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.

Measure Square Yards

Introduction to Square Yard:-
The square yard is an imperial/US customary (non-metric) unit of area, formerly used in most of the English-speaking world but now generally replaced by the square metre outside of the U.S., Canada and the U.K. It is defined as the area of a square with sides of one yard in length. I like to share this Surface Area of a Trapezoidal Prism with you all through my article.

Measure Square Yards-solved Problems:-

area = `a^2` .

a = length of side.

Here a = 4.

By plugging it in to the formula we get

`area = 4^2 `  = 16.

So the area of the square is 16 square yards.

Problem 2

Measure the area of the rectangle which has the length of 5 yard and breadth 3 yard.

Solution:-

Given the length and breadth of the rectangle is 5 yard and 3 yard respectively.

The formula used to find the area of rectangle is   length * breadth.

Area= Length*breadth

By plugging in the given values in to the formula we get

Area = 15 * 9

= 145

Area of the given rectangle is 145 square yards.

Problem 3

Measure the area of the circle which has the radius of 6 yards.

Solution:-

Given the radius of the circle is 6 yards.

The formula used to find the area of the circle is `pi r^2` .

Area = `pi r^2` .

By plugging in the given values in to the formula we get

Area = `pi 6^2`

= 36 `pi` .

Area of the given circle is 36 pi squareyards. Please express your views of this topic completing the square equation by commenting on blog

Measure Square Yards-practice Problems:-

Problem: - 1

Measure the area of the rectangle which has the length of 10 yards and breadth 8 yards.

Answer: - 80 square yards.

Problem: - 2

Measure the area of the square with side measure of 4 yard find the it by applying the conversion of yard to yards.

Answer: - 144 square yards.

Problem: - 3

Measure the area of circle which has the radius of 2 yards.

Answer: - square yards.

Naming Lines in Geometry

Introduction for naming lines in geometry:
Lines are one dimension straight geometry figure and in solid geometry lines are used in designs.A lines are start with the one end and end with one direction then it said to be line segment.Lines are classified into many types which depends upon the line projection.Line segment is denoted with a connected piece of line.line segments names  has two endpoints and it is named by its endpoints. In this article contains naming lines in geometry I like to share this Area of a Rhombus Formula with you all through my article.

Naming Lines in Geometry:

In naming lines geometry section we have many types of lines which has propertyof its own.Lines are classified into following types.

Parallel lines:
In geometry parallel lines are mostly aplicable in design section, two lines which does not touch each other are called parallel lines.

Perpendicular lines:
In geometry Perpendicular lines are mostly aplicable drawing section,Two line segment  that form a L shape are called perpendicular lines.

Intersecting lines:
If two lines intersect at a point, these lines are called intersecting lines.

Concurrent lines:
The three or more lines passing through the same point are called concurrent lines. Understanding math help live is always challenging for me but thanks to all math help websites to help me out.

Problems in Naming Lines Geometry:

Example 1:
Find co-ordinate of the mid point of the line segment joining given points A(-4,1) and B(5,4)

Solution:
The required mid point is
Formul a   `((x_1+x_2)/2 ,(y_1+y_2)/2)` here,  (x1, y1) = (-4,1),(x2, y2) = (5,4)

=  `((-4+5)/(2))``((1 +4)/(2)) `

= `(-1/2) ` ,  ` (5/2)`


Example 2:
Find the slope of the lines given (1,-3) and (-1,3)

Solution:
(x1,y1)= (1,-3), (x2,y2)= (-1,3).
We know to find slope of line,m=` (y_2-y_1) /(x_2-x_1)`

=`(3+3)/(-1-1)`

m =`6/-2` = -3

Example 3:
Find the equation of the line having slope 5 and y-intercept -1.

Solution:
Applying the equation of the line is y = mx + c
Given,       m =5 ,c = -1
y = 5 x -1

or  y = 5x - 1
or  5x- y +1 = 0.