Area of a Semi Circle

Introduction:-
In mathematics (more specifically geometry), a semicircle is a two-dimensional geometric shape that forms half of a circle. Being half of a circle's 360°, the arc of a semicircle always measures 180°. A triangle inscribed in a semicircle is always a right triangle.


The semi circular looks liek the following image.

The formula used to calculate the area of semicircle is `pi/2 r^2` .

Solved Problems:-

Problem 1:-

Calculate the area is semi circle, which has the radius of 6 centimeter.

Solution:-

The formula used to calculate the area of the semi circle is `pi/ 2 r^2` .

Given the radius of the semi circle is 6 centimeter.

Area of semi circle = `pi/ 2 r^2.`

Plug-in the value of radius in the formula we get

`= pi/ 2 6^2` .

`6^2` can be written as 6 * 6 = 36.

So the area of the semi circle is `18 pi` .


Problem 2:-

Calculate the area is semi circle, which has the radius of 12 centimeter.

Solution:-

The formula used to calculate the area of the semi circle is` pi/ 2 r^2` .

Given the radius of the semi circle is12 centimeter.

Area of semi circle =` pi/ 2 r^2` .

Plug-in the value of radius in the formula we get

= `pi/ 2 12^2` .

`12^2` can be written as 12 * 12 = 144.

So the area of the semi circle is `72 pi` .

Problem 3:-

Calculate the area is semi circle, which has the diameter of 4 centimeter.

Solution:-

The formula used to calculate the area of the semi circle is` pi/ 2 r^2` .

Given the diameter of the semi circle is 4 centimeter.

Radius is half of the diameter so the radius is `4 / 2 = 2` .

Area of semi circle = `pi/ 2 r^2.`

Plug-in the value of radius in the formula we get

= `pi/ 2 2^ 2` .

`2^2` can be written as 2 * 2 = 4.

So the area of the semi circle is `2 pi.`

Formula for Volume of a Triangular

Formula for volume of a triangular:
It have three angles and it forms like a triangle is called triangular figure  The volume of a triangular depends on their sides. A triangle pyramid has a triangle for a base. That is shown below.

Triangular prism

Triangular pyramid


Formula for volume of a triangular prism:

Volume = `1/2` * length * width * height.

L –   Length

W –  Width

H –   Height

Formula for volume of a triangular pyramid

Volume of Pyramid = (1/6)abh

a – apothem length

b – Side

h – Height

Example Problems Regarding Formula for Triangular Prism:

Example 1:

The rectangular prism has its length 12 cm , width 9 cm and height 18cm. find the volume of triangular prism.

Solution :

Given that Length  = 12

Width   =  9

Height = 18

Formula for volume of a triangular = `1/2` * length*width*height.

= `1/2` * 12 * 9 * 18

= 0.5 * 1944

= 972

The solution for the triangular prism is = 972

Example 2:

The rectangular prism has its length 10 cm , width 8cm and height 15cm. find the volume of triangular prism.   

Solution :

Given that Length  = 10

Width   =  8

Height = 15

Formula for volume of a triangular = `1/2` * length*width*height.

= `1/2` * 10 * 8 * 15

= 0.5 * 1200

= 600

The solution for the triangular prism is = 600


Example Problem Regarding Formula for Volume of a Triangular Pyramid:

Example 3:

The pyramid has its apothem length 8 cm , base side 4cm and height 12cm. find the volume of triangular pyramid.

Solution:

Given that  a - 8 cm

b – 4 cm

h – 12 cm

Formula for volume of a triangular pyramid = (1/6)abh

a – apothem length

b – Side

h – Height

=  `( 1 / 6 )` * 8 * 4 * 12

= 8 * 4 * 2

=  64

The solution for the triangular pyramid is = 63.744

Example 4:

The pyramid has its apothem length 9 cm , base side 6cm and height 13cm. find the volume of triangular pyramid.

Solution:

Given that  a - 9 cm

b – 6 cm

h – 13 cm

Formula for volume of a triangular pyramid = (1/6)abh

a – apothem length

b – Side

h – Height

=  `( 1 / 6 )` * 9 * 6 * 13

= 9 * 13

=  117

The solution for the triangular prism is = 116.532

Inverse of Non Square Matrix

Introduction to inverse of non square matrix:

Matrix has a list of data. In math matrix is a rectangular arrangement of the elements. The elements are shown in the rows and the columns. In math array elements are put in the parenthesis or square brackets. In math matrix is represented by capital letters for example A, B, C…… Square matrix has equal number of rows and equal number of column. If matrix has not equal number of rows and columns called as non square matrix.

Inverse of Non Square Matrix:

Square matrix:

Square matrix has equal number of rows and equal number of columns.

Example:

`[[a,b],[c,d]]` 

The above matrix has equal number of rows and equal number of columns. So it is called as square matrix. The order of square matrix is represented by n `xx` n or m `xx` m. The order of above matrix is 2 `xx` 2.

Non-square matrix:

Non square matrix has not equal number of rows and columns.

Example:

`[[a,b],[c,d],[e,f]]` 

The above matrix has 3 rows and 2 columns. The number of rows and number of columns of given matrix is not equal so it is a non-square matrix.

Inverse of non square matrix:

Here we see additive inverse of non-square matrix. The additive inverse of matrix X is –X.

In additive inverse put – sign to all the positive numbers in the given matrix and put + sigh to all the negative number in the given matrix. The addition of normal and inverse matrix is zero matrixes.

Additive rules for inverse of non-square matrix.

X + (-X) = (-X) + X = 0

Example:

Matrix A:

A= `[[2,1],[6,4],[9,7]]`

Inverse of matrix A:

-A = `[[-2,-1],[-6,-4],[-9,-7]]`

Example Sums for Inverse of Non Square Matrix:

Example 1:

A= `[[-5,2],[3,-8],[-1,-4]]`

Find inverse of matrix A

Solution:

Given A=  `[[-5,2],[3,-8],[-1,-4]]`

In additive inverse put – sign to all the positive numbers in the given matrix and put + sigh to all the negative number in the given matrix. The addition of normal and inverse of a matrix is zero matrixes.

-  A= -  `[[-5,2],[3,-8],[-1,-4]]`

- A =  `[[5,-2],[-3,8],[1,4]]`

Example 2:

Prove

X + (-X) = (-X) + X = 0

Solution:

Let take X = `[[1,2],[3,4],[5,6]]`

- X = -  `[[1,2],[3,4],[5,6]]`

- X= `[[-1,-2],[-3,-4],[-5,-6]]`

X + (-X) =  `[[1,2],[3,4],[5,6]]` + `[[-1,-2],[-3,-4],[-5,-6]]`

X + (-X) = 0

(-X) + X is similar to X + (-X)

Therefore X + (-X) = (-X) + X = 0 is proved

Introduction to subset and proper subset

Introduction to subset and proper subset:

SET:  A set is a collection of distinct objects, considered as an object in its own right.

Example:   A = { 4,9,6,9 } , B = {blue, green , red}

SUBSET:

In mathematics, especially in set theory, a set A is a subset of a set B if A is "contained" inside B. A and B may coincide. The relationship of one set being a subset of another is called inclusion or sometimes containment.

Example : A = { 1,5,3,8,}    , B = { 3,5} ,Here B is subset of A.  That is B `sube`

Proper Subset:

If  A and B are two sets means, A contains all the elements of  B and some additional elements that are not in B.

Example : A = { 3,5,8,10} and B ={ 3,5} .Here B is proper subset of A.

An empty set is always a proper subset of all sets.That is empty set {} is always a proper subset.

This can be denoted as ,  `O/` `subs`

Problems on Subset and Proper Subset :

Problem 1: Find the possible subsets of the set  A = { green ,Yellow,Blue }

Solution:

Given A = { green,yellow,Blue,Black}

We know that empty set is subset of every set.

So subsets of a given set are ,

B = {}

C = {Green,yellow,Blue}

D = { Green,yellow}

E=  { Yellow,Blue}

F = {Green, Blue}

G= {Green}

H = { yellow }

I = {Blue}

The above sets are the subsets of the given set A.

Problem 2 : Find the parent set of the following subsets

B = { 3,8} ,C = { 15,7}, D = { 34,15,8} , E = { 3,7,8}

Solution:

Given B = { 3,8} ,C = { 15,7}, D = { 34,15,8} , E = { 3,7,8}

We know that Subsets are the sets that contains some elements of the Parent set.

So The parent set  might be A = { 3,8,15,7,34,7 }

Problem 3: Express the following sets in-terms of  Venn diagram.

A = { -6 ,8 ,9,0 ,2 } , B = { 0,2,6} , C = { 2,6 } and D = { 34,67,89 }

Solution:

Given A = { -6 ,8 ,9,0 ,2 } , B = { 0,2,- 6} , C = { 2,-6 } and D = { 34,67,89 }

Venn diagram:

Introduction to circumference of a sphere

Introduction to circumference of a sphere

               The junction of a plane with a sphere is a circumference if the plane intersects the sphere: this circumference can have various radiuses depending on the detachment between the plane and the center of the circumference; the most circumferences are obtained while the plane contains the middle of the sphere. The lines that go round or encompass a circular figure; a periphery is called circumference.

              The workings and properties of a sphere are analogous to those of a circle. A diameter is more than a few line segments between sphere and transient from side to side its centre.

Calculating Circumference of a Sphere

Formula for calculating circumference of a sphere

The circumference of a sphere is:

                                         C=2*p*r

Where

                   C is the circumference

                    r is the radius of sphere

The value of p is 3.14

The circumference of a sphere is the space around a circle, or the outer rim. The method is quite simple but will need you to understand the different parts of a circle first. There are two corresponding formulas for arriving at the circumference of a sphere. The formula for circumference of sphere is 2 x p x Radius.

A three dimensional surface all points of which are central from a fixed point.

If two intersecting planes go beyond from side to side its center of sphere then they will subdivide the sphere into four lines or bi angles, the verticals of which all coincide with the antipodal points lying on the line of intersection of the planes.

Is this topic live math help hard for you? Watch out for my coming posts.

Circumference of a Sphere : Examples

1) Calculate the circumference of sphere of radius 17cm.

Sol

               Circumference of sphere=`2*pi*r`

Here r=17

So

                  `2*3.14*17` =106.76

Result=106.76 cm

2) How to calculate circumference of sphere with radius 4.16?

Sol

Formula for calculating surface area of sphere is `2*pi*r`

So

`2*3.14*4.16` =26.1248

Result=26.1248

Introduction to practice prime number

Introduction to practice prime number:

To Practice perfect number, we have to know the perfect number. Perfect number is obtained by adding all possible factors of this number, excepted itself but it includes 1.In other words, the addition result of factors is equal to the perfect number. The factor of the perfect number is called proper divisor. The perfect number is denoted as,

N=s(n)

Where,

N denotes perfect number.

S(n) denotes sum of all possible factors.
Types of Perfect Number to Practice:

Perfect number is classified into two different types. They are

Even perfect number
Odd perfect number

Even perfect number:

Even perfect number is invented by Euclid. The formula to practice even perfect number is 2p-1(2p-1). Where p denotes prime number.

Example to practice even perfect number:

When p=1: 21-1 ( 21 – 1) = 20 (21 – 1 ) = 1 ( 2 – 1 ) = 1 ( 1 ) = 1

When p=2: 22-1 ( 22 – 1) = 21 (22 – 1 ) = 2 ( 4 – 1 ) = 2 ( 3 ) = 6

When p=3: 23-1 ( 23 – 1) = 22 (23 – 1 ) = 4 ( 8 – 1 ) = 4 ( 7 ) = 28

When p=4: 24-1 ( 24 – 1) = 23 (24 – 1 ) = 8 ( 16 – 1 ) = 8 ( 15 ) = 120

When p=5: 25-1 ( 25 – 1) = 24 (25 – 1 ) = 16 ( 32 – 1 ) = 16 ( 31 ) = 496

Odd perfect number:

Euclid invented the perfect number, but he stated that there is no odd perfect number. But Euler took care on this. Up to this odd perfect number is unknown. If there is any perfect number that should satisfy

N>10300

Where N is in the form of,

N =qa p2ek 1 …..p2ekk
Where
q,p1......pk  denotes distinct prime numbers ( defined by Euler)
q=a=1 ( by Euler)
The least prime factor to N is smaller that (2k + 8) / 3

qa =1020
N < 24k+1
The most prime factor of  N is larger that 108
The second prime factor is greater that 104

Example to Practice of Perfect Number:

The perfect number is 6.

1 + 2 + 3 = 6

Where 1,2 and 3 are proper divisor. When we add these proper divisor or factor we will get perfect number.

Introduction to surface area of a pyramid

Surface area of a Pyramid is an important concept of study under Geometry. Geometry deals with the study of 2 dimensional shapes and 3 dimensional solids. Pyramids are solids with a base and one vertex. Students can learn about working problems associated to solids and they can get help with Geometry problems from the online tutors.

Introduction about find surface area of a pyramid:

Pyramid is a three dimensional object. There are different types of pyramid. They are square pyramid, rectangular pyramid, hexagonal pyramid and pentagonal pyramid and bases for these pyramids are square, rectangle, hexagon and pentagon respectively. The total surface area of a pyramid can be calculated by adding lateral area and base area. If the base is rectangle, we have to find the area of rectangle. The lateral area is perimeter time side length and whole divide by 2.

Surface Area of a Pyramid Formulas

Pentagonal pyramid:

Formula:

Surface area of Pentagonal pyramid =5/2 x apothem x side + 5/2 x side x slant height

Problem 2:

Example: Find the surface area of pentagonal pyramid with the given apothem length 7 meter, side 5 meter, height 6 meter and the slant height 9 meter.

Solution:

Given:

Apothem = 7 meter, side = 5 meter and slant height = 9 meter.

Surface area of pentagonal pyramid               = 5/2x 7 x 5 + 5/2 x 5 x 9

= 5/2 x 35    + 5/2 x 45

= 87.5 + 112.5

= 200

The surface area of pentagonal pyramid is 200 square meters.

Formula:

Surface area of a triangular pyramid = 1/2 x apothem x side + 3/2 x side x slant height

Example : Find the Surface area of a triangular pyramid with the given apothem length 4 meter, side 5 meter, height 6 meter and the slant height 8 meter.

Solution:

Given:

Apothem = 4 meter, side = 5 meter, slant height = 8 meter

Surface area of a triangular pyramid              = 1/2 x 4 x 5 + 3/2 x 5 x  8

=1/2 x 20    + 3/2 x 40

= 10      +    60

= 70

The surface area of triangular pyramid is 70 meter2.

Area of a Pyramid with a Square Base

Square based pyramid:

Formula:

Surface area of a square pyramid = side2 + 2 x side x length

Example :Find the surface area of a square pyramid with the given side 5 cm, height 6 cm, and the slant height 9 cm.

Solution:

Given:

side = 5 cm, length = slant height =  9cm

Surface area             = (5)2 + 2 x 5 x  9

= 25 + 90

= 115

The Surface area of pyramid is 115 square cm

Students can get help with Geometry homework associated with Pyramids and problems from the online tutors.