Base Pentagon Prism

Introduction:-
In geometry, the pentagonal prism is a prism with a pentagonal base.

Pentagonal prism is a type of heptahedron.

If faces are all regular, the pentagonal prism is a semi regular polyhedron and the third in an infinite set of prisms formed by square sides and two regular polygon caps.

A pentagonal prism has                                                                                                                                                                         Source:- Wikipedia.

7 faces
10 vertices
15 edges.
The pentagonal prism looks like .


Formulas For Pentagonal Prism:-


`Base area = 5/ 2 * a * s`
`Base perimeter = 5s`
`Prisms Surface area = 5as + 5sh.`
`Volume of Prism = (5/2)ash`
`here`
`a = apothem Leng th,`
`s = side,`

`h = height.`

Please express your views of this topic Congruence of Triangles by commenting on blog.

Problems on Prism:-


Problem 1:-

Find the base area  and base perimeter of the pentagonal prism if the apothem length is 4 and the side is 2.

Solution:-

Given

The apothem, length is 4

The side is 2.

The formula that is used to calculate the area of the base is `5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2` * 4 * 2.

By crossing 2 and 2 in the above equation we get

Area of the base = 5 * 4 = 20.square units.

The formula that is used to find the base perimeter is  `5s` .

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 2 = 10.

Problem 2:-

Find the base area  and base perimeter of the pentagonal prism if the apothem length is 6 and the side is 4.

Solution:-

Given

The apothem, length is 6

The side is 4.

The formula that is used to calculate the area of the base is` 5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2 * 6 * 4.`

By crossing 2 and 4 in the above equation we get

Area of the base = 5 * 12 = 60.square units.

The formula that is used to find the base perimeter is  `5s.`

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 4 = 20.

Problem 3:-

Find the base area and base perimeter of the pentagonal prism if the apothem length is 5 and the side is .

Solution:-

Given

The apothem, length is 5

The side is 7.

The formula that is used to calculate the area of the base is `5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2 * 5 * 7.`

Area of the base = `175/ 2` square units.

The formula that is used to find the base perimeter is  5s.

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 7 = 35.

angles at a point

Introduction (Angles at point):

In geometry an angle is the figure produced by two ray’s distribution a common endpoint, called the vertex of angle. The degree of the angle is the quantity of revolution that separates the two waves, and deliberate by considering the length of circular curve is out when one ray is rotate regarding the vertex to correspond with the other. The angle along with a line and a curve or along with two intersecting curve.


Positive and negative angles at a point:

In mathematical script is that angles specified a sign are positive angles if considered anticlockwise and negative angles ? is efficiently the same to a positive angle of one full rotation less ?. if considered clockwise, from a known line. If no line is specified, that can be understood to be the x-axis in the Cartesian plane. In many geometrical situations a pessimistic angle of ?? is efficiently the same to a positive angle of one full rotation less ?.

Example, a clockwise rotation of 45° (angle of ?45°) is efficiently the same to an anticlockwise rotation of 360° ? 45° (angle of 315°).

Types of Angles:

Right angle
Acute angle
obtuse angle
reflex angle
Vertical opposite angles
Co-responding angles and Alternative angles
Interior angle
Identifying angles:

Angles may be recognized by the labels involved to the three points to identify them. Example, the angle by vertex A with this by the rays AB and AC.

Potentially, an angle denoted,  ?BAC may refer to any of four angles: the clockwise angle from B to C, the anticlockwise angle from B to C, the clockwise angle as of C to B, or the anticlockwise angle as of C to B, wherever the way in that the angle is deliberate determines its sign.


Examples for angles at a point:


Example 1:

Find the value of x.

Solution:

x + 80° + 2x + x = 180° (contiguous angles on a straight line)

4x = 180° - 80°
= 100°

x = 100°
4
The answer of x = 25°

Example 2:

Find the value of x.

Solution:

48° + 90° + 120° + x = 360° ( Angles at a point )
x = 360° - (48° + 90° + 120° )
= 360° - 258°
The answer of x= 102°

Solve Allied Angles Axiom

Introduction to solve allied angles axiom:

The allied angles are nothing but the co-interior angles where they are transversely cuts the two parallel lines and the allied angles will be formed. The supplementary angles present in the geometry figures are also called as one of the types of allied angles. The allied angles total measurements are about 180 degrees. The tutors will describe the concepts to students with some example problems. Now we see how to solve allied angles axiom with the help of the tutor.

About How to Solve Allied Angles Axiom:

Now we see about how to solve for the allied angles axiom and its concepts with the help of the tutor. The allied angles are nothing but the angles where they are formed when two parallel lines are cut by a transverse line and the interior angles forms are called as the allied angles. The allied angles measurements are be about 180 degrees.

From the above figure it is given that as there are two parallel lines such as L1 and L2. These two lines can be cut by the traversal line T and the allied angles forms in the figure are given as a, b, c, d.

`angle a` +`angle d` = 180 degrees.

`angle b`  +`angle c` = 180 degrees.

The angle measures are given and we can determine the other angle with the total measurement of the angles are about 180 degrees.

Now we see some of the problems on allied angles axiom with the help of the tutor. Understanding Alternate Interior Angle is always challenging for me but thanks to all math help websites to help me out.

Problems to Solve Allied Angles Axiom:
Example:

Calculate the allied angles from the given figure?

Solution:

Now we see how to solve the allied angles measurement as follows,

The allied angles measurements are about 180 degrees.

From the given diagram, S1and S2 are the two parallel lines and cut by the transverse line T.

The allied angle are thus formed between the parallel lines.

For calculating the allied angle we have to do as follows,

Y + 70 = 180

Y = 180 - 70

Y = 110

Thus, the allied angles for 70 is about 110 degrees.

11th Grade Geometry

Introduction of 11th grade geometry :-

In grade (11) means eleventh grade is a year of education in all over the world. The eleventh grade is the final year of the secondary school. Students are usually 16 - 17 years old. Geometry is concerned with size, shape, relative figures etc. In 11th grade geometry lessons study about the angle, circle , quadrilateral etc.

Example Problems for 11th Grade Geometry :-

Problem 1:-

Determine the equation of the straight line passing through the points (1, 2) and (3, − 4).

Solution:

The equation of a straight line passing through two points is

`(y - y1)/ (y1 - y2)` = `(x - x1)/( x1 - x2)`


Here (x1, y1) = (1, 2) and (x2, y2) = (3, − 4).

Substituting the above, the required line is

`(y - 2)/(2 + 4)` = `(x - 1)/(1 - 3)`

`(y - 2)/6` = `(x - 1)/(- 2)`

`(y-2)/3` = `(x-1)/(-1)`

y − 2 = − 3 (x − 1)
y − 2 = − 3x + 3

3x + y = 5 is the required equation of the straight line.Is this topic AAA Postulate hard for you? Watch out for my coming posts.

Problem 2:-

Find the equation of the straight line passing through the point (1,2) and making intercepts on the co-ordinate axes which are in the ratio 2 : 3.

Solution:-

The intercept form is

`x/a +y/b` = 1 … (1)

The intercepts are in the ratio 2 : 3  a = 2k, b = 3k.

(1) becomes

`x/(2k) +y/(3k)` = 1     i.e. 3x + 2y = 6k

Since (1, 2) lies on the above straight line, 3 + 4 = 6k i.e. 6k = 7

Hence the required equation of the straight line is 3x + 2y = 7


Problem 3:-

Find the distance between the parallel lines 2x + 3y − 6=0 and 2x + 3y + 7 = 0.

Solution:-

The distance between the parallel lines is

`|(c_1 - c_2)/sqrt(a^2 + b^2)|` .

Here `c_1` = − 6, `c_2` = 7, a = 2, b = 3

The required distance is

`|(- 6 -7)/sqrt(2^2 + 3^2)|` = `| (-13)/sqrt(13)|`

= `sqrt(13)` units.


Practice Problems for 11th Grade Geometry :-

Problem 1:-

Find the equation of the straight line, if the perpendicular from the origin makes an angle of 120° with x-axis and the length of theperpendicular from the origin is 6 units.

Answer: The required equation of the straight line is x − `sqrt(3)` y + 12 = 0


Problem 2:-

Find the points on y-axis whose perpendicular distance from the straight line 4x − 3y − 12 = 0 is 3.

Answer: The required points are (0, 1) and (0, − 9).

Sum of Two Squares

Introduction to sum of two squares

In algebra, we have some formulae to expand squares.

They are:

`( a + b ) ^2 = a^2 + 2ab + b^2`
`( a ** b ) ^2` = `a^2 ** 2ab + b^2`
`( a + b ) ^2 + ( a ** b ) ^2` = `2 ( a^2 + b^2 )`
`( a + (1/a) ) ^2` = `a ^2 + (1/a^2) + 2`
`( a ** 1/a ) ^2` = `a ^2 + 1/a ^2 ** 2`
`( a + (1/a) ) ^2` + `( a ** 1/a ) ^2`   = `2 ( a ^2 + (1/a ^2))`
`a + b = sqrt (( a ** b ) ^2 + 4 ab)`
`a ** b ` = `sqrt (( a + b ) ^2 ** 4 ab)`


Keeping these formulae in mind, we can break up the square values to get the final answer. Now let us see few problems on sum of two squares. I like to share this Example of Obtuse Angle with you all through my article.

Example Problems on Sum of Two Squares

Ex 1: Find the value of a ^2 + b^2, If  a + b = 7 and ab = 7.

Soln: By using the above formulae, `a^2 + b ^2 = (1/2) [ ( a + b ) ^2 + ( a ** b ^2]`

Therefore  a – b = `sqrt (( a + b ) ^2 **4 ab)`

a – b = `sqrt (7 ^2 ** 4 ( 7 ))` = `sqrt (49 **28)` = `sqrt 21`

Therefore `a ^2 + b ^2` = `(1/2)[ ( a+ b ) ^2 + ( a ** b) ^2]`

= `(1/2) [ 7^2 + ( sqrt21)^2]`   =  `(1/2) [ 49 + 21 ]`

Therefore  `a^2 + b^2 = 35`

Ex 2: Find the value of A^2 + b^2, if a – b = 7 and ab = 18.

Soln: By using the above formula, `a + b = sqrt (( a ** b) ^2 + 4 ab)`

= `sqrt ((7) ^2 + 4 (18)) = sqrt (49 + 72)`

= `sqrt 121`    = 11

Therefore `a^2 + b^2` = `(1/2) [( a + b ) ^2 + ( a ** b ) ^2]`

= `(1/2) [ 11^2 + 7 ^2 ] = (1/2)[ 121 + 49 ]`

= `(1/2)` [ 170 ]  =  85

Therefore `a^2 + b^2 = 85`

Ex 3: If `a + (1/a) = 6` , find the value of `a ^2 + (1/a^2)` .

Soln: `a ^2 + (1/a^2) = (a + (1/a) ) ^2 **2 = ( 6 ^2) ** 2 = 34`

Therefore `a^2 + (1/a^2) = 34` [By using formula in 4]

Ex 4: If `a ** (1/a) = 8` , find the value of `a^2 + (1/a ^2)`

Soln: Therefore `a^2 + 1/a^2` = `(a ** (1/a)) ^2` + 2

= `8^2 + 2`

= 64 + 2 = 66.

Therefore `a^2 + (1/a^2)` = 66.   [By using formula in 5]

Ex 5: If a^2 – 5a – 1 = 0, find the value of `a^2 + (1/a^2)`

Soln: Given:  a2 – 5a – 1 = 0

`rArr` a – 5 – (1/ a) = 0    [Divide throughout by]

`rArr` `a **(1/a)`  = 5

Therefore `a^2 + (1/a^2)`  = `(a ** (1/a) ) ^2` + 2 =  `5^2` + 2 = 27. Understanding Area of Hexagon is always challenging for me but thanks to all math help websites to help me out.

Practice Problems on Sum of Two Squares

1. If a + 1/a  = 2, find a^2 + 1/a^2

Ans: 2

2. If a + b = 9 and ab = -22, find the values of a ^2 + b^2.

[And: 125]

3. If a^2 – 3a + 1 = 0, find the value of a^2 + 1/a ^2.

[Ans: 7]

Set-builder Notation Online Help

Introduction:

Set-builder notation is a mathematical data for concerning a set by stating the property that the members of the set must suit. The terms "things" in a set is known as the "elements", and are listed inside curly braces. In online study, many websites provides free help on set builder notation questions. In online study, students can learn more about all topics. Practicing set-builder notation problems help students to get prepared for test preparation and exam preparation. Practicing these problems helps student to get good scores in test and exams. In this article, we are going to see about, set-builder notation online help.I like to share this Surface Area Triangular Prism with you all through my article.

Set-builder Notation Online Help: - Representation of Set-builder Notation

The set {x: x < 3} is stated as, "the set of all x such that x is less than 0." The other form of representation of set-builder notation is the vertical line, {x | x < 3}.

General Form: {formula for elements: restrictions} or {Method for elements| restrictions}

{X: x ≠ 5} the set of all real numbers except 5

{X | x < 15} the set of all real numbers less than 15

{X | x is a positive integer} the set of all real numbers which are positive integer

{n + 1: n is an integer} The set of all real numbers (e.g. ..., -2, -1, 1, 2, 3...)

Understanding help on math homework for free is always challenging for me but thanks to all math help websites to help me out.

Set-builder Notation Online Help: - Examples

Example 1:

Express the following in set-builder notation: Y = {45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60}

Solution:

The set-builder notation for the given problem is given as,

Y = {45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60}

{ Y| Y `in` R, 44 < Y < 61}

Example 2:

Write the set of even numbers between 10 and 30 in set-builder notation.

Solution:

The set-builder notation for the given problem is given as,

X = {10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30}

{X|X `in`   even numbers, 10 `<=` X `<=` 30 }

Vertex Cycle Cover

Introduction to vertex cycle cover:

Vertex cycle cover is defined as the number of cycles, which are having the vertices and edges. The vertices are represented as G. Vertex graph are also having the subgraph.This subgraph also represented using the letter G.In this vertex cycle cover, each cover of the cycle are having only one cycle. The length of the cycle are also mentioned in this vertex cover cycle.

Explanation for Vertex Cycle Cover:

Vertex cycle cover are having the subgraph and the vertices. In this vertex cycle cover, if no vertices are present in common means, then that cycle is called vertex-disjoint cycle.If the cycles are having no edges present means, then that cycle is called as the edge-disjoint cycle. Vertex cycles covers having short cycles covers are used to represent the cubic graph. This can also having the applications in the permanent and the minimum cycles.

Properties of Vertex Cycle Cover:

The properties of the vertex cycle graph are mentioned below the following,

1. Vertex cycle cover is a permanent one among the remaining vertex covers.

The permanent vertex cycle cover are having both the directed graph and also the adjacent matrix. Both of them are mentioned in the vertex cycle cover.

2.Vertex cycles covers are having only minimum disjoint cover cycles.

This vertex cycle graph are having only minimum disjoint cycles.Because these are mentioned in the problem of finding the complexity of vertex.

3. Vertex cycles covers are having only minimum weight cover cycles.

This vertex cycle cover having the minimum weight covers are denoted by using the weighted graph present in the vertex gaph.This minimum weight cover cycles are having the sum of weights for the respective vertices.

4.Vertex cycles covers are having only double cover cycles.

This vertex cycle cover problem having the double cycle cover are denoted by using the open cycles. The set of vertices representing the open cycles.