Area Ratio Calculator

Introduction to area ratio calculator

Ratio of areas of two surfaces is called the area ratio. For example, a rectangle is constructed inside a circle such that the circle lies on the four vertices of the rectangle. The ratio of the area of the circle to the area of the rectangle is known as the area ratio.

This is especially useful in calculating the Floor Area Ratio (FAR). The floor area ratio is the ratio of the sum of the area of all the floors in a building to the area of the plot on which the building is constructed. The floor area ratio is used to limit the number of floors that a building can have.

Calculating Floor Area Ratio (far)

For example, a building is built on a plot of area 12000 square feet, and there are two floors in it. The area of each floor is 10000 square feet. Then the floor area ratio is given by

Floor area ratio = Sum of areas of all floors/area of plot

`= (2 * 10000)/12000`

`= 20000/12000`

`= 1.66`

The floor area ratio is known by different names in different countries. In Australia, it is known as “Floor Space Ratio (FSP)” and “Plot ratio”; in India is it known as “Floor Space Index (FSI)”; in UK, it is known as “Plot ratio” or “Site ratio”.

The floor area ratio is used widely in zoning and city planning. It is used to limit the number of floors. An architect has to adhere by a particular floor area ratio for a particular site and construct a building in accordance to it.

General real estate appraiser at Analytic Real Estate Services, Edward Ptacek noted in 2009 that the maximum allowable floor area ratio of a particular place affects the value of the land. The greater the floor area ratio, the more is the value of a particular land.

The concept of floor area ratio has been widely used in Japan since the 1970's to build the residential colonies. However, it is arguable whether this has proved beneficial or not. Some say it is beneficial, and some say that it is not so. I am planning to write more post on How to do Math Problems, factoring polynomials word problems. Keep checking my blog.

Use of Floor Area Ratio in Zoning

Zoning is the method of dividing land into various parts each part intended for a particular usage like residential, commercial, industrial, recreational. The local governments of each city make a zoning map which shows the exact division of land use in that city.

The main factors that affect the zoning of a city is the population, land-usage, existing settlements, etc. By using the concept of floor area ratio in zoning, a city can be efficiently zoned in accordance with the above mentioned factors.

Different values for the floor area ratio are predetermined for different parts of land in the city. This helps to limit the amount of construction on a particular piece of land. For example, the floor area ratio of a region in the vicinity of an airport has to be less so as to inhibit the height of buildings in that area.

An architect can make buildings of different dimensions to adhere by the limit of the floor area ratio. For example, a building built on a large plot having only two floors and a building built on a smaller plot having three or four floors will have the same floor area ratio. Thus, an architect can choose the buildings' dimensions accordingly.

Possible Rational Zeros

Introduction to possible rational zeros:

All polynomials in one variable of degree n, n > 0, has at least one real or complex zero. This is the Fundamental Theorem of Algebra.To find all the possible rational zeros of a polynomial we use the Rational Root Test. Understanding Properties of Rational Exponents is always challenging for me but thanks to all math help websites to help me out.

Problems on Rational Root Test:
If a polynomial function has integer coefficients, then all the rational zeros  will be of the form p/q where p is a factor of the constant and q is a factor of the leading coefficient.

Lets look at a few examples.

Ex 1:  Find all the possible rational zeros of 2`x^(2)` + 6x +4

Here p = 4 and q = 2

The factors of 4 are 4,2 and 1

similarly the factors of 2 are 2 and 1

So all the possible rational zeros will be of the form : `( +-4, +-2, +-1)/( +-2, +-1)`

Now let us list out all the possible rational zeros: 4, 2, -4, -2, 1, -1,  `(1)/(2)` , `(-1)/(2)` .

Ex 2:  Find all the possible rational zeros of 2`x^(4)` - 7`x^(3)` +3x + 6.

Here p = 6. The factors of 6 are 3, 2, 1

q = 2. The factors of 2 are 2,1

So all the possible rational zeros will be of the form :`( +-3, +-2, +-1)/( +-2, +-1)`

Therefore the possible rational zeros are :


` ( 3)/(2)` ,  `(-3)/(2)` , 3, -3, 2, -2, 1, -1, `(1)/(2)` ,  `(-1)/(2)`


Is this topic how to calculate p value hard for you? Watch out for my coming posts.

Things that we should be Careful while Applying the Rational Root Test:

Always make sure the polynomial has integer coefficients. We have to multiply to get rid of fractions or decimals and later divide by the same number.
This method gives us only the possible rational zeros. All of these might not be rational zeros of the polynomial and also there might be non-rational / complex zeros apart from these.
The numbers that we obtain from the Rational Root Test are possible zeros, it is not necessary that all of them are zeros of the polynomial.

Solving Equations Using the Distributive Property

Introduction for distributive property:

An operation is distributive if the result of applying it to a sum of terms equals the sum of the results of applying it to the terms individually.
a ( b + c ) = ( a x b ) + ( a x c ).
Here, ‘a’ is multiplied with the sum of two terms ‘b and c’ in the left hand side which, gives the same answer when ‘a’ is multiplied individually with ‘b’ and ‘c’ and then added. Solving equations using the distributive property is very easy. Here we are going to see solving equations using the distributive property.

Distributive Property:

The types of distributive property used in solving equations are

a (b + c) = (a x b) + (a x c).

a (b - c) = (a x b) - (a x c).

These two are used for solving the equations.

Example Problems for Solving Equations Using the Distributive Property:

Example 1 for solving equations using the distributive property.

Solve the equation 10 (2 + 3) = (10 x 2) + (10 x 3).

Left hand side:

First take 10 (2 + 3)
By adding 2 + 3 we get 5
And now it is 10 (5)
So the answer for the Left hand side is 50

Right hand side:

Now we take (10 x 2) + (10 x 3)
By multiplying 10 with 2 and 10 with 3 we get (20) + (30)
And now we add 20 with 30
So the answer for Right hand side is 30
Hence proved Left hand side is equal to right hand side. Is this topic irrational number definition hard for you? Watch out for my coming posts.

Example 2 for solving equations using the distributive property.

Solve the equation 10(4 – 6) = (10 x 4) - (10 x 6).

Left hand side:

First take 10 (4 - 6)
By adding 4 - 6 we get -2
And now it is 10 (-2)
So the answer for the Left hand side is -20

Right hand side:

Now we take (10 x 4) - (10 x 6)
By multiplying 10 with 4 and 10 with 6 we get (40) - (60)
And now we subtract 40 with 60
So the answer for Right hand side is -20
Hence proved Left hand side is equal to right hand side.


Example 3 for solving equations using the distributive property.

Solve the equation 4(5 – 3) = (4 x 5) - (4 x 3).

Left hand side:

First take the 4(5 – 3)
By subtracting 5 – 3 we get 2
And now it is 4 (2)
So the answer for the Left hand side is 8

Right hand side:

Now we take (4 x 5) - (4 x 3)
By multiplying 4 with 5 and 4 with 3 we get (20) - (12)
And now we subtract 20 with 12
So, the answer for Right hand side is 8
Therefore, Left hand side is equal to right hand side.
Hence proved

Straight Line Products

Introduction to Straight Line Products:
In geometry, line is one dimensional shape that has no end points.Line is a fundamental concept in the mathematics. Line is representing the straight line object. Geometrical object that is straight, infinitely long is called line. Line is the one dimensional figure. Lines have the no height and  then no width. In this article we see detailed about the straight line products. In this article we are discuss the straight line products.

Example Problems for Straight Line Products:

Straight line products – Example: 1

Find the equation of the straight line passing through the points (1,1),(2,3)

Solution:

The equation of the straight line passing through two points is

`y-y_1=m(x-x_1)` where m is a slope.

`y-1=\frac{3-1}{2-1}(x-1)`

`y-1=2x-2`

`y-2x+1=0`

The equation of the straight line is `y-2x+1=0.`

Straight line products – Example: 2

Find the equation of the straight line passing through (-4,5) and cutting off equal intercepts on the coordinate axes.

Solution:

Let the intercept be a

Now the equation of the line is

`\frac{x}{a}+\frac{y}{a}=1`

`x+y=a`

Since the point (-4,5) passes throough this line,

we get

`a=1`

Hence the equation of the line is

`x+y=1`

Straight line products – Example: 3

Find the equation of the straight line passing through (3,-4)  and making intercepts in the ratio 2:3.

Solution:

Let the intercepts be 2k,3k

Then the equation of the line is

`\frac{x}{2k}+\frac{y}{3k}=1`

`3xk+2yk=6`

Since the line passes through (3,-4)

`9k-8k=6`

Hence the equation is

`\frac{x}{12}+\frac{y}{18}=1`

`3x+2y=36`


Practice Problems for Straight Line Products:

1. Find the equation of the straight line passing through (4,-3)  and perpendicular to the line joining (1,1),(2,3).

`Answer: 2y-x+2=0`

2. Show that the equation of the straight line passing through `(x_1,y_1)` and making an angle of \theta  with the X-axis in positive direction is `\frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin \theta}.`

3. Find the equation of the straight line which makes `135^\circ`   with X-axis and passing through (3,-2)

`Answer: y+x-1=0`

Perpendicular Bisector Compass

Introduction to  perpendicular bisector compass:-

We can draw perpendicular to a line segment at any point on it. However, when we draw a perpendicular on the line segment such that it divides the line segment in two equal halves, it is called perpendicular bisector of the line segment. Any point on the perpendicular bisector of a line segment is always at equal distance from the two ends of the line segment.

As can be seen from the given figure that dotted line is the perpendicular bisector of the line segment AB. If you take any point on the dotted line and measure its distances from A and B, you will find that the distances are same.

How to Draw Perpendicular Bisector Using a Compass

We can draw perpendicular bisector of a line segment using a compass. Use a properly sharpened pencil in the compass and adjust it in the compass properly. Suppose we have to draw perpendicular bisector of line segment of length 6 cm.

Draw line segment AB of length 6 cm. Taking A as center and radius equal to more than half of AB, draw arcs on both the sides of AB. Again taking B as center and radius equal to more than half of AB, draw arcs on both the sides of AB, intersecting the previous arcs at P and Q as shown in the figure given below. Join PQ. PQ is the required perpendicular bisector of the line segment AB.

If you measure two parts of the line segment AB, you will find that each part is 3 cm. Having problem with free 6th grade math problems keep reading my upcoming posts, i will try to help you.

Examples of Perpendicular Bisector Compass

Q 1 :       Why compass is used while draw perpendicular bisector of a line segment? Please pick a correct answer.:

A.            to draw the line segment AB

B.            to draw arcs

C.            to measure length of line segment

Sol :

We draw arcs using a compass while drawing perpendicular bisector of a line segment.

Q 2:    Perpendicular bisector of line segment AB of length 12 cm intersects it at O. What is the length of AO?

Sol :

We know that perpendicular bisector divides the line segment into two equal parts. Therefore, length of AO = 6 cm.

Factoring Trinomial Squares

Introduction to factoring trinomial squares:

In math, factoring trinomials square method is one significant in basic algebra. It is definite because the operation of sum of three monomials. Given term consists of three expressions that are in the standard form of ax^2 + bx + c. Factoring trinomial is the operation to make arithmetic operation of multiplication in a reverse way. Let us resolve some instance problems for factoring trinomials are x^2 +6x – 5.

Factoring Trinomial Squares:

A trinomial is a polynomial representing of three terms. A trinomial is an equation concerning three expressions. A trinomial is the extension of a power of a computation of three expressions into monomials. The expansion is specified by

`(a+b+c)^(n) = sum_(i,j,k) ((n),(i,j,kk))a^ib^jc^k`     where n is a non negative integer also the computation is in use over all grouping of nonnegative index i, j, and k such to i+j+k = n. The trinomial coefficients are known with  `((n),(i,jk)) = (n!)/(i!j!k!)`

This method is a particular case of the multi nominal method for m = 3. The number of expressions of an extended trinomial is     `((n+2)(n+1))/(2)` Where n is the exponent.

Examples for Factoring Trinomial Squares:

Example 1:

how to solve factoring trinomial square x^2+8x+15

Solution:

Step 1: the given trinomial is x^2+8x+15

Step 2: to factoring this equation is (x+3)(x+5)

so the solution is (x+3)(x+5)

Example 2:

how to solve factoring trinomial square (x+6)2

Solution:

Step 1: the given equation is (x+6)2

Step 2: (a+b)2 = (a2+2ab+b2)

Step 3: using this formula in the given equation

(x+6)2  = (x^2+12x+36)

so the solution is (x^2+12x+36)

Example 3:

how to solve factoring trinomial square (x-2)2

Solution:

Step 1: the given equation is (x-2)2

Step 2: (a-b)2 = (a2-2ab+b2)

Step 3: using this formula in the given equation

(x-2)2  = (x^2-4x+4)

so the solution is (x^2-4x+4)

Example 4:

how to solve factoring trinomial square 9-16x^2

Solution:

Step 1: the given equation is 9-16x^2

Step 2: a2-b2= (a+b)(a-b)

Step 3: using this formula in the given equation is

32-4x^2 = (3+4x) (3-4x)

so the solution of the given equation is   (3+4x) (3-4x)

Descartes Cartesian Coordinate System

Introduction for descartes Cartesian coordinate system:

In Cartesian coordinates system,name Cartesian is derived from the name of french mathematician and philosopher René Descartes (Latin: Cartesius) from the Euclidean geometry.Cartesian coordinate system which consist of analytic geometry, calculus, and cartography. Descartes introduces the new  point  or plane  on a surface by two intersecting axes.In analytical geometry solving we have point coordinates with the formulas for finding the required parameters and in this section we have the points and the problems.

Descartes Cartesian Coordinates System:

In descartes Cartesian coordinates system we have the point geometry and the formulas for finding the parameters.In a Cartesian coordinates we have many points such as collinear point coordinates, equidistant point coordinates and mid point coordinates.

Collinear points:
Collinear points are a point when three or more points lies on same line.

Midpoint:
Mid point is a halfway point where the line segment divided into two equal parts.

Equidistant point:
In a line segment a point is equal length from other points which are in congruent then the point are equidistant point.I like to share this second order differential equation with you all through my article.

Problems in Descartes Cartesian Coordinates System:

Example 1:
Find the Cartesian coordinates distance between the points A(2,6) and B (2,3).

Solution:
Let assume "d" be the distance between A and B.           (x1,y1)= (2,6), (x2,y2)= (2,3)

Then d (A, B) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

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

= `sqrt(0^2+(-3)^2)`

= `sqrt(0+9)`

=`sqrt9`

=3
Example 2:
Determine the mid point coordinates  of a line segment joining given points A(1,8) and B(-1,-5)

Solution:
The required mid point is
Formula = ` ((x_1+x_2)/(2))`,`((y_1+ y_2)/(2))` here, (x1, y1) = (1,8),(x2, y2) = (-1,-4)

= `((1-1)/(2))``((8-4)/(2)) `

= ` (0/2)`` (4/2)`

= (0,2)

Example 3:
Find the centroid coordinates of a triangle whose vertices's points are given (0, -3), (0,6) and (3,3).

Solution:
(x1 y1) = (0,-3),  (x2 y2) = (0,6), (x3 y3) =(3,3)

Formula for centroid =   ` ((x_1+ x_2+ x_3)/(3))` ,`(( y_1+ y_2+ y_3)/(3))`

The centroid of the triangle  = `((0+0+3)/3)`,`((-3+6+3)/3)`

=`(3/3),(6/3)`

=(1,2)

Computing Rational Expressions

Definition of computing rational expressions:
A rational expression is simplified, or reduced to lowest terms, if its numerator and denominator have no common factors other than 1 and -1. If a rational expression does contain common factors, we use the properties of the real number system to write

`ac/bc`=`a/b``xx``c/c`=`a/b``xx`1=`a/b`   (a, b, c are real number, and bc`!=`0.)

This process is often called “canceling common factors.” To indicate this process, we often write

`a/b`=`a/b`

Rules for Computing Rational Expressions:
Rule 1: Steps for computing rational expressions

Step 1: Set the terms containing the identical variable collectively in algebraic exponential expressions.

Step 2: Accomplish the operation inside the parentheses for the variable and other.

Step 3: Revise the rational expressions and to simplifying rational expressions.

Step 4: To make sure rational expressions, if there is able to simplify rational expressions and then repeat the step 1 to 4.

Rule 2: Order of operation for computing rational expressions

In long math problems with +,-,x,%,(), and exponents in them, you have to identify what to do first. Without follow the similar rules, you may get unlike answers. You can easily keep in mind the silly sentence, Big Elephants Destroy Mice And Snails, you can commit to memory the order of operations, and you must follow.

Big                     “B” means Brackets. We need to carry out operation in side parentheses first.

Elephants         “E” means an exponent; you must calculate exponents next in expressions by j.

Destroy             “D” means division

Mice                  “M” means multiply begin on the left of the equation and perform all divisions and multiplication in the order in which they appear.

And                    “A” means addition

Snails               “S” means subtract. For all time on the left hand side and accomplish additions and subtractions operation.

Rule 3: study for computing rational expressions with exponents.

The rules are given by

`x^m` `xx` `x^n` = `x^(m+n)`

`((x^(m))/(x^(n)))` =  `x^(m-n)`

`((x^(m))^n)` = `x^(mn)`

`(x y)^m` = `x^m` `y^m`

`(x/y)^n` = `(x^(n))/(y^(n))`

`x^(-n)` = `1/(x^(n))`

`(x/y)^(-n)` = `(y/x)^(n)`

Where quantities in the denominator are taken to be nonzero in computing rational expressions, Special cases include

`x^1`=`x`

And

`x^0=1`

For  x`!=`0. The definition `0^o`=1 is sometimes used to simplify formulas, but it should be kept in mind that this equality is a definition and not a fundamental mathematical truth.

Example Problem for Computing Rational Expressions:

To solving computing rational expressions using above rules, ( 8`V^(2)` `-:``2V)+5`Y^(2)``xx``Y^(4)`

Solution:

Step 1: Is to group like terms. Set the terms containing the same variable jointly. Set constants collectively and brackets.

( 8`V^(2)` `-:``2V)+5`Y^(2)``xx``Y^(4)`

Step 2: Is to accomplish the operation inside the parentheses for the variable V.

( 8`V^(2)` `-:``2V)=(8`-:`2)`xx``(V^(2)``-:`V^(1))`=(`8/2`)`xx``(V^(2-1))`=4`xx``V^(1)`=4V

Step 3: Is to accomplish the operation inside the parentheses for the variable Y using expression with exponents rule.

5`Y^(2)``xx``Y^(4)`=5`xx``(Y^(2+4))`=5`xx``Y^(6)`=5`Y^(6)`

Step 4: Is to revise the given problem.

4V+5`Y^(6)`

Since the left over terms are not like terms, the problem cannot be any further.

The answer is 4V+5`Y^(6)`

Venn Diagram Problem Solution

Introduction to Venn diagram problem solution:

In set theory, the concept of Venn diagram is an essential part of set. The Venn diagram show all theoretically like rational relations among limited groups of sets. These are used for solving the set problems in use of diagrams. Another name of Venn diagram is known as set diagram. Solving the set problems with use of Venn diagram is very easy. We will see few problems on Venn diagram with solutions.

Example Problems:

Problem 1:

In a class of 64 students, 26 students participating in Chess, 35 participating in Tennis, and 8 take both Chess and Tennis. Find how many students in the class are not participating in either Chess or Tennis with use of venn diagram?

Solution:

Given, Total number of students in class = 64

Number of students participating in Chess = 26

Number of students participating in Tennis = 35

Number of students participating in both Chess and Tennis = 8

The venn diagram for these information is,



Here, a = Number of students not participating in either Chess or Tennis.

That is, 18 + 8 + 27 + a = 64

53 + a = 64

a = 64 - 53

a = 11

Therefore, the number of students not participating in either Chess or Tennis is 11 students.Understanding differential calculus is always challenging for me but thanks to all math help websites to help me out.

Problem 2:

There are 70 houses in a city. In that, 48 houses have Washing machine and 32 houses have Refrigerator. Find how many numbers of houses have both Washing machine and Refrigerator with use of venn diagram?

Solution:

Let W = {Number of houses having Washing machine}

R = {Number of houses having Refrigerator}

WnR = {Number of houses having both Washing machine and Refrigerator}

Consider, n(WnR) = a



Therefore, Number of houses having both Washing machine and Refrigerator is,

48 - a + a + 32 - a  = 70

80 - a = 70

-a = 70 - 80

-a = -10

a = 10

Answer: Number of houses having both Washing machine and Refrigerator = 10 houses.

These are all few example problems of using venn diagram with solution.

Algebra Polynomials Function

Introduction of algebra polynomials function:

In the algebra polynomials functions the algebraic expression is look y=an is said to be monomial in y here a is the constant number, y is the variable  and n is the positive integer. The number a is said to be the coefficient of yn and n is the degree of monomial.

Example of polynomial:  7x3 is a monomial in x of degree 3 and 7 is the coefficient of x3. In monomial the sum of finite number in x is called a polynomial in x.

Algebra polynomial function is denoted by the function p(x), here p is the function and x is the variable.

Example of algebra polynomials function:

P(x) = ax^5+bx^4+cx^3+dx^2 +ex + f here we need to find the algebra polynomials function of p(5). X is the variable  and  other alphabet are constant.

Problem on Algebra Polynomials Function:

Problems1:. Find the functions of f(3). f(x) = x^3 +2x^2 + 2x + 4

Solution

Given function f(x)

f(x) = x^3 +2x^2 + 2x + 4 find the f(3).

Here the value of x is given as 3

f(3) = 33 + 2*32 + 2*3 +4

f(3) = 27 +18+ +6 +4 In this step 33 is calculated  as 27 and 3 square is 9

f(3) = 55.

The answer for algebra polynomials functions  f(3) = 55.


Problems Using the Algebra Polynomials Function with High Powers:

Problems 1: Find the fucntions f(5). f(x) = x5+x^4 +x^3 +2x^2 + 2x + 4

Solution

Given the function of  f(x)

f(x) = x5+x^4+x^3 +2x^2 + 2x + 40 find the  function of f(5).

Here the value of x is given as 5

f(5) =55+54+ 53 + 2*52 + 2*5 +40.

f(5)=3125+625+125+50+10 +40 In this step5 power 5 is 3125, 5 power 4 is 625, 5cube is 125 and calculated in the function.

f(5) = 3975.

The answer for algebra polynomials functions f(5) = 3975.

Scientific Method Math

Introduction to scientific method math:
In this article shall we discuss about scientific method math. Here, scientific method explains to the organization of system to examine fact, attain new information, or precise and integrate preceding information. To be appearance scientific, a method of question have to be based on gathering visible, experimental and computable proof subject to precise principles of analysis. Scientific methods include the set of numbers through study and testing, and the formulation and testing of theory.

Steps for Scientific Method Math:

A step for scientific method math is given below that:

The scientific method math is an approach to ask and answer scientific questions through creation comments and doing experiment.

The steps for scientific method math are:

Inquire a Question

Perform Background Research

Build a Hypothesis

Observe Your Hypothesis through Doing an Experiment

Observe Your Data and sketch a Conclusion 

Communicate your outcome

It is significant for your explore to be a fair test. A "fair test" occurs when you modify just one factor (variable) and stay all other situation the same.



Overview for scientific method math:

The scientific method math for testing that is used to find out comments and answer questions. Scientists use the scientific method to seem for reason and result relations in nature. In other words, they plan a research so that changes to one thing cause something else to vary in an expected way.

Now as it does for a specialist scientist, the scientific method will help you to center your science fair scheme question, construct a hypothesis, plan, perform, and estimate your experiment.

Application of Scientific Method Math:

The application of scientific method math is given below that:

How many facts do you know about the sun?

Mass: 3.28 x 1027 tons

Temperature: 2.7 x 106 degrees Fahrenheit

Energy generated per minute: 3.5x 104 horsepower

Application of solving scientific method math:

All the numbers now are writing as products of a number between 1 and 10 and a suitable power of 10. This is also denoted as scientific method. When written in standard method, these numbers are

3,280,000,000,000,000,000,000,000,000

2,700,000

35,000

Regular Polygon Definition

Definition of polygons:
Polygons are the two dimensional closed figure that is made up of three or more than three line segments.

Types of polygons:

regular polygons
irregular polygons
convex polygons
concave polygons
Crossed polygons.

Basic Criteria that Helps in Identifying Polygons:

Identification of regular polygon:

In this polygons

All the sides of the polygons are equal in length

All the angles of the polygons are equal in degrees.

Regular polygons are convex

Identification of irregular polygon

In this polygons

No the sides of the polygons are equal in length

All the angles of the polygons are different in degrees.

Irregular polygons are convex or concave.

Identification of convex polygon:

Here

The internal angle of the polygon is equal to 180 or less than 180 degrees

Identification of concave polygon:

Here

The one or more internal angle of the polygon is greater than 180 degrees

Formula to identify the regular polygons:

Internal angle of regular polygon= `(((M-2)*180)/M)` degrees
External angle of the regular polygon = 180- internal angle of regular polygon.
Number of diagonals in a polygon= `(M (M-3))/2`


Here M is the number of sides of the regular polygon.

Area of regular Polygon = 1/2× M × Radius2 × sin (2 × p/n)
Area of regular Polygon = 1/4 × M× Side2 / tan (p/n)

I like to share this geometric probability formula with you all through my article.

Using this Formula, we can Find the Identify the Type of the Polygon.

Model problem 1:

Help to identify the polygon whose sides are equal and the number of the sides is 6 and internal angle is 120 degrees:
Solution:

Number of the sides of the polygon is 6

Internal angle is 120 degrees

All the sides are equal.

From the given condition, we can conclude the given polygon is regular hexagon.

Here the internal angle is less than 180 degrees

So it is a convex one

The polygon is regular hexagon convex polygon.

2.Help to identify the polygon whose sides are equal and the number of the sides is 8 and internal angle is 135 degrees:

Solution:

Number of the sides of the polygon is 8

Internal angle is 135 degrees

All the sides are equal.

From the given condition, we can conclude the given polygon is regular octagon

Here the internal angle is less than 180 degrees

So it is a convex one

The polygon is regular octagon convex polygon.

Solving Volume of a Cylinder

Introduction to solving volume of a cylinder:

Volume is how much three-dimensional space a substance or shape occupies or contains, often quantified numerically using the SI derived unit, the cubic meter. The volume of a container is generally understood to be the capacity of the container . Here we are going to learn about how to solving the volume of a cylinder and its examples.                                                                             (Source from Wikipedia)


Formula for solving the volume of cylinder:



Volume of cylinder = `pi` r 2 h cubic units

Solving Volume of a Cylinder - Example Problems

Example: 1

A cylinder with the radius 14 meter and height is 18 meter. calculate the volume of the cylinder.

Solving steps:

We know that formula for figure out volume of the cylinder is `pi` r2 h

Here the given is h = 18 m, r = 14 m, `pi ` = 3.14

Substitute all the value in the above formula we get

Volume = 3.14 * (14 2) * 18

Simplify the above we get

=3.14 * 196 * 18

= 11077.92

Therefore the volume of tank is 11077.92 m3

Example: 2

A cylinder with the diameter 40 meter and height is 14 meter. Calculate the volume of the cylinder

Solving steps:

Volume of cylinder =  `pi` r2 h

Here the given is diameter so we have to find the radius value

Radius = `("diameter" / 2)`

r = `38/2`

r = 20 m

Now we calculate the volume

r = 20 and h = 14 m substitute the formula we get

Volume = 3.14 * (202) * 14

Simplify the above we get

= 3.14 * 400 * 14

= 17584 m3

Therefore the volume of the cylinder is 17584  m3

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Solving Volume of a Cylinder - Example: 3

A cylinder with the radius 7.9 cm and height is 8.1 cm. calculate the volume of the cylinder

Solving steps:

Formula:

Volume of cylinder = `pi` r2 h

`pi` = 3.14, r = 7.9 cm, h = 8.1 cm substitute this value into the above formula we get

= 3.14 * (7.92) * 8.1

Simplify the above we get

= 3.14 * 62.41 * 8.1

= 1587.33 cm3

Therefore the volume of the tank = 1587.33 cm3

Bernoulli Trial Probability

Introduction to Bernoulli trial probability

Bernoulli Trails probability: Trails of a random experiment are called Bernoulli trails, if they satisfy the following conditions:

(i)   There should be a finite number of trails.

(ii)  The trails should be independent.

(iii) Each trail has exactly two outcomes: success or failure.

(iv) The probability of success remains the same in each trail.

Examples on Bernoulli Trial Probability

For example :  Throwing a die 50 times is a case of 50 Bernoulli trails, in which each trail results in the success (say an even number ) or  the failure ( an odd number ) and the probability of success (p) is same for all 50 throws. Obviously, the successive throws of the die is independent experiments. If the die is fair and have six numbers 1 to 6 written on six faces, then p =`(1)/(2)`  and q=1-p = 1-`(1)/(2)` =`(1)/(2)`  = probability of failure.

Solved Problems on Bernoulli Trial Probability

Question based on Bernoulli trail probability:

Qu:  Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether or not the trails of drawing balls are Bernoulli trails when after each draw the ball is draw the ball drawn is

(i)  Replaced into the urn.

(ii) Not replaced in the urn.

Solution:(i) The number of trails is finite. When the drawing is done with replacement, the probability of success (say, red ball) is p=`(7)/(16)`  which is same for all six trails (draws). Hence, the drawings of balls with replacement are Bernoulli trails.

(ii) When the drawing is done without replacement, the probability of success (i.e. red ball) in first trail is `(7)/(16)` ; in 2nd trail is `(6)/(15)` if the first ball drawn is red or `(7)/(15)` if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trails; hence the trails are not Bernoulli trails.

Number of Possible Combinations

Introduction for number of possible combinations:

In combinatorial mathematics, a k-combination of a finite set S is a subset of k distinct elements of S. Specifying a subset does not arrange them in a particular order; by contrast, producing the k distinct elements in a specific order defines a sequence without repetition, also called k-permutation. As an example, a poker hand can be described as a 5-combination of cards from a 52-card deck: the 5 cards of the hand are all distinct, and the order of the cards in the hand does not matter.

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Formula for Number of Possible Combinations:

Number of possible Combinations formula is commonly used in probability and statistics. Number of possible combination formula solving is the number of possible combinations of several items from a set of items. We can write in a number of ways similar to:

nCr , Where,

r => number of possibilities in the thing,

n => complete possibilities in the thing.

Also it can be understand as “n select r”.

The formula for the number of possible combinations are given as follow:

n!          n(n-1) (n-2) (n-r+1)
nCr = --------- = ----------------------------
r!(n-r!)                  r!

Example Problems for Possible Number Combinations:

1) From the group of 15 employees how many combinations are formed as an union of 3 employees?

Solution:

Solving Here n= 15 employees,

r = 3 teams

n!
nCr = ---------
r!(n-r!)


15!
15C3 = --------------
3! (15 – 3!)


15!
= -----------
3! (12)!


15*14*13
= --------------
3*2*1

= 5 * 7 * 13

15C3 = 455 possible number of combinations

2) From a group of 8 students the teacher chooses 4 students to give a demonstration. How many combinations are possible?

Solution:

n = 8 students

r = 4 students

Formula:

n!
nCr = ---------
r!(n-r!)

8!
8C4 = -----------
4! (8-4)!

8!
= -----------
4! (4)!


8*7*6*5
= -------------
4*3*2*1

15C4 = 70 possible number of combinations

3) In how many techniques can 5 girls be chosen this leaves four more girls to be chosen from the remaining 17 girls.

Solution:

The number of ways of arranging four boys from seventeen is 17x 16x 15x 14x 13

The number of methods of arranging the four among themselves is 5x4x3x2x1.

Therefore the number of ways of choosing the five girls,

17 x 16 x 15 x 14 x 13
= --------------------------
5 x 4 x 3 x 2 x 1

= 6188 combinations.

Quotient Identities

Introduction to quotient identities

In trigonometry, the following are called the quotient ratios. They are:

1. tanx = `sinx/cosx`

2. cotx = `cosx/sinx`

3. secx = `1/cosx`

4. cscx = `1/sinx`

The quotient identities are formed by the two trigonometric ratios sine and cosine functions. These quotient identities are very helpful in proving certain identities in trigonometry. These quotient identities can be manipulated in different form so that based on the need we can make use of them to prove the identities. They also can be used in different application problems to find the heights, distances and angle of elevations and angle depressions etc. Though the above things can be achieved by the two basic trigonometrical ratios sine and cosine functions, but it may take too long time to evaluate the problems involved in it. Hence it is better to know and understand the usage of the quotient trigonometric identities. We can also see some relations among them which will be of very helpful not only in trigonometry but it helps in evaluation of certain problems in calculus as well.

We have the following important identities:

1.(i) 1 + tan^2x = sec^2x

(ii) sec^2x - tan^2x = 1

(iii) tan^2x = sec^2x - 1

2. (i) 1 + cot^2x = csc^2x

(ii) csc^2x - cot^2x = 1

(iii) cot^2x = csc^2x - 1

The above identities are used to check whether the given trigonometric expressions or trigonometric equations are true or not.

Now let us solve few problems on quotient identities.

Example Problems on Quotient Identities

Ex 1: Prove the following identity:

`tanx/(secx ** 1)` +` tanx/(secx+1)` = 2cscx.

Sol: LHS =` tanx/(secx **1) + tanx/(secx+1)`

= tanx `[1/(secx ** 1) + 1/(secx +1)]`

= tanx`[( secx + 1 + secx ** 1)]/[(secx ** 1)( secx + 1)]`

= tanx `[(2 secx)/( sec^2x ** 1)]`

= `(2 tanx secx)/(tan^2x)`

= `[2secx]/tanx`

= 2 cscx = RHS.

2. Prove that (cosecx - sinx) ( secx - cosx) = `1/[tanx+cotx]`

Proof: LHS = (cosecx - sinx) ( secx - cosx)

= `(1/sinx - sinx)(1/cosx - cosx)`

= `(1- sin^2x)/sinx xx (1-cos^2x)/cosx`

= `[cos^2x]/sinx xx [sin^2x]/cosx`

= sinx cosx = `[sinx cosx]/[sin^2x + cos^2x]`

= `1/[[ sin^2x]/[sinx cosx] + [cos^2x]/ [sinx cosx]]`

= `1/[tanx + cotx] ` = RHS.

Hence the proof.

3. Show that tan^2x + cot^2x + 2 = cose2x sec^2x.

Proof: LHS = tan^2x + cot^2x + 2

= tan^2x + cot^2x + 2tanx cotx

= (tanx + cotx)2

= `[sinx/cosx + cosx/sinx]^2`

= `[sin^2x + cos^2 ]/ [ sin^2x cos^2x]`

= `1/[sin^2x] xx 1/[cos^2x] ` = cosec^2x sec^2x = RHS.

Hence the proof.

Practice Problems on Quotient Identities

Prove the following identities:

1. (cosec x - cotx )2 = `[1-cosx]/[1+cosx]`.

2. sec4x - sec^2x = tan4x + tan^2x.

Optimal Probability Equation

Introduction to Optimal Probability Equation:

Science which goes into details, at times sounds like a jargon and resembles a game. A kind of theory of word puzzles. This introduction includes a lot of jargon for little practical purposes. Theory of probability without a doubt  has its jargon. It also contains a big number of equations and formulas.

Brief Explanation of Optimal Probability Equation

The principal investigator will study several problems in optimal deterministic or stochastic control and probability, using the theory of nonlinear partial differential equations. The project consists of three parts. The optimal control of a singularly controlled Brownian motion will be studied using the results for free boundary problems. This approach has yielded regular results when the dimension is equal to two and it is the first multi-dimensional result which yields the construction of the optimal process. It now appears that it can be generalized to higher dimensions. The second part of this project is devoted to the perturbation theory of infinite dimensional Hamilton-Jacobi equations. As a case study, the principal investigator will study an asymptotic problem related to simple exclusion processes. Finally, an application of the finite dimensional perturbation theory to a problem of production planning is described. The model to be studied has failure-prone machines, and the goal is to construct approximate optimal policies by exploiting the hierarchical structure of the model.

Example of Optimal Probability Equation

Let us take example of Optimal probability, suppose, the maximum possible sanction is 500, so optimal deterrence could be achieved with a probability of 2 percent. Alternatively, we could employ a sanction of 100 and a probability of 10 percent. However, half the individuals overestimate the probability by one percentage point and the other half underestimate it by the same amount.

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Summary : Optimal Probability Equation

Sampling methodology in probability equation can be profitably used to estimate characteristics of a population of interest with less cost, faster speed, greater scope and even greater accuracy compared to a complete enumeration. We discuss basic ideas, underlying assumptions and general considerations of sampling by surveying sampling methods.

Greatest Prime Factor

Introduction to greatest prime factor:

In this section we will study about greatest prime factor. Greatest prime factor is also called as greatest common factor. It is used us to calculate the greatest prime factors of two or more numbers. In this section we will see some solved problems for greatest common factor. Let us study greatest prime factor.

Example Problems for Greatest Prime Factor:

Example problem 1: What is the greatest common factor of 4 and 2?

Solution:

First we have to write the prime factorization for 4 and 2. 2 is a prime number.

4 = 2 × 2

Next, find the common factors shared by both of the numbers.

4 = 2 × 2

2 = 2

The only common factor of 4 and 2 is 2, so the greatest common factor is 2.

Answer: The greatest common factor is 2.

Example problem 2: Dante has 12 math books and 18 science books. If he wants to distribute them evenly among some bookshelves so that each bookshelf has the same combination of math and science books, with no books left over, what is the greatest number of bookshelves Dante can use?

Solution:

Write the prime factorization for 12 and 18.

12 = 2 × 2 × 3

18 = 2 × 3 × 3

Next, find the common factors shared by both of the numbers.

12 = 2 × 2 × 3

18 = 2 × 3 × 3

Finally, multiply the common factors to find the greatest common factor.

2 × 3 = 6

The greatest common factor of 12 and 18 is 6. That means that the greatest possible number of bookshelves is 6, because 12 math books could be put onto 6 bookshelves with 2 math books each and 18 science books could be put onto 6 bookshelves with 3 science books each.

The greatest number of bookshelves Dante can use is 6.

Answer: The greatest number of bookshelves Dante can use is 6.

Practice Problems for Greatest Prime Factor:

Practice problem 1: Zahra and Raul are training for a marathon. Zahra runs 9 miles at a time while Raul prefers to run in blocks of 8 miles. At the end of a month, they realize that they have run same total number of miles. What is the smallest number of miles that each must have run?

Practice problem 2: The city of Hillsdale is honoring 15 mothers and 20 fathers as winners of its "Best Parent" contest. The plan is to take several group photographs, each with the same combination of mothers and fathers and no parents left out. What is the greatest number of photos that can be taken?

Solutions for greatest prime factor:

Solution 1: Each of them must have run 72 miles.

Solution 2: The greatest number of photos that can be taken is 5.

Binomial Theorem Probability

Introduction to Binomial theorem probability:-

In a binomial theorem probability, we deal with two outcomes.  They are called 'success' and  'falure'.  These are two mutually
disjoint outcomes.
We denote success by the symbol p and the failure by the symbol q.  Obviously p+q=1
Formula  for binomial distribution is P[X=x] = {ncx.pxqn-x}  x= 0,1,2 ....... n
The two independent constants n and p are called the parameters of the distribution
Mean = np
Variance = npq
Standard deviation =vnpq

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Problems on Binomial Theorem Probability.

Problem 1:- Eight coins are tosses simultaneously.  Find the probability of getting atleast 6 heads.
Solution:-
Number of trials n = 8
Probability of getting a head is 1/2   That is p= 1/2  that mean q = 1/2
Formula  p(x) = P(X-x) = ncxpx. qn-x and x= 0,1,2 ..... n
= 8cx (0.5)x.(0.5)n-x
= 8cx(0.5)8
Probability of getting at-least 6 heads is P(X=6) = P(X=6) + P(X=7)+P(X=8)
= p(6) +p(7)+p(8)
= (0.5)8[ 8c6 + 8c7 + 8c8}
= (0.5)8[ 8c2 + 8c1  + 1]   since 8c6 = 8c2 and  8c7=8c1
= (0.5)8[ 28 +8+1]
=  1(37)           37
----           =  -----    or   0.14 
256              256
The answer is 0.14

Problem 2 :  Given n = 6 and 9P(X=4)= P(X = 2)  ,  find p

Formula  for probability of randon variable is  P(X=x) = ncx.px.qn-x   x = 0,1,2,3........n
Here n= 6  q = p-1   and P(X=x) = 6cx pxq6-x; x=0,1,2 .....6
x=4 and x=2   hence P(X=4) = 6c4p4q2= 6c2p4q2  (since 6c4 = 6c2)
P(X=2) = 6c2p2q4
It is given 9P(X=4) = P(X=2)
hence      9.6c2p4q2 = 6c2p2q4 =   9 p2 =q2  

Let us take positive square root of both sides.
Then we get 3p= q  which can be written as 3p = 1-p
Transposing p to the other side we get        4p = 1    or  p = 1/4 = 0.25
Hence p =¼  or 0.25 

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Practice Problems on Binomial Theorem Probability:-


1.In a Binomial disribution the mean is 12 and standard deviation is 2,  Find n and p ( Answer  n=18,  p=2/3)

2.Ten coins are tossed simultaneously.Find the probability of getting
a) atleast 7 heads                                                                                                                     
b) exactly 7 heads                                                                                                                    
c) atmost 7 heads     
Answer  (a) 11/64     (b) 15/128   (c)=121/28     

3.A pair of dice is thrown 4 times.  Getting a doublet is considered a success, find the
probability of 2 success.   (answr 25/216)   

4) For a binomial distribution, mean = 7 and variance is 16.  Is this possible?  ( Answer  impossible)

5) Find the mean of a binomial distribution where  n = 10 and p = 3/5  (Answer   6)