplifying this we get v80

WY= `sqrt(((-2-2)^2 + (1+3)^2))`

Simplifying this we get v32

XZ = `sqrt(((-6-6)^2 + (-7-5)^2))`

Simplifying this we get v288

YZ = `sqrt(((-6+2)^2 + (-7-1)^2))`

Simplifying this we get v80

WZ= `sqrt(((-6-2)^2 + (-7+3)^2))`

Simplifying this we get v80

Therefore WX=XY=YZ=ZW

WY is not equal to XZ

Therefore WXYZ is a rhombus but not a square.

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Area of a Cone Shape

Introduction :
                    Cone is the one type of three dimensional shapes. It has one base and one vertex. The base of the cone is circle in shape. It surface is tappers smoothly towards the top. The distance between center of the base and any point on the boundary of base is called radius. In this article we shall see how to calculate the volume and surface area of the cone.


Area of a Cone Shape – Formula:

Formula for volume of the cone (v) =`1/3` p r^2 h cubic units

                                                    v - Volume of cone

                                                    r – Radius

                                                    h – Height 

Surface area of cone (A) = lateral surface area of cone + area of base 

                                              = p r s    + p r^2 square unit 

                                                s – Slant height

Area of a Cone Shape – Example Problems:

1. The cone has the radius = 7 cm, height = 14 and slant height = 15.6 cm. Find the volume of the cone.

Solution:

 Given:

                        Radius (r) = 7 cm

                        Height (h) = 14 cm

                      Slant height (s) = 15.6 cm

Formula:

Total surface area:

Total Surface area of cone (A) = lateral surface area of cone + area of base

Lateral surface area (L.S.A) = p r s    square unit.

Substitute the r and s value in formula and simplify,

                                                           = p x 7 x 15.6

                                                            = 3.14 x 7 x 15.6

                                                            = 342.88 cm3

       Lateral surface area (L.S.A) = 204.1 cm^2

                          Area of base = p r^2 square unit

                                                    = 3.14 x 72

                           Area of base = 153.86 cm^2

Total Surface area of cone (A) = lateral surface area of cone + area of base

                                                    = 204.1 +153.86

                                                    = 357.96 cm^2

   2. The cone has the radius = 8 cm, height = 15 and slant height = 17 cm. Find the volume of the cone.

Solution:

 Given:

                        Radius (r) = 8 cm

                        Height (h) = 15 cm

                     Slant height (s) = 17 cm

Formula:

Total surface area:

Total Surface area of cone (A) = lateral surface area of cone + area of base

Lateral surface area (L.S.A) = p r s    square unit.

Substitute the r and s value in formula and simplify,

                                                     = p x 8 x 17

                                                     = 3.14 x 8 x 17

                                                     = 427.04

Lateral surface area (L.S.A) = 427.04 cm^2

                          Area of base = p r^2 square units

                                                  = 3.14 x 8^2

                         Area of base = 200.96 cm^2

Total Surface area of cone (A) = lateral surface area of cone + area of base 

                                                         = 427.04 + 200.96

                                                         = 628 cm^2
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Word Problems based on Ratio and Proportion

Introduction :

A ratio is the comparison of two quantities by division. It is a relation that one quantity bears to another with respect to magnitude. If A and B are two numbers, than the ratio of A to B is A/B and is denoted by A:B. The ratio does not have any unit.

The equality of two ratios is called Proportion. If (A/B) = (C/D) , then A,B,C,D are said to be in proportion and can be written as

A : B :: C:D

The Ratio and Proportion world problems are a kind of problems in which , the relation between different quantity has to be determined and then using the definition of ratio and proportion, the unknown has to be calculated.
Here are some of the ratio and proportion word problems:

Ratio and Proportion: Word Problems.

Problem: Find the value of k that must be added to 7, 16, 43, 79 so that they are in proportion. (Answer: 5)

Problem: Find the fourth proportional to the numbers 60, 48, 30. (Answer: 24)

Problem: The Income of Alex and Bob are in the ratio of 3:2 and their expenditure in the ratio of 5:3. Find the income of Alex if each saves dollars 1000. (Answer:  $6000)

Problem: A mixture contains alcohol and water in the ratio of 12:5. On adding 14 litres of water, the ratio of alcohol to water becomes 1:1. Find the quantity of alcohol in the mixture. (Answer: 24 litres)

Ratio and Proportion: Multiple Choice Word Problems:

Problem: If the ratio of ages of Alex and Bob is 6:5 at present and fifteen years from now, the ratio will get changed to 9:8, then find Alex's age.

(A). 24 years

(B) 30 years

(C) 18 years

(D) 33 years

(Answer: (B) 30 years)

Problem: If dollars 58 is divided among 150 children such that each girl and each boy gets 25 dollars and 50 dollars respectively. Then how many girls are?

(A) 52

(B) 54

(C) 68

(D) 62

(Answer: (C) 68)

Problem: The number that must be added to each of the numbers 8, 21, 13 and 31 to make the ratio of first two numbers equal to the ratio of last two numbers is

(A) 5

(B) 7

(C) 9

(D) None of these.

(Answer: 5)