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

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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.

Solving Centroid Formula

Introduction to solve centroid formula:

In general, Centroid formula is a point on a given body or shape at which the entire mass of the body acts (center of gravity of the mass), it might also be the center of area for certain shapes. For a triangle, solving centroid is the point at which the medians of the triangle intersect; they intersect at the ratio 2:1. In the case of polygons the Centroid is found using the boundary co-ordinates solving.

Solving Centroid Formulae:

In this case the Centroid of the triangle is taken and the formula used to find out solving the centroid of a triangle is,

G (x1+x2+x3)/3 , (y1+y2+y3)/3
Where,
(x1, y1)
(x2, y2)
(x3, y3)
are the co-ordinates of the triangle.

In general, for any shape in the x-y plane the Centroid formulae can be generalized to,
G (x1+x2+x3+….+xn)/3n , (y1+y2+y3+….+yn))/n
Where,
(x1, y1)
(x2, y2)
(x3, y3)
(........)
(........)
(xn, yn) are the co-ordinates of the given shape.

Example Problems on Solving Centroid Formula:

1. Calculate the Centroid of a triangle whose co-ordinates are (3, 6) (4, 2) (3, -4)

Sol:
The given points are (3, 6) (4, 2) (3, -4),
Therefore solving,
(x1, y1) is (3, 6)
(x2, y2) is (4, 2)
(x3, y3) is (3, -4)

Formulae for the Centroid of triangle is,
G (x1+x2+x3)/3 , (y1+y2+y3)/3
(3+4+3)/3 , (6+2-4)/3
(10)/3 , (8-4)/3
10/3 , 4/3
3.33 , 1.33
Therefore the Centroid is (3.33, 1.33)

2. Calculate the Centroid of a triangle whose co-ordinates are (4, 8) (3, 2) (5, -4)
Sol:
The given points are (4, 8) (3, 2) (5, -4),
Therefore solving,
(x1, y1) is (4, 8)
(x2, y2) is (3, 2)
(x3, y3) is (5, -4)

Formulae for the Centroid of triangle is,
G (x1+x2+x3)/3 , (y1+y2+y3)/3
(4+3+5)/3 , (8+2-4)/3
(12)/3 , (10-4)/3
12/3 , 6/3
4 , 2
Therefore the Centroid is (4, 2).

3. Calculate the Centroid of the quadrilateral, whose co- ordinates are (3, 2) (5, -4) (4, 2) (3, -4)

Sol:
The given points are (3, 2) (5, -4) (4, 2) (3, -4),
Therefore,
(x1, y1) is (3, 2)
(x2, y2) is (5, -4)
(x3, y3) is (4, 2)
(x4, y4) is (3, -4)

Formulae for the Centroid is
G (x1+x2+x3+….+xn)/3n , (y1+y2+y3+….+yn))/n,
(3+5+4+3)/4 , (2-4+2-4)/4,
(15)/4 , (4-8)/4,
3.75 , -4/4
3.75 , -1
Therefore the Centroid is (3.75, -1)

Three Horizontal Lines

Introduction to three horizontal lines:

Three horizontal lines are nothing but three lines parallel to x – axis or three lines perpendicular to y – axis. Three horizontal lines mean their slopes will be zero. Because the slope of x – axis is 0. We know if there is any two lines are parallel their slope s will be equal. We will some example problems for graphing three horizontal lines. If the line is horizontal their y value is constant.

Examples for three Horizontal Lines:
If the line is parallel to x – axis we can say those lines are horizontal lines. The slopes of the horizontal lines are zero and the y value of the line is constant. So the equation of the horizontal lines are like y = some constant value.Having problem with geometric probability formula keep reading my upcoming posts, i will try to help you.

Example 1 for three horizontal lines:

Graph the following lines y = 1, y = 5 and y = -1.

Solution:

Here the line equations are y = 1, y = 5 and y = -1

The slope intercept form general equation is y = mx + c

If we compare the given equation with general form we can get the slope of the lines are 0.

If we graph these equations we will get the graph like the following.



More Examples for three Horizontal Lines:
Example 2 for three horizontal lines:

Graph the following lines y = 2, y = 3 and y = -2.

Solution:

Here the line equations are y = 0, y = 3 and y = -2

The slope intercept form general equation is y = mx + c

If we compare the given equation with general form we can get the slope of the lines are 0.

If we graph these equations we will get the graph like the following.



In this the line y = 0 is lies on the x axis.

These are some of the examples for three horizontal lines. From the above we can understand how to graph the three horizontal lines and slopes of the horizontal lines.

Line Segment Circle Intersection

Introduction:
A line can intersect a circle at two points or it can touch the circle at one point or never pass throught.

To find the point of intersection of line segment with a circle, we need to solve both equations. Thus we can get two points of intersection of the line with the circle.

Let us see few problems of this kind.

Example Problem on Line Segment Circle Intersection.

Ex 1: Find the point of intersection of the line 2x + y = 1 and x 2 + y 2 = 1.

Soln: Given: The line is 2x + y = 1 ……….. (1)

The circle is x 2 + y 2 = 1…………… (2)

(1) `=>` y = 1 – 2x

Therefore (2) = x 2 + (1 – 2x) 2 = 11 `=>` x 2 + 1 + 4x 2 – 4x = 11

`=>` 5x 2 – 4x – 10 = 0

Therefore x = `[4+- sqrt [(-4) ^2 ** 4(5) (-10)]] / [2 (5)]`

=` [4 +- sqrt [16 + 200]] / 10`

= `[4 +- sqrt 216] / 10`

=` [4 +- 6 sqrt 6] / 10` 

= `[2 +- 3 sqrt 6] / 5`

Therefore y = `1 ** [2 [(2 + 3 sqrt 6) / 5]]`

= `[5 ** 4 ** 6 sqrt 6] / 5`

= `[1 ** 6 sqrt 6] / 5`

When x = `[2 + 3 sqrt 6] / 5` , y = `[1 ** 6 sqrt 6] / 5`

When x = `[2 ** 3 sqrt 6] / 5` , y = `1 ** [2 [2 ** 3 sqrt 6] ]/ 5`

y =`[ 5** 4 + 6 sqrt 6 ]/ 5` = `[1 + 6 sqrt 6 ]/ 5`

Therefore the point of intersections are given by (`[2 + 3 sqrt 6] / 5` , `[1 ** 6 sqrt 6] / 5` ) and (`[2 ** 3 sqrt 6] / 5` , `[1 + 6 sqrt 6] / 5` )


Example Problem on Line Segment Circle Intersection.

Ex 2: Find the point of intersection of the following line and the circle. x – y = 1, x 2 + y 2 + 4x + 2y + 2 = 0.

Soln: Given: x– y = 1   =  x = y + 1 ……….. (1)

x2 + y 2 + 4x + 2y + 2 = 0 ………... (2)

By using (1) in (2), we get (y + 1) 2 + y 2 + 4 (y + 1) + 2y + 2 = 0

`=>` y 2 + 2y + 1 + y 2 + 4y + 4 + 2y + 2 = 0

2y 2 + 8y + 7 = 0

y =`[ -8 +- sqrt[ 8 ^2 ** 4 (2) (7)]] / [2 (2)]`

= `[-8 +- sqrt 64 ** 56] / 4`

=` [-8 +- 2 sqrt 2] / 4`   =`[-4 +- sqrt2]/2`

Therefore y = `[- 4 + sqrt 2 ]/ 2` , (1)  `=>` x =` [-4 + sqrt 2 ]/ 2` + 1

= `[- 2 + sqrt 2] / 2`

y =` [-4 ** sqrt 2] / 2` , (1) `=>` x =` [-4 ** sqrt 2] / 2` + 1 = `[-2 ** sqrt 2 ]/ 2`

Therefore the points are (`[-2 + sqrt 2] / 2` , `[-4 ** sqrt 2] / 2` )

Geometric Solids Patterns

Introduction to geometric solids patterns:

Geometry is a topic which deals with the shapes and sizes of any objects. The shapes are classified as solid and plane surface shapes. The geometric solids are the shapes that can be classified as cubes, cone, sphere, rectangular prism, hemisphere, etc. The geometric solid patterns can be explained with their properties. Here we see some of the solids pattern in geometry.

Classifying Solids Patterns:

Solids patterns can be classified into many shapes. We see some of them with properties.

Sphere:

It is a 3-dimensional solid shape pattern which does not have a base point. It is rounded in shape, as it is spherical. The parameters used in the solid sphere are calculated using the following formulas,

•    Volume of a sphere = 4/3 ?r3

•    The surface area of a sphere = 4?r2

It is rounded in shape, as well as spherical.

Cube:

The cube is also one of the three-dimensional solid shapes. It is made up of six equal sides with 12 edges and 8 vertices. Some parameters like volume of cube and surface area are calculated using the given formulas.

Volume of a cube can be found as a3

•    Area of a cube is given as 6a2

Rectangular prism:

The rectangular prism is a 3-dimensional solid shape which is having six numbers of sides.

The formulas used to find the parameters of a rectangular prism are given below:

•    The surface area of rectangular prism = 2(lb + bh + hl)

•    Volume of the rectangular prism can be given as lbh

Where,

l is the length, b is the breadth and h is the height of the rectangular prism.

These are some basic geometric solid patterns.

Problems to Geometric Solids Patterns:

We can solve some example problems for geometric solids patterns.

Example1:

Find the volume of the rectangular prism if its length, breadth and height are given as 6 cm, 4 cm and 2 cm.

Solution:

Formula to find volume of rectangular prism is lbh.

On substituting the given values in formula, we get

lbh = `6 xx 4 xx 2`

= `24 xx 2`

= 48

Hence the volume of the rectangular prism is found to be 48 cubic cm.

Example 2:

Find the volume of a cube if the side is measured as 3 cm.

Solution:

Given that,

Side length = 3 cm

We know the formula to find volume of cube as,

Volume of the cube is a3 = 27

Hence the volume of the cube is given as 27 cm3.

Thus these are some examples to geometric solids patterns