Distinct Points

Introduction for distinct point:

The distance between any two different points (x1, y1) and (x2, y2).  The distance between two different points is basic concept in geometry. We now give an algebraic expression for the same.  Let P1 (x1, y1) and P2(x2, y2) be two distinct points in the Cartesian plane and denote the distance between P1 and P2 by d(P1, P2) or by P1P2. Draw the line segment P1P2. There are three cases are following.

Cases for Distinct Point

Case (i):

The segment `bar (P_(1)P_(2))` is parallel to the x-axis.  Then y1 = y2. Illustrate P1L and P2M, perpendicular in the direction of the y-axis. Then d(P1, P2) is equal to the distance between L and M.  But L is (x1, 0) and M is (x2, 0). So the length LM = |x1 – x2|.  Hence d(P1, P2) = |x1 – x2|.

Case (ii):

The segment `bar (P_(1)P_(2))` is parallel to the y-axis.  Then x1 = x2. Illustrate P1L and P2M, perpendicular in the direction of the y-axis. Then d(P1, P2) is equal to the distance between L and M.  But L is (0, y1) and M is (0, y2). So the length LM = |y1 – y2|.  Hence d(P1, P2) = |y1 – y2|.

Case (iii):

The line segment `bar (P_(1)P_(2))` is neither parallel to the x-axis nor parallel to the y-axis. Draw a line through P1 parallel to x-axis and a line through P2  parallel to y-axis. Let these lines intersect at the point P3. Then P3 (x2, y1). The length of the line segment P1P3 is |x1-x2| and the length of the segment P3P2 is |y1-y2|. We observe that the triangle ΔP1P3P2 is a right triangle.

Formula for distinct point:

`sqrt((x_(2) - x_(1)^(2)) + (y_(2) - y_(1))^(2))`

Problems for Distinct Points:

Let us some problems of distinct points:

Problem 1:

Find the distance between the points A(10, 5) and B(4, 8).

Solution:

Let d is the distance between the two points A and B.

Formula for distinct point:

`sqrt((x_(2) - x_(1))^(2)) + (y_(2) - y_(1))^(2))`

` = sqrt((4 - 10^(2)) + (8 - 5)^(2))`

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

`= sqrt (36 + 9)`

` =sqrt ( 45)`

` = 3sqrt ( 5)`

So, the dietance is `3sqrt(5)`

Problem 2:

Find the distance between the points A(7, 11) and B(20, 10).

Solution:

Let d is the distance between the two points A and B.

Formula for distinct point:

`sqrt((x_(2) - x_(1)^(2)) + (y_(2) - y_(1))^(2))`

`= sqrt((11 - 7^(2)) + (20 - 10)^(2))`

`= sqrt( ((4)^(2)) + (10)^(2))`

`= sqrt ( 160)`

` = 4sqrt ( 10)`

`These are problems of distinct points.`

Least-squares Line

Introduction to least squares line:

There are many methods avilable for curve fitting. The most popular method of curve fitting is the principle of least squares line method. Curve fitting is a process of finding a functional relationship betweent the variables. It is useful in the study of correlation and regression.

Definition of least Squares Line :

Let (xi, yi) be the observed set of values of the variables (x, y), where i = 1, 2, 3,…,n. Let y = f(x) be  a functional relationship between x and y. Then di = yi - f(xi) which is the difference between the observed value of y and the value of y is determined by the functional relation is called the residuals. The priniciple of least squares states that the parameters involved in f(x) should be chosen in such a way that `sum`  di2 is minimum.

Fitting a Straight Line Using least Squares Method

Consider the fitting of the straight line y = ax + b to the data (xi, yi), i = 1, 2, 3, …, n. The residual

di is given by di = yi - (axi + b).

Therefore, `sum` di2 =`sum`  (yi - axi - b)2 = R (say).

Since we are using the principle of least squares, we have to determine the value of a and b so that R is minimum.

Determine the Parameters of a and B Using Leats Squares Line Method:

Since R is minimum, `(del R)/(dela)`   = 0 `=>` - 2 `sum` (yi - axi - b)xi = 0

`=>`  `sum` (xiyi - axi2 - bxi) = 0.

Therefore, a`sum`xi2 + b `sum`xi =  `sum`xiyi  ————(1)

`(del R)/(del b)`   = 0 `=>` - 2 `sum` (yi - axi - b) = 0

Therefore, a`sum`xi + nb =  `sum`yi  ————(2)

Equations (1) and (2)  are called normal equations from which a and b can be found.

Note:  If the given data is not in linear form it can be brought to linear form by some suitable transformation of variables. Then using the priniciple of least squares the curve of best fit can be achieved.

Polar Coordinates R

Introduction :

The polar coordinates R system is an option for rectangular system. In polar coordinate system, instead of a using (x, y) coordinates, a point is represented by (r, θ). Where r delineate the length of a straight line from the point to the origin and θ delineate the angle that straight line makes with the horizontal axis. The θ as the angular coordinate and r is generally referred to as the radial coordinate. From the origin the distance of a point P is consider by a point r (an arbitrary fixed point provided by the symbol Q).

Equations for Polar Coordinates R:

Consider θ =angle between the radial line from point P to Q and the given line “θ = 0”, a kind of positive axis for polar coordinates r system. Polar coordinates r are referred in terms of ordinary Cartesian coordinates through the transformations

x = r cos θ
y = r sin θ

Where r ≥0 0≤ θ < 2π.

From these relation we can see that the polar coordinates r of point P delineates the Relation x2 + y2 = r2 (cos2 θ + sin2 θ) ⇒ x2 + y2 = r2 (so that, as we indicated, P(x, y) point is on a circle of radius r centered at Q), other hand, we can find θ by calculating the equation

tan θ = y/x =⇒ θ = arctan (y/x),

for θ in the interval 0 ≤ θ < 2π.

Examples of Polar Coordinates R:

1) The following are typical “slices” in polar coordinates r (see the margin):

Radial slice = {(r, θ): θ = π/4, 1 ≤ r ≤ 2}

Radial slice = {(r, θ): θ = 3π/2, 0.5 ≤ r ≤ 0.8}

Circular slice = {(r, θ): r = 1.2, π/4≤ θ ≤ π/2}

Circular slice = {(r, θ): r = 3, 3π/4≤ θ ≤ π}

Now we can start describing regions using slices.

2) The ideas in Example 6 show that the circumference, C, of the circle x2 + y2 = R2 can be described by both in polar coordinates r.

C = {(r, θ): r = R, and 0 ≤ θ < 2π},

Along with the Cartesian description

C = {(x, y): |y| = R2 − x2, and − R ≤ x ≤ R}.

Example of Congruence

Introduction to congruence:

Two objects are congruent if they consist of the similar shape with size. The given two triangles are congruent if their equivalent sides are equal within length also their equivalent angles are the same in size. Assume the triangles DEF and RST is congruent. These can be written as ? DEF ? ? RST. In geometry two congruent triangles contains the equal corresponding angles.

Example of Congruence

The followings are the important congruence test

ASA congruence

The two angles with the integrated faces of one triangle are equal to the corresponding two angles with the integrated faces of another triangle.

SAS congruence

The two faces with the included angle of triangle are the same to two faces with the included angle of another triangle.

AAS congruence

The two angles along throughout a non integrated side of one triangle are congruent to the identical measurements of a different triangle.

SSS congruence

Three sides of one triangle are identical to corresponding three sides of another triangle.

Example

Consider the following two triangles.




The triangle IJK is congruent to the triangle LMN

Angle I = Angle L

Angle J = Angle M

Angle K = Angle N

Length IJ = Length LM

Length JK = Length MN

Length KI = Length LN

Understanding how do you simplify fractions is always challenging for me but thanks to all math help websites to help me out.

Examples for Congruence

Example 1 for congruence

In the following triangles are congruent then find the length of sides a, b, c.



Solution

The given triangles are congruent. Therefore the lengths of the sides of the triangles are equal.

Length EG = 52

Therefore the length VT = a = 52

Length FG = 48

Therefore the length UT = b = 48

Length EF = 50

Therefore the length UV = c = 50

Thus the a = 52, b = 48, c = 50.

Example 2 for congruence

Prove that triangle LMN is congruent to triangle PQR.



Solution

Given figure the angle L and angle P are the same.

Angle L = Angle P = 75 degree

Given figure the angle N and angle Q are the same.

Angle N = Angle Q = 65 degree.

Line segment LM is equal to the line segment PR.

Line LM = Line PR = 40 cm.

Therefore ? LMN and ? PQR are congruent through AAS congruence.

Unit Circle Equation

Introduction :

A unit circle is defined as a circle with the radius value is one. Particularly in the trigonometry the unit circle with radius one is pointed at (0, 0) that is the origin in Euclidean plane of the Cartesian coordinate system. The unit circle is represented as S1. The higher dimension of the unit circle is called as the unit sphere.

Formula for Unit Circle Equation:

If the point (x, y) is on the first quadrant of the unit circle, then the point x and y are the lengths of the right triangle and the hypotenuse length is one. By using the Pythagorean Theorem, the equation of the unit circle is,

` x^2 +y^2=1`

consider` x^2=(-x)^2` for all the value of x ,it gives the positive x value and the reflection of any point of x and y axis of the unit circle is provides the unit circle equation that is ` x^2 +y^2=1` and this is not only for the first quadrant of the unit circle.

The unit circle coordinates:

The unit circle having the angle theta and also having the radius one for the unit circle. The unit circle coordinates are,(x,y) that is,

`x=cos theta or cos theta=x/1=x`

`y=sin theta or sin theta=y/1=y`

By using the Pythagorean Theorem, the equation of the unit circle is,

`cos^2 theta + sin^2 theta=1`

Example 1 for Unit Circle Equation:

To check whether the following points are on the unit circle equation or not.

i) ` (1/ 2, sqrt3/2)`

Solution:

Take the unit circle equation is,

`cos^2 theta + sin^2 theta=1`

`x^2 +y^2=1`

` x=1/2 and y=sqrt 3/2`

put x and y values in the unit circle equation

`(1/2)^2+(sqrt 3/2)^2 =1/4 +3/4 =1`

Therefore these two points are situated on the unit circle equations.

`cos theta =1/2`

`theta =cos^(-1) (1/2) =pi/3 =60^@`

Therefore these two points are situated on the unit circle equations with the angle `theta=60^@.`

Example 2 for unit circle equation:

To check whether the following points are on the unit circle equation or not.

i) ` (0, 1)`

Solution:

Take the unit circle equation is,

`cos^2 theta + sin^2 theta=1`

`x^2 +y^2=1`

here x=0 and y=1

put x and y values in the unit circle equation

`(0)^2+(1)^2 =0 +1 =1`

Therefore these two points are situated on the unit circle equations.

`cos theta =0`

`theta =cos^(-1) (0) =pi/2 =90^@`

Therefore these two points are situated on the unit circle equations with the angle `theta=90^@.`

Three Dimensional Pyramid

Introduction:

In geometry, we have two-dimensional shape and three dimensional shape. Pyramid is a thee dimensional shape. It is also called as Polyhedron. The base of pyramid may be square, rectangle, triangle or any other polygon shape. The sides of pyramid look triangular faces. In Pyramid, all vertices of base are connecting a point to above, it is called apex of pyramid.

We have various kind of three dimensional pyramids. They are following.

For example, we have Square based pyramid, Rectangular pyramid, Triangular Pyramid, Pentagonal pyramid, Hexagonal pyramid and etc.

In this article, we will see about volume of all kinds of three dimensional pyramid.

Three Dimensional Pyramid: Square Pyramid, Rectangle Pyramid

Square based pyramid:

It is a three dimensional pyramid shape having a base is square. In square pyramid, we have five vertices, four triangular faces.

Formula:

Volume of square pyramid = `1/3*base^2*height`

Example 1:

Find the volume of square pyramid for the given base side is 7 meter and height is 10 meter.

Solution:

Given:

Base side = 7 m

Height = 10 m

Volume of square pyramid           = `1/3*base^2*height`

= `1/3*7^2*10`

= 0.333 * 49 * 10

= 163.17

Therefore, Volume of square pyramid is 163.2 cubic meter.

Rectangular pyramid:

It is a three dimensional pyramid shape having a base is Rectangle. In Rectangle pyramid, we have five vertices, four triangular faces.

Formula:

Volume of Rectangular pyramid = `1/3*LenGth*width*height`

Example 2:

Find the volume of Rectangular pyramid for the given length is 10 meter , width is 8 meter and height is 12 meter.

Solution:

Given:

Length = 10 m

Width = 8 m

Height = 12 m

Volume of rectangular pyramid          = `1/3*LenGth*width*height`

= `1/3*10*8*12`

= 0.333 * 960

= 319.68

Therefore, Volume of Rectangular pyramid is 319.7 cubic meter.

Three Dimensional Pyramid: Triangular Pyramid, Hexagonal Pyramid

Triangular pyramid:

It is a three dimensional pyramid shape having a base is Triangle. In triangular pyramid, we have four vertices, four triangle faces.

Formula:

Volume of triangular pyramid = `1/6*base*height*HEIGHT`

Example 3:

Find the volume of triangular pyramid for the given base is 6 meter , height is 10 meter and HEIGHT is 12 meter.

Solution:

Given:

Base = 6 m

Height = 10 m

HEIGHT = 12 m

Volume of Triangular pyramid      = `1/6*base*height*HEIGHT`

= `1/6*6*10*12`

= 0.167 * 720

= 120.24

Therefore, Volume of triangular pyramid is 120.2 cubic meter.

Hexagonal pyramid:

It is a three dimensional pyramid shape having a base is Hexagon. In Hexagonal pyramid, we have seven vertices, six triangle faces.

Formula:

Volume of Hexagonal pyramid = `Apothem*Side*Height`

Example 3:

Find the volume of Hexagonal pyramid for the given side is 6 meter , height is 10 meter and Apothem is 7 meter.

Solution:

Given:

Side = 6 m

Height = 10 m

Apothem = 7 m

Volume of Hexagonal pyramid     = `Apothem*Side*Height`

= `7*6*10`

= 420

Therefore, Volume of Hexagonal pyramid is 420 cubic meter.

Solving Geometry Explanation

Introduction :-

In geometry, an arc is a segment of a differentiable curve in the two-dimensional plane; for example, a circular arc is a segment of the circumference of a circle. If the arc segment occupies a great circle (or great ellipse), it is considered a great-arc segment.I like to share this Math Pythagorean Theorem with you all through my article.
(Source : Wikipedia)

Example Problems for Solving Geometry Explanation

Problem 1:-

Solving geometry explanation to find the volume of cone with radius 7 cm and height 8 cm.

Solution:

Given: Radius = 7 cm

Height = 8 cm.

Volume of cone = (`1/3` ) * `pi` * radius2 * height

= (`1/3` ) * 3.14 * 72 * 8  ( multiplying these values)

= 0.33 * 3.14 * 49 *9  ( multiplying the values)

= 456.96 cubic cm.

The volume of cone is 456.96 cubic cm.

Problem 2:

Solving geometry explanation to find the Perimeter of Parallelogram for the side of a 8 and side of b is 6.

Solution:

Given: Side a = 8

Side b = 6

Perimeter of Parallelogram P = 2 * 8 + 2 * 6  ( multiplying the values)

P = 16 + 12

P = 28

The Perimeter of Parallelogram is 32

Problem 3:

Solving geometry explanation to find the circle area and circumference radius with 6 cm.
Solution:

Given: Radius = 6 cm

Area of Circle = `pi` * radius2          `pi` = 3.14

= 3.14 * 62

= 3.14 * 36   ( multiplying the values)

= 113.04 square cm.

The Area of Circle is 113.04 square cm

Circumference of Circle = 2 * `pi` * radius

= 2 * 3.14 * 6   ( multiplying the values)

= 37.68cm.

The Circumference of Circle 37.68 cm

More Example Problems for Solving Geometry Explanation

Problem 1:

Solving geometry explanation to find the Area of Triangle with height 3 cm and Base 7 cm.

Solution:

Given: Height = 3 cm

Base = 7 cm

Area of Triangle = (½) * height * base

= 0.5 * 3* 7   ( multiplying these values)

= 10.5 square cm.

The Area of Triangle 10.5 square cm

Problem 2:

Solving geometry explanation to find the Area of rhombus whose diagonal lengths are 5 cm and 8 cm.

Solution:

Area of Rhombus = (½) * Length of the diagonal 1 * Length of the diagonal 2

= `1/2` * 5* 8 ( multiplying these values)

= 20 square cm.

The Area of Triangle 20 square cm