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]