Solve Regular Polygons

If all sides of the polygon is same length in all of its sides and all of its angles are equal, then the polygon is known as regular polygon.  The regular polygon cannot be a concave polygon. Most regular polygons are convex or star. We will see about Solve regular polygons in this article.


Regular polygon examples:
Few examples of Regular polygons are,
                      Square                                 Equilateral triangle                  Pentagon                                    
               

                        Hexagon

                    

To solve regular polygons formulas:

     The following formulas are used to solve regular polygons,

• Interior angle of each side = `(180(n-2))/n ` degrees 

• Exterior angle of each side = `360/n ` degrees 

• Diagonal=` (n(n-3))/2` 

• Area = n x area of triangle 

      = ½ * (apothem * perimeter)  (Or)

• Area  = ½ *( n* s*r) ( or )

    A= `(s2 n)/ (4tan (pi/n))`
    And there is lot of formulas for area.
 Where, n = number of sides, s = side length, r = radius or apothem

Solve regular polygons Example problems:

Example 1:
The regular hexagon has the apothem 7 cm and side is 5 cm. Calculate the area.
Solution:
Now, let us solve regular polygons using the first formula (above mentioned)
By the formula,
Area = (½) * (apothem) *(perimeter)

Perimeter of hexagon = Length of the side * Number of side
                                    = 5 * 6
                                    = 30cm
   Area of the hexagon = (1/2) * 7 * 30
                                     = 105cm²
Example 2:
The octagon has the apothem of 9 cm and the side length is 6 cm. Find its area.
Solution:
Given, n=8, s=6,  r = 9
    Let us solve regular polygons using the second formula of area (above mentioned)
           By the formula
                                    Area = (½) * n * s * r
                                            = (1/2) * 8 * 6 *9
                                            = 216cm²
Example 3:
A regular pentagon has the side of 5 inches. Determine its area.
Solution:
Given,
N= 5(pentagon), s= 5 inches
We can use the third formula now,
                                      Area= `(s^2 n)/ (4tan (pi/n))`
                                       = ` (5^2 * 5) / ( 4tan(pi/5))`

                                      = `125/ 2.906`
                                     =  43.01inches²   


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Example 4:
 An equilateral triangle has the side of 5inches length. Find its area and perimeter.
Solution:
Length = a= 5 inches
By formula,
     Area of equilateral triangle =   `sqrt(3)/4` a²  
                                                     = `sqrt(3)/4` * 25
                                                     = 10.825 sq.inches
    Perimeter of the triangle= a+b+c
                                               = 5+5+5

                                               = 15 inches.

Surface Normals

Let us see about surface normals. A surface normal also called the simple normal, to a flat surface is a vector that is vertical to that surface.A line normal to a flat, the normal factor of the force, the normal vector, etc. The concept of routine generalizes to orthogonality.A normal is used for computer graphics.

Calculating a surface normals

For a convex polygon such as a triangle, a surface usual will be calculated as the vector cross product of two (non-parallel) edges of the polygon.
For a plane given by the equation ax + by + cz + d = 0.
For a plane is represent the following equation r = a + αb + βc.
where a is a vector to get onto the level surface and b and c are non-parallel vectors lying on the plane, the normal to the plane defined is given by b × c .
For a hyperplane in n+1 dimensions equation given from following equation r = a₀ + α₁a₁ + α₂a₂ + ... + α_na_n,
where a₀ is a vector to get onto the hyperplane and a_i for i = 1, ... , n are non-parallel vectors two-faced on the hyperplane, the normal to the hyperplane can be approximated by (AA^T + bb^T) ^{− 1}b where A = [a₁, a₂, ... , a_n] and b is a random vector in the space not in the linear span of a_i.

Cross product partial derivation


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In surface normal,a surface S is given completely   as the set of points (x,y,z) it satisfying F(x,y,z) = 0, so the normal point (x,y,z) on the plane.

For a surface S given clearly as a function f(x,y) of the independent variables x,y (e.g., f(x,y) = a₀₀ + a₀₁y + a₁₀x + a₁₁xy).

The normals can be identified in at least two equivalent ways.

The first getting its contained from F(x,y,z) = z − f(x,y) = 0, from the normal follows readily.

Angle Sum Theorem

In a triangle the sum of all the three interior angles will be equal to 180^o.  If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles. Here we are going to see about the angle sum theorem.

Understanding Angle Bisector Theorem is always challenging for me but thanks to all math help websites to help me out.

Angle sum theorem:

Angle sum theorem of a triangle is equal to two right angles, i.e., 180 degrees
Given:
ABC is a triangle
To Prove
Angle A + Angle B + Angle ACB = 180^o
Produce BC to D. Through C draw CE || BA.

Proof of angle sum theorem of a Triangle:

Statement	Reason
1. Angle A = Angle ACE	Alternate angles angles BA is parallel to CE
2. Angle B = Angle ECD	Corresponding angles BA is parallel to CE
3. Angle A + angle B = Angle ACE + Angle ECD	statements (1) and (2)
4. Angle A + angle B  = Angle ACD	statement (3)
5. Angle A + Angle B + Angle ACB = Angle ACD + Angle   ACB	adding Angle ACB to both sides
6. But Angle ACD + Angle ACB = 180o	linear pair
7. Angle A + Angle B + Angle ACB = 180 °	statements (5) and (6)

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Corollary of Theorem on Angle theorem of a Triangle

If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles.
Given:
In Triangle ABC, BC is produced to D.
To Prove Corollary of Sum of Angles of Triangle:
Angle ACD = Angle A + Angle B



Proof:

Statement	Reason
1. Angle ACB + Angle ACD = 180o.	linear pair
2. Angle A + Angle B + Angle ACB = 180o	sum of the angles of a triangle = 180
3.  Angle ACB + Angle ACD = Angle A + Angle B + Angle ACB	statements (1) and (2)
4. Angle ACD = Angle A + Angle B Reason	statement (3); Angle ACB is common

Hard Fraction Problems

Hard fraction problems:

This articles discusses hard fraction problems solving. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Hard fraction problems solve for proper fraction, improper fraction and complex fraction problems.

Example for Hard fraction problems:

Example 1:
Subtract the fractions `4/5` – `3/4`

Solution:
The denominator (bottom number) is different so we have to take least common denominator (lcd).
LCD = 5 x 4 = 20
`(4 xx 4)/ (5 xx 4)` = `16/20` and `(3 xx 5)/ (4 xx 5)` = `15/20`
`16/20 ` – `15/20`
The denominators are equals
So subtracting the numerator directly = `(16-15)/20`
Simplify the above equation we get = `1/20`
Therefore the final answer is` 1/20`

Example 2:
Subtract the mixed fractions for given fractions,` 4 7/5` – `5 8/5`

Solution:
The given two mixed fractions are `4 7/5` – `5 7/5`
We need convert to mixed fraction to improper fraction `27/5` – `33/5`
The same denominators of the two fractions, so
                                         = `27/5` – `33/5`
Subtract the numerators the 27 and 33 = 27 - 33 = - 6.
The same denominator is 5.
                                         = `-6/5`
The subtract fraction solution is -`6/5` .

Example 3:
Multiply the mixed fractions for given two fraction,` 4 2/4` x `5 2/6`

Solution:
The given two mixed fractions are `4 2/4` x `5 2/6`
We need convert to mixed fraction to improper fraction `18/4` x `32/6`
Multiply the numerators the 18 and 32 = 18 x 32 = 576.
Multiply the denominators the 4 and 6 = 4 x 6 = 24
                                         =` 576/24`
The multiply fraction solution is 24

Example 4:
Convert `16/ (5/4)` to a simple fraction and reduce.

Solution:
The given complex fraction `16/ (5/4)`
Can be written as `16/1 -: 5/4`
First we have to take the reciprocal of the 2^nd number, and then multiply with the second one
Reciprocal of `5/4 ` is `4/5`
`16/1` x `4/5`
Multiply the numerator and denominator
`(16 xx 4) / (1 xx 5)` = `64/5`
Therefore complex fraction solution is `64/5`

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Practice for hard fraction problems:

Problem 1: Convert `5/ (5/4)` to a simple fraction and reduce.
Solution: 4.
Problem 2: Adding the mixed fractions for given two fraction,` 4 2/3` + `5 2/3`
Solution: `31/3`
Problem 3: Subtract the mixed fractions for given fractions,` 4 7/3` – `5 8/3`

Solution:`-4/3`

Written Translation

Introduction:
In word problem we can use the English words for translating words into mathematical expressions. In the translation we can use the math equation. This is used in the math problem to convert in the fairly simple problems. But we can’t do the figuring problem. This translating method is help to solve the word problems.

1. Working clearly will help you think clearly, and
2. Figuring out what you require will help you convert your final answer back into English.

Written translation - Translation methods:

• Addition
• Subtraction
• Multiplication
• Division
• Power

Written translation - Addition
Words that mean adding

words that mean adding	example
the sum of	the sum of a number and 6	y + 6
more than	6 more than a number	y + 6
plus	a number plus two	y + 2
added to	6 added to a number	y + 6
increased by	a number increased by 3	y + 3

Written translation - Subtraction

Words that mean subtracting

words that mean subtracting	example
less than	6 less than a number	y - 6
less	6 less a number	6 - y
minus	6 minus a number	6 - y
subtract from	subtract 6 from a number	y - 6
reduced by	a number reduced by 6	y - 6
decreased by	a number decreased by 6	y - 6
diminished by	6 diminished by a number	6 - y

Written translation - Multiplication

Words that mean multiplying

Words that mean multiplying	example
multiply	multiply a number by 3	3y
multiplied by	a number multiplied by 3	3y
times	4 times a number	4y
double	double a number	2y
twice	twice a number	2y
triple	triple a number	3y

Written translation - Division

Words that mean dividing

Words that mean dividing	example
divided by	a number divided by 6	y/6
divide into	divide 6 into a number	y/6
the quotient of	the quotient of a number and 3	y/3
half of	half of a number	y/2
one third of	one third of a number	y/3

Power

Word that mean rising to the n^th power

words that mean raising to the nth power	example
a number raised to the nth power	a number raised to the fifth power	y6
raise a number to the nth power	raise a number to the fifth power	y6
the nth power of a number	the fifth power of a number	y6
squared	a number squared	y2
cubed	a number cubed	y3

Binomial Theory

In algebra, a binomial theory is nothing but the study of binomial. The binomial is defined as the polynomial with sum of two monomial terms. Some of the binomial term is given as,The binomial also consist of distributions, coefficient, variables etc.some of the binomial terms are given as,

3x + 1 = 0

4x² + 5 = 0

9x³ + 15 = 0

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).



Properties of binomial:

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

A sum of binomial coefficients of even term is equal to a sum of binomial coefficients of odd terms, and it is equal to

2n-1

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

Examples

Example1: find the value of 5²- 3² and prove it?

Solution:

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).

52 − 32 = (5 + 3) (5 − 3).

= (8)(2)

= 16.      ---------- (1)

Proof:

a2 − b2 = 52 − 32

= 25 – 9

= 16       ---------- (2)

From 1 and 2 it proved.

Example 2: find the value of (3x + 1) and (2x + 3)?

Solution:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

(3x + 1)(2x + 3) = 3*2x2 + 3*3x + 1*2x + 1*3.

= 6x2 + 9x + 2x + 3.

Therefore, (3x + 1) (2x + 3) = 6x2 + 9x + 2x + 3

Example 3: Prove that (a + b) ³ = 8?

Solution:

From the property of binomial,

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

(a + b) ³ = (1 + 1) 3 = 2 3

= 8

Hence it is proved.

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Practice problem:

Problem1: find the value of 9²- 4² and prove it?

Answer is 65

Problem2: find the value of (2x + 1) and (x + 3)?

Answer is 2x2 + 6x + 1x + 3.

Solving Sums Integrals

Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function f. In this case it is called an indefinite integral. Integral can be classified as definite and indefinite integral. (Source: Wikipedia)

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Examples problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function ∫ (234x2 + 132x4 - 5x) dx.

Solution:

Given ∫ (234x2 + 132x4 - 5x) dx

Integrate the given function with respect to x, we get

∫ (234x2 + 132x4 - 5x) dx = ∫ 234x2 dx + ∫ 132x4 dx - ∫ 5x dx.

= 234 (x3 / 3) + 132 (x5 / 5) - 5 (x2 / 2) + c.

= 78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Answer:

The final solution is  78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Solving sums integral problem 2`:`

Find the value of the integration

`int_2^5(x^6)dx`

Solution:

Integrate the given function with respect to x, we get

`int_2^5(x^6)dx`  = `(x^7 / 7)`52

Substitute the lower and upper limits, we get

= `((5^7 / 7) - (2^7/ 7))`

= `((78125 / 7) - (128 / 7))`

= `(77997 / 7)`

Answer:

The final answer is `(77997 / 7)`

Solving sums integral problem 3:

Integration using algebraic rational function ∫ 7dx / (17x + 37)

Solution:

Using integrable function method,

Given function is ∫ `(7dx) / (17x + 37)`

Formula:

∫ [L / (ax + c)] dx = (L / a) log (ax + c)

From given, L = 7, a = 17, and c = 37

Integrate the given equation with respect to x, we get

= `(7 / 17)` log (17x + 37)

Answer:

The final answer is `(7 / 17)` log (17x + 37).

Practice problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function using integrable function ∫ `(14 / (11x + 12))` dx

Answer:

The final answer is `(14 / 11)` log (11x +12)

Solving sums integral problem 2:

Integrate the given function using integrable function ∫ `(15 / (21x + 92))` dx

Answer:

The final answer is `(15 / 21)` log (21x + 92)

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Solving sums integral problem 3:

Integrate the given function ∫ (7.9x2 - 12.9x) dx

Answer:

The final answer is `(7.9 / 3)` x3 - `(12.9 / 2)` x2