Addition Postulate Definition : 2

20 May

Angle Addition Postulate defines that the  if a position D lies in the inside of ∠ABC, then ∠ABD + ∠DBC = ∠ABC.

angle addition postulate

  • An angle is supposed to be a complementary angle; If the sum of two angles are equal to 90° then it is known as complementary angle
  • An angle is said to be supplementary angles then the sum of two angles must be equal to 180, then it is known as supplementary angle
  • The angles which are sharing a same side are known as ‘adjacent angles’.

 

Definition of Angle Addition Postulate

 

Example 1:

Find the m∠DBC [ ∠ABD = 50° and ∠ABC = 75° ]

angle addition postulate

Solution:
m∠ABD + m∠DBC = m∠ABC  [ By Angle Addition Postulate.]

  m∠DBC = m∠ABC – m∠ABD

Now we can substitute the given angles in the equation.
so, we get ⇒ m∠DBC = 75 – 50
m∠DBC = 25

The m∠DBC is found as 25o by using the addition postulate.

 

Example 2:

Find m∠DBC. [Given ∠ABD = 40° and ∠ABC = 70°.]

angle addition postulate

Solution:
m∠ABD + m∠DBC = m∠ABC  [ By Angle Addition Postulate.]

  m∠DBC = m∠ABC – m∠ABD

Now we can substitute the given angles in the equation.
so, we get ⇒ m∠DBC = 70 – 40
m∠DBC = 30o

The m∠DBC is found as 20o by using the addition postulate.

 

Example 3:

By using the angle addition postulate, find the missing angle in the given figure.

 angle addition postulate

Solution:

Here we have to find the missing angle (i.e.) x

m∠SQR = X ( so we have to find m∠SQR )

from the given figure we can find the values of m∠PQS and m∠PQR.

m∠PQS = 35o and m∠PQR = 85o

By using the angle addition postulate we can find the missing angle (i.e. x)

m∠PQR = m∠PQS + m∠SQR

m∠PQR = m∠PQS + x

so, x = m∠PQR – m∠PQS

x = 85 – 35

x = 50o

The missing angle x is found as 50o by using the angle addition postulate.

 

Practice Problems in addition postulate definition:

 

Example 1:Find the m∠DBC [ ∠ABD = 30° and ∠ABC = 95° ]

Answer:65°

Example 2:Find m∠DBC. [Given ∠ABD = 60° and ∠ABC = 80°.]

Answer:20°

Average Total Product

17 May

Mathematics is a field that deals with the large numbers of calculations and the solutions. Product of the solution is the total of the given data. The average of the solution is that equal dividend to the number of the terms. The average of the data are been generally calculated in branch of statistics. In the field of statistics the calculations are been based on the median, mode, and the mean. The product of the average and the number of terms gives the total of the terms.

Problem for the average total product:

A mobile company sold its product for ten years and got different profit of the products in all sorts of years. The product profit differs from the production of the company. The production of the company and its profit on various years is given below.

               Year                       Production in thousands                  Profit in lakhs
                2000                                     526                                             58

2001                                     556                                             65

2002                                     530                                             74

2003                                     585                                             53

2004                                     575                                             64

2005                                     512                                             57

2006                                     606                                             62

2007                                     574                                             79

2008                                     562                                             43

2009                                     545                                             59

These are the product table that are been given by the company.

Now to calculate the total of the production in thousands we add all the ten years production:

= 526 + 556 + 530 + 585 + 575 + 512 + 606 + 574 + 562 + 545

= 5571.

Thus the total production of the company is 5571.

Now to calculate the total profit in lakhs we add all the ten years profit.

= 58 + 65 + 74 + 53 + 64 + 57 + 62 + 79 + 43 + 59

= 614.

Now to find the total average of the production;

= 5571 / 10

= 557.1

Now to calculate the total average of the profit:

= 614 / 10

= 61.4

Now we get the average total product by

= 5571 / 614

= 9.07

Thus the average total product of the company is 9.07

 

Focus on the average total product:

The quantity of the total output given per unit of the total input gives the average total product. It is just a simple calculation to the arithmetic mean.

Average product = Total product / Variable input.

Introduction To Construction

16 May

Construction is one of the clear explanations of the geometric problems. The diagram, which were required to prove a theorem or solving exercises were not necessarily precise. They were drawn only to give a motion for the position and as an aid for proper reasoning. At times one needs an accurate figure, for example to sketch a chart of a building to be constructed, to design tools, and different parts of a machine to draw road maps etc. Geometry box is used to construct the angles. It contains the following items,

1. A pair of set squares, one with angle 90°, 60°, 30° and other with angle 90°, 45°, and 45°.

2. A pair of divider with adjustments

3. A protractor.

 

introduction to construction Problem1:

 

Construct the vertical bisector of a row segment.

Steps of construction:

introduction to construction

1. Taking A and B as centers and radius new ½ of AB, draw arcs on equally sides of the line segment AB.

2. Let these arcs intersect every other at C and Q join CD.

3. Let CD cross AB at the point M. Then line CMD is the necessary at right angles bisector of AB.

Join A and B to both C and D to form AC, AD, BC and BD.

In triangles CAD and CBD,

AC=BC

AD=BD

CD=CD

CAD= CBD

ACM=BCA

Now in triangles CAM and CMB,

AC=BC

CM=CM

ACM=BCM

CMA= CMB

AM=BM and

CMA=CMB

CMA+CMB=180°

We get CMA=CMB=90°

CM, that is, CMD is the vertical bisector of AB.

I am planning to write more post on Mutually Exclusive Events with example, systems of linear equations in two variables. Keep checking my blog.

 

introduction to construction Problem 2:

 

Construct an angle of 60 degree at a point of a given ray.

Solution:

Let us take a ray AB primary point A. Assemble a ray AC such that angle CAB = 600. One angle of doing is given below

introduction to construction1

Steps:

1. Taking A as centre and a few cm as radius, draw an arc of a circle, which intersects AB, say at a point D.

2. Taking D as centre and with the equivalent radius as previous to , sketch an arc intersect the previously drawn arc, say at a point E.

3. Draw the ray AC passing through E

Then, AE=AD=DE

Therefore, EAD is an equilateral triangle and the EAD, which is the same as CAB is equal to 60°.

How To Do Rectangular Division

14 May

Introduction on how to do rectangular division:

The division of rectangle simply dives of two types. One is internal rectangular division and another one is external rectangular division. Internal rectangular division is nothing but the line segment two joining two pints in rectangular in the ratio internally. External rectangular division is nothing but the line segment two joining two pints in rectangular in the ratio externally. Let we see the formula how to do the division of rectangular internally and externally.

 

Formula – how to do the Rectangular Division

 

How to do the division of rectangular internally:

By using this formula, we can know how to find the coordinates in the internal rectangular division.

How to do the division of rectangular externally:

                      

By using this formula, we can know how to find the coordinates in the external rectangular division.

    

 

Example problems – How to do the rectangular division

 

How to find the coordinates of the points A (-4, -5), B (-9, 8) which divides the line segment joining the points A and B in the given ratio 6:8

i) Internally and

ii) Externally.

Solution:

Given in the problem:

X1 = -4      Y1 = -5

X2 = -9      Y2 = 8

m = 6        n = 8

How to find internal division:

X = { 6 ( -9 ) + 8 ( -4 ) }/(6 + 8) = ( -54 – 32)/14 = -86 / 14

Y = { 6 ( 8 ) + 8 ( -5 ) }/(6 + 8) = ( 48 – 40)/14 = 8 / 14

The required point is { ( -86/14 ), ( 8/14 ) }

I am planning to write more post on Finding Reciprocals with example, Counting Numbers. Keep checking my blog.

How to find the external division:

X = { 6 ( -9 ) – 8 ( -4 ) }/(6 + 8) = ( -54 + 32)/14 = -22 / 14

Y = { 6 ( 8 ) – 8 ( -5 ) }/(6 + 8) = ( 48 + 40)/14 = 88 / 14

The required point is { ( -22/14 ), ( 88/14 ) }

Now clear about how to do the division of of rectangular in internal and external.

Online Polar Coordinates r Help

13 May

We take the certain angle and distance from positive horizontal axis. These r and teta are polar coordinates of system. Polar coordinates are expressed as (r,teta).The equation of polar  are defined as, x= r cosΘ, y=r sinΘ. The ‘r’ represents the radial distance and ‘teta’ represent angle in clockwise.  Polar coordinates are used to plot the polar curve and polar equation.

polar coordinates

 

Online help for find polar coordinates

 

Following equations are used to determine the polar coordinates:

r = `sqrt(x_(2)+y_(2))`

Θ = tan-1`y/x`

Online polar equations are expressed by polar coordinates. Using polar coordinates, we can plot the curve. Polar coordinates have two types. They are spherical polar coordinates and cylindrical polar coordinates. Circular coordinates are used to refer the polar coordinates.

I am planning to write more post on Properties of Multiplication with example, Right Angled Triangle. Keep checking my blog.

Spherical polar coordinates:

Two angle and radial distance are used in spherical polar coordinates. Latitude and longitude angles are used in this system. The spherical coordinates are used in three dimensional polar coordinate systems. The distance and angles are represented as (ρ,Φ,Θ).

1). The radius is calculated from origin to point P(0`<=` ρ).

2). The angle between origin and z axis(0`<=`Φ`<=` 180°)

3). The angle between origin and x axis(0`<=` Θ`<=`360°).

Higher dimensional space is extending by spherical coordinates. It supports different transformation. Spherical coordinates are also defining the hyperspherical coordinate system. Online coordinate systems need help from these kinds of coordinates.

 

Cylindrical coordinates :

 

Cylindrical coordinates:

It also defines the three dimension polar coordinate systems. The triple value is defined as(r,teta,h). Using these triple values, explain the entire system.

1). The radius from point P to z axis(0`-<` r).

2). The angle between origin and x axis(0`<=` Θ`-<` 360°).

3). Signed distance is represented as h.

The above conditions are help when it is in critical position in solve the equation of system. Redundancy is involved in this. If r=0, teta lose the significance.

If we want to analyse the system, use cylindrical coordinates because it define the symmetrical axis. We take online help for plot the polar coordinates. We can convert polar coordinates into cartesian coordinates and cartesian coordinates into polar coordinates using online help.

Introduction Congruent Triangles

10 May

Introduction:

We know that a closed form created by three interconnecting lines is called a triangle. Two triangles are said to be congruent if and only if their corresponding sides and angles are equal. That is two triangles are congruent if the sides and slants of one triangle are equal to the subsequent sides and angles of the other triangle.

let us see about the introduction to congruent triangles theorem.

 

Introduction for congruent triangles theorem

 

Introduction Theorem for congruent triangles using ASA congruence rule: If two angles of the triangles and the two enclosed sides of the triangles are equal, then the two triangles are said to be congruent.

Proof : We are given two triangles PQR and STU in which:

                                         ∠ Q = ∠ T, ∠ R = ∠ U

and                                  QR = TU

We need to prove that Δ PQR ≅ Δ STU

For proving that the two triangles are congruent, we have three cases. Congruence of the two triangles sees that three cases arise.

Case (i) : Let PQ = ST

Now what do you observe? You may observe that

PQ = ST (Assumed)

∠ Q = ∠ T (Given)

QR = TU (Given)

So, Δ PQR ≅ Δ STU (By SAS rule)

 

Case (ii) : Let if possible PQ > ST. So, we can take a point M on PQ such that

                       MQ = ST. Now consider Δ MQR and Δ STU

Observe that in Δ MQR and Δ STU,

                      MQ = ST (By construction)

                     ∠ Q = ∠ T (Given)

                    QR = TU (Given)

So, we can conclude that:

                   Δ MQR ≅ Δ STU, by the SAS axiom for congruence.

Because, the triangles are congruent, their subsequent parts will be equal.

So,            ∠ MRQ = ∠ STU

But, we are given that

                 ∠ PRQ = ∠ STU

So,          ∠ PRQ = ∠ MRQ

Is this possible?

This is possible only if M coincides with P.

or,              QP = TS

So,            Δ PQR ≅ Δ STU (by SAS axiom)

congruent trianglecongruent triangle

Case (iii) : If PQ < ST, we can choose a point A on ST such that AT = PQ and repeating the arguments as given in Case (ii), we can conclude that PQ = ST and so,

Δ PQR Δ STU.

Incase, at this instant in two triangles two pairs of angles and one pair of consequent sides are alike but the side is not consist between the subsequent equivalent pairs of angles. Are the triangles still congruent? You observe that they are congruent.

I am planning to write more post on Converting Decimal to Fraction with example,Kite Geometry. Keep checking my blog.

The reason why the triangles are congruent?

We know that the addition of the three angles of a triangle is 180°. So if two couples of angles are equal, the third pair is also equal.  Two triangles are similar if any two couple of angles and one pair of corresponding sides is equal. It may said to be AAS Congruence Rule.

Draw triangles with angles of 40°, 50° and 90°. How many triangles can you able to draw like this?In fact, we can able to draw as many triangles as we want with different lengths of sides. Examine that triangles may or might not be similar to each other.

Therefore equality’s of three angles are not satisfactory for congruence of triangles.

 

Example illustrating introduction to congruent triangles:

 

Example problem for congruent triangle: Line-segment PQ is parallel to another line-segment RS. O is the mid-point of PS. Show that (i)ΔPOQ ΔSOR (ii) O is also the mid-point of QR.

congruent triangle

Solution: (i) Consider Δ POQ and Δ SOR.

PQO = ∠ SRO               (Alternate angles as PQ || RS and QR is the transversal)

POQ = ∠ SOR               (Vertically opposite angles)

OP = OS                              (Given)

Therefore, ΔPOQΔSOR (AAS rule)

(ii) OQ = OR (CPCT)

So, O is the mid-point of PQ.

Free Centimeter Graph Paper

9 May

Let us know about free centimeter graph paper, centimeter graph paper means in the graph both axis are represents in the centimeter measurement, the scale intervals are same, centimeter graph paper are used to draw the picture or chart, if we draw the diagrams on the graph paper means it will be look like a very neat and the exact measure.

Types of free centimeter graph paper :

 

There are many types of the free graph paper; here we are listed out the some of the graph paper it will be shown below,

  • Centimeter graph paper
  • Inch graph paper
  • Isometric graph paper
  • Trigonometric graph paper
  • Rectangular graph paper
  • Number lines

Image of the graph paper:

In the above diagram image of the graph paper will be shown, (the graph will be divided into the equal interval) using the graph paper we can draw the shapes like triangle, rectangle, square, circle and curve like that. It will be explained through the below graph paper

graph paper

 

Drawing square using free centimeter graph paper

 

Here we are going to draw the square using the centimeter graph paper Requirements of drawing  square we need 4 co ordinates ,sides of the squares are equal (all the sides)and angle of the square is 90 degree so it will be shown below,

Step 1

In Free centimeter graph paper we can draw two axis named as x and y

Then both the axis are separated in equal interval (here x axis =1cm and y axis =1cm)

I am planning to write more post on Horizontal Lines with example, 6th grade math problems with answers. Keep checking my blog.

Step 2:

Using the coordinate’s value we can draw the square. co ordinates are (4,0)(0,0)(0,4)(4,4).

Graph

Step 3:

In that square all sides are equal measure.

Drawing triangle using free centimeter graph paper:

Here we are going to draw the equilateral triangle  using the centimeter graph paper Requirements of drawing  triangle we need 3 co ordinates ,sides of the equilateral triangle are equal (all the sides)and angle of the equilateral triangle is 60 degree so it will be shown below,

Step 1

In Free centimeter graph paper we can draw two axis named as x and y

Then both the axis are separated in equal interval (here x axis =1cm and y axis =1cm)

Step 2:

Using the coordinate’s value we can draw the equilateral triangle , coordinates are (0, 0) (4, 0) (2, 4).

Graph

Step 3:

In that equilateral triangle all sides are equal measure and the angle also 60 degree.