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

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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

Line Segments in a Pentagon

Introduction to line segments:

The division of a line with two end points is called a line segment. Line segment RS which we denoted by the symbol `bar(RS)` .



Note: We shall denote a line segment `bar(RS)` by RS only.

From the above figure, we call it a line segment RS. The points R and S are called end-points of the line segment RS.

We can also name it as line segment RS.

A line segments:

(a) A line segment has a definite length.

(b) A line segment has two end-points

Line Segments in a Pentagon:
Find the line segments of the given pentagon. The pentagon shown below figure,



Solution:

Given:

Pentagon EFGHI

To find the line segments in a pentagon:

We know that the line segments are consisting of two end points. Here, the pentagon has five end points, such as E, F, G, H, and I. The five end points to form the line segments in a pentagon by connecting these end points shown in figure, such line segments are EF, FG, GH, HI, and IE. These line segments are represented by `bar(EF)` , `bar(FG)`, `bar(GH)` , `bar(HI)` , and `bar(IE)` . Therefore, the given pentagon has five line segments.Please express your views of this topic Converting Fractions to Percents by commenting on blog.

Line Segments in a Solid Pentagon:

Find the line segments of the given solid pentagon. The solid pentagon shown below figure,



Solution:

Given:

Solid pentagon ABCDEFGHIJ

To find the line segments in a solid pentagon:

We know that the line segments are consisting of two end points. Here, the pentagon has ten end points, such as A, B, C, D, E, F, G, H, I, and J. The ten end points to form the line segments in a solid pentagon by connecting these end points shown in figure, such line segments are AB, AD, AJ, BC, BF, CD, CE, DI, EF, EG, FH, GH, GI, HJ, and IJ. These line segments are represented by `bar(AB)` , `bar(AD)` , `bar(AJ)` , `bar(BC)` ,` bar(BF)` , `bar(CD)` , `bar(CE)` , `bar(DI)` , `bar(EF)` , `bar(EG)` , `bar(FH)` , `bar(GH)` , `bar(GI)` , `bar(HJ)` , and `bar(IJ)` . Therefore, the given solid pentagon has fifteen line segments.

Radius of a Circle from Circumference

Radius of a circle from circumference:
The terms radius, diameter and circumference are related to two-dimensional geometric shape named circle. Circle is a two dimensional closed shape with curved edges. The distance between the center of the circle and any point on  the circle  is always same. Circumference of the circle is 2pi r  .where, r is the radius of the circle .

Here radius is the distance between the center of the circle to any point on the circles. Circumference is the total distance around the circle. Let us discuss about the radius from the circumference of the circle,



Example Problem to Find Radius from Circumference:

Example 1:

Find radius of the circle if circumference is 34cm.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =2`pi` r =34cm

Simplify it for radius (r) we get ,

Radius r=`34/2pi`

We know that `pi ` =3.14 or` 22/7`

Therefore, r= `34/(2*3.14)`

=5.41cm

Therefore value of radius from circumference is 5.41cm

Example 2:

Find radius of the circle if circumference is 23cm.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =`2pir` =23cm

Simplify it for radius (r) we get ,

Radius `r=23/2pi`

We know that pi =3.14 or `22/7`

Therefore,` r= 23/(2*3.14)`

=3.66cm

Therefore, value of radius from circumference is 3.66 cm

Example 3:

Find radius of the circle if circumference is 72.4cm.

Solution:

The classic formula for circumference is` 2 pi r`

Therefore,

Circumference =`2pir ` =72.4cm

Simplify it for radius (r) we get ,

Radius r=`72.4/(2pi)`

We know that `pi` =3.14 or `22/7`

Therefore, r=`72.4/(2*3.14)`

=11.52 cm

Therefore, value of radius from circumference is 11.52cm

Example 4:

Find radius of the circle if circumference is 11cm.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =`2pir` =11cm

Simplify it for radius (r) we get ,

Radius r=`11/2pi`

We know that pi =3.14 or `22/7`

Therefore, r= `11/(2*3.14)`

=1.75cm

Therefore, value of radius from circumference is 1.75 cm

Example 5:

Find radius of the circle if circumference is 28inch.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =`2pir ` =28inch

Simplify it for radius (r) we get ,

Radius` r=28/(2pi)`

We know that `pi` =3.14 or `22/7`

Therefore, r= `28/(2*3.14)`

=4.45inch

Therefore, value of radius from circumference is 4.45inch

Is this topic how to measure volume hard for you? Watch out for my coming posts.

Practice Problem to Find Radius from Circumference:

1) Find radius of the circle if circumference is 12cm.

Answer:1.91cm

2)Find radius of the circle if circumference is 44cm.

Answer:7cm

Trisecting a Line Segment

Introduction for line segment:

A line segment is the basic and fundamental topic in geometry and math subject. Generally in math, a straight and long line is divided with two definite end points on both sides are known as line segments. Here in this article we have to brief explain about line segment and trisecting a line segment. And use dome example figures for how to do trisecting line segment.

Line Segment General Definition:

In math, a line segment is can be defined as one small part or distance between the two endpoints of a long line.
Line segment is also shape like as ‘a straight line’, which is joining the two points with coordinates, and the line is infinity after that, the end points.
Example figure for general line segment:



In this figure xy is the infinity line, and A, B are the two end points, and the line segments are` bar (AB)` .
The length of the line segments AB would be written as `bar (AB)` . And the line segments have used the name as to be two end points AB.

Trisecting a Line Segment:

Trisecting a line segment is the one of the process of dividing the line segments with a new line and makes the new line segments.  It is simply defined as which is one line segment, is trisected.

First we have to draw a line segment, and then bisect the line segment with a new line, and then we have to get trisecting line segment.

Generally trisecting a line segment, we use compass and ruler and makes easy.

Step by step process for trisecting line segment:

Step 1:

First we have to draw a line segment with two end points A, B. And name of the line segment is AB.

Stwp2:

Then next, draw a new dotted line through endpoint A, but not coincident with AB, draw that line for our convenient and put an end point with the name of  C and D.

Step 3:

Here the line segment distances of AC and AD are equal.

Step 4:

Then again draw a dotted line through End point B, same length and put end points and named as E and F.

This E and F are opposite lines for AB from points C and D. distance BE = EF.

Step 5:

Connect CF and DE; those two lines will cut AB into third lines equal.

Example figure for trisecting line segment:



The above example figure and explanations will make clears for the trisecting a line segment.