Basic geometric shapes

Geometry is a module of math which involves about the study of shapes, lines, angles, dimensions, etc. it plays vital role real time application like elevation, projection. here some content have given about basics geometry shapes. The basics geometric shapes are circle, triangle, rectangle, square.In practical geometry, we do not construct exactly the geometrical shapes but draw rough sketches of the shapes to give support to our logical reasoning.

Basic Geometric Shapes: Circle

When a point P moves in a plane in such a way that its distance remains constant from a fixed point (say C). The figure thus formed is called a circle.it is one of the basics shapes in geometry. The fixed point is called the centre of the circle and the distance between the moving point (P) and the centre (C) is called the radius of the circle.

Basic Geometric Shapes: Triangle

The triangle is a closed figure formed by three line segments. Make three points A, B, C not on the Same line. Draw the line segments AB, BC and AC .The figure formed is called triangle ABC. The symbol Δ is used to denote the triangle. Thus ΔABC is read as triangle ABC. The points A, B, C is called the vertexes of the triangle ABC.

Basic Geometric Shapes: Rectangle

In a quadrilateral, if opposite sides are equal and all the four angles are equal to right angle then it is called a rectangle. The rectangle is also one type of parallelogram, whose one of the angles is a right angle. In the figure ABCD is a rectangle AB and CD as also AD, and BC are its two pairs of opposite sides, and A, C and B, D are its two pairs of opposite Angles.

Basic Geometric Shapes: Square

In a quadrilateral if all the four sides are equal and if all the angles are right angle then the quadrilateral is called a square. a square is a rectangle, with a pair of adjacent sides equal . ABCD is a square. Since, it is also a rectangle. Therefore, AB = CD and AD = BC .A = B = C = D = 90° AC = BD Since its adjacent sides are equal.

Introduction to base of a triangle

The base of a triangle is that side of the 3 sides of a triangle which one is the perpendicular to the altitude.
There are many kind of triangles, they are categorized as Equilateral triangle (3 of whose sides are equal in length and each angle is 60o   ), Isosceles triangle (Two of whose sides are equal in length), Scalene (none of whose sides or angles are equal)

Now, Isosceles triangles are also have further divided into some categories, those are – Acute angled, right angled, obtuse angled. Scalene triangles are also have further divided into some categories - those are – Acute angled, right angled, obtuse angled.

Various Kinds of Triangles and the Circulation of their Bases

Right angled triangle and its base :

First we will  deal with the easiest one -  the right angled triangle . the picture below describes the base of a triangle

b - base of the triangle.

Base of a EquilateralTriangle :

This kind of triangle has a typical story . Any one  of the 3  sides can be taken as a base if we do  rotate the triangle keeping any  of the sides perpendicular to the altitude.

Base of a Scalenetriangle:

The side perpendicular to the base is called the base of the triangle.

Here b is the base of the triangle.

Example Problems - Base of a Triangle:

1)Find the base of the equilateral triangle shown below

solution for base of a triangle:

given side = 5 and Height = 4

consider one half of the triangle,

by using Pythagoras theorem,

a^2=b^2+c^2

b =`sqrt(a^2-c^2)`

b = `sqrt(25 -16)`

b = `sqrt(9)`

b = 3.

Therefore the base 2b = 6

6 is the base of the triangle.

2) Find the base of the triangle shown below

The given figure is a right angled triangle.

So we use Pythagorean theorem,

a^2=b^2+c^2

b =`sqrt(a^2-c^2)`

b = `sqrt(625 -46)`

b = `sqrt(576)`

b = 24.

Median-Measure of Central Tendency

Median in Statistics a branch of Mathematics is one of the measures of central tendency. We can define Median as the middle value of the given list when arranged in the numerical order.  For instance, let us consider the data, 10, 10, 9, 9, 8, 10, 7. Let us first arrange the given data in the numerical order that gives us 7,8,9,9,10,10,10. There are 7 data values, so the middle value would be the fourth value. Here the fourth value is 9 and hence the median.

To find the Median of a given set of data, first we need to arrange the given data list in the numerical order, then count the number of values in the given list. This is important as to know whether the count is even or odd. If the count is odd then we do not need to worry as the middle value in the arranged list would be the median, but if the count is even then we will have two middle values.  So, how do we go about, the middle of the two middle values in simple words the average of the two middle values gives us the median.  For instance, consider the data, 8, 7, 4, 3, 9, 10, 9, 8. First we arrange the given data in the increasing order or numerical order, 3, 4, 7, 8, 8, 9, 9, 10.

Now we count the data values, there are 8 data values which is an even number and hence we have two middle values. The two middle values are the fourth and fifth values which are 8, 8. To find median when the count of the data values is even is to find the average of the two middle values which would be (8+8)/2 = 16/2 = 8. So, the median is 8.

We can calculate median of the given data list using the Median Formula which is one more than the total number of data values divided by 2. Median Formula is given by (n+1)/2 , the value got is the position of the data value that would be the median when the list is arranged in the numerical order. For instance if there are 7 data values in the list, then (7+1)/2 = 8/2 = 4, which means the fourth value will be the median. If there are 8 values that is even number of values in the list, then (8+1)/2 = 9/2 = 4.5, a bit confusing, as we do not have 4.5th value. Here we take the average of the fourth and fifth values to calculate median as 4.5 lies between 4 and 5.

A short note on how to do standard deviation

To find the standard deviation we make a standard deviation chart. We shall try learning how to make a standard deviation chart using an example.

Example: The runs scored by a cricket player in 6 innings are 60, 45, 25, 40, 70 and 30. Find the standard deviation.
Solution: First we find the mean of the data given to us. We see that the number of observations is 6. Therefore our n = 6.

The sum of observations = ∑x_i = x_1 + x_2 + x_3 + x_4 + x_5 + x_6
∑x_i = 60 + 45 + 25 + 40 + 70 + 30 = 270
Therefore the mean = x ̅ = ∑x_i/n = 270/6 = 45

Next we make the chart for the standard deviation calculation. The first column of the chart would have the observations x_i. The second column would be x_i - x ̅ for each of the observations. The third column would be (x_i - x ̅)^2 for each of the observations. The table would look as follows for this problem:
x_i x_i - x ̅ (x_i - x ̅)^2
60 60-45 = 15 (15)^2 = 225
45 45-45 = 0 (0)^2 = 0
25 25-45 = -20 (-20)^2 = 400
40 40-45 = -5 (-5)^2 = 25
70 70-45 = 25 (25)^2 = 625
30 30-45 = -15 (-15)^2 = 225
Next we sum up the third column so we have:
∑(x_i - x ̅)^2 = 225 + 0 + 400 + 25 + 625 + 225 = 1500
Lastly we use the formula to find the standard deviation which is like this:
Standard deviation = σ = sqrt[(∑(x_i - x ̅)^2)/n] = sqrt[1500/6] = sqrt [250] = 15.81
The above steps remain the same for any number of observations.

Standard deviation graph: A standard deviation graph would look like the picture below. This is a typical graph for a normal distribution. A normal distribution is symmetric about the mean. In a normal distribution, about 68% of the observations would lie between one standard deviation from the mean. That means 68% of the observations would lie between the range x ̅ – σ to x ̅ + σ. 95% of the data would lie within 2 standard deviations of the mean, that is in the range between x ̅ – 2σ to x ̅ + 2σ. 99.7% of the data would lie with 3 standard deviations of the mean.

My forthcoming post is on Frequency Table will give you more understanding about Statistics

P a B Conditional Probability

Conditional Probability Definition:-
If B is any arbitrary event in a sample space and P ( B ) > 0. The probability of A given B is defined by
P ( A / B ) = P ( A n B ) / P ( B ).
This is called conditional probability of A given B and is written as P ( A / B ).

Proof of Conditional Probability
Let A and B be any two events in a random experiment associated with the sample space "S" then
if n ( S ) = number of elements in S
n ( B ) = number of elements in B
n ( A n B ) = number of elements in A n B.
P ( B ) = n ( B ) / n ( S )
P ( A n B ) = n ( A n B ) / n ( S )

Now for the conditional event A after B the favorable outcomes must be one of the sample points "B"
In other words, the sample space for the event A after B is B.
P ( A / B ) = n ( A ) / n ( B )
P ( A / B ) =n ( A n B ) / n ( B )                [ A n B = A ]
P ( A / B ) =  [ P ( A n B ) / n ( S ) ] / [ P ( B ) / n ( S ) ]
P ( A / B ) = P ( A n B ) / P ( B )
 Hence Proved.

Example of Conditional Probability
If the probability that a communication system will have fidelity is 81 and the probability that it will have fidelity and selectivity is 18. What is the probability that a system will high fidelity will also have high selectivity?
Solution:-
Let the event B will be the communication system to have high fidelity
P ( B ) = 81
Let the event to have high fidelity and selectivity is A n B
P ( A n B ) = 18
By the definition of Conditional probability
P (A / B) = P (A n B) / P (B)
Probability to have selectivity if it will have fidelity
P (A / B) = 18 / 81
P (A / B) = 2 / 9  = 0.22.....

I have recently faced lot of problem while learning statistics, But thank to online resources of math which helped me to learn myself easily on net.

Statistics Answers Explained

In Statistics, a branch of mathematics there are numerous Statistics Problems for which we can calculate Statistics Answers. The following are some of the Statistics math Answers which use stepwise explanation for the given problems.

Consider the scores obtained by 10 students in mathematics test is given as, 55 36 95 73 60 42 25 78 75 62. Calculate and give the statistics Answer for the mean, median, mode and range of the given scores. First we need to arrange the data in the ascending order (preferably). That gives us, 25 36 42 55 60 62 73 75 78 95. The mean of the scores is given by adding the scores and dividing it by the number of scores, which would be, [25+36+42+55+60+62+73+75+78+95]/10 = 601/10=60.1, the median is the middle value in the ordered data set. Here, the data has even number of scores and hence there are two middle values, the median in such case would be the mean of the two middle values, [60+62]/2= 61. The mode is the number that occurs the most number of times, here there is no such score and hence there is no mode for the given data set. Range is the difference of the highest value and the lowest value of the ordered data set, range = 95-25= 70

Find the mean deviation about the mean for the following data: 12,3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0,  5
Let us first find the mean of the (x bar) of the given data
(xbar) = 1/20*summation(i=1to20)[x(i)]= 200/20 = 10
The respective absolute values of the deviations from the mean , modulus[x(i) – xbar]are
2, 7, 8,7, 6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5
Summation[(i=1to20)[modulus(x(i) – xbar)] = 124
So, the mean deviation about the mean would be, M.D.(x bar) = 124/20 =6.2

Given the frequency table of the monthly salaries of 20 employees of a company:
Monthly salary(in dollars)   frequency of number of employees[f(i)]
     3500 5
     4000 8
     4200 5
     4300 2
Find the statistics answer of the mean of the salaries of the 20 people and also calculate the standard deviation
Let x(i) be the i th salary and f(i) be the corresponding frequency
Mean of the grouped data can be calculated as mu= summation [x(i)f(i)]/summation[f(i)] which gives us
= [3500x5 + 4000x8 +4200x5 +4300x2]/5+8+5+2
= [17500+32000+21000+8600]/20
= $3955
Now, let us find the standard deviation of the 20 people which can be calculated by the formula
Sqrt of {[summation [x(i)-mu]^2 times f(i)]/summation[f(i)]}.
= sqrt {[5*(3500-3955)^2+ 8*(4000-3955)^2+5*(4200-3955)^2+2*(4300-3955)^2]/20} on simplification, we get 282, rounded to the nearest unit.

Solving indefinite integrals

The following theorems and standard forms are used for finding indefinite integrals:
Theorem I: If f and g are differentiable on (a,b) and f’(x) = g’(x) for all x belonging to (a,b), then f(x) = g(x) + c, where c is a constant called the arbitrary constant.