Number of Quadrilaterals

Introduction to number of quadrilaterals:

A quadrilateral is a polygon with four sides or edges and four vertices or corners
Quadrilaterals are either simple (not self interesting).simple quadrilaterals are either convex or concave.
Plane is a shape which consist of four sides, and consequently four angles.

Description about Number of Quadrilaterals :
Quadrilaterals are classified into the following types

They are:

Trapezium

Parallelogram

Rhombus

Rectangle

Square

Kite

Number of Quadrilaterals with Example:
Square :

The square is one of the best example for quadrilaterals. It is defined by sides are equal, and its sum of  interior angles of all right angles should be (90°). and its opposite sides will be parallel.
we can say square is a specific case of regular polygin, in this case there are  4 number of sides. All the facts and properties described for regular polygons apply to a square.
Rectangle:

The rectangle is somwhat similar to the square, and this is one of the most generally known quadrilaterals. It is defined by having all four interior angles 90° (right angles).
Parallelogram

A quadrilateral has both pairs of opposite sides parallel and equal in length.
Parallelogram is defined by opposite sides are parallel and it should be congruent. It is the the root way  of some other quadrilaterals, which are consisting by adding some restrictions of various techniques:
A rectangle is a parallelogram but all the angles fixed at 90°
A rhombus is a parallelogram but with all sides equal in length
A square is a parallelogram but with all sides equal in length and all angles fixed at 90°.
Trapezoid

A quadrilateral should have  at least one pair of parallel sides
Rhombus:

In rhombus, the number of sides are four and its all equal.
Rhombus is looking like a special type of parallelogram. Iin a parallelogram, each sets of opposite sides are equal in length. but in rhombus, all four sides are the same length.so the properties are also same.
Rhombus is somewhat light similar ro square but the angles will not be 90o
Kite

A quadrilateral which contains  two distinct pairs of equal adjacent sides.
kite is a part of the quadrilateral relations, and while easy to understand visually, is a little tricky to define in precise mathematical terms. It has two pairs of equal sides. Each pair must be adjacent sides (sharing a common vertex) and each pair must be distinct. That is, the pairs cannot have a side in common.

Base of a Polygon

Introduction to base of polygon:

Let us discuss the base of polygon. The base of the polygon is declaring the lowest part. The base is commonly known as bottom line of the shape. Solid objects are placed on a plane by the bottom line of the surface. The straight side shape is called  the base of a polygon. Next we see the polygon base. For example in triangle we take one of the sides as base (from three sides). Similarly in square we take one of the sides of a base (from four sides).

Base of the Polygon

The two dimensional polygons are any side can be declare a base. There is the polygon “sits” bottom side is normally known as the base. The triangle is also known as the polygon. The three side polygon called as the triangles. We can say any side of triangle is base of the triangle. The all type of triangle contains three bases. The following diagram is representing the base of the polygons.

`Base = (2A)/(h)`           

The formula for the base of the triangle polygon is b = 2A / h. The b is representing the base of   the triangle. A is represent the area of the triangle. The h is representing the height of the triangle. Next we see the base of rectangle.

The rectangle is another part of the polygons. The rectangle polygons is represented the four sides. The base of the rectangle is declaring the bottom side of the diagram. The general formula for base of rectangle is b = A / h. the b is represents the base of the rectangle polygon. The A is represents the area of the rectangle polygon. The h is represents the height of the polygons.

`Base = (A)/(h)`

I am planning to write more post on How to Construct a Triangle, Properties of Quadrilaterals. Keep checking my blog.

Other Polygon Base

The next polygon declares the parallelogram. The side of the parallelogram is four. The base of the parallelogram is specifying the bottom side of the diagram. The general formula for base of parallelogram is b = A / h. the b is represents the base of the parallelogram polygon. The A is represents the area of the parallelogram polygon. The h is represents the height of the parallelogram polygons.

`Base = (A)/(h)`

These Form the Bases of a Prism

Introduction :

The prisms are the shapes that exist in the three dimensions. All the prisms are formed by the two bases and the bases of the prism formed by the faces of the prism. Regular polygons form the bases of a prism. In the following article we will discuss more about the online Volume of Right Prism help in detail.

More about the Topic these Form the Bases of a Prism

As we described before the prisms are the shapes formed by the two bases the upper base and the lower bases. All bases of the prism are the regular polygons and these form the bases of the prism. In the right regular prisms the bases of the prism are equal. The area of the bases is calculated as the product of the perimeter fo the base and the height of the prism. And the Formula for the area of the bases of the prism and the area of the total prism including the faces of the prism with the base made of n sides and side length S are,

Area of a base of the prism = `[n*S^2*cot (pi/n)]/4`

Total surface area = `[n*S^2*cot (pi/n)]/2 + S*H*n`



Example Problems on these Form the Bases of a Prism:

1. Calculate area of the base and the total area of the hexagonal prism with the height 10cm and side length of the base 5cm.

Solution:

Area of a base of the prism `= [n*S^2*cot (pi/n)]/4`

`= [6*5^2*cot (pi/5)]/4`

`= 43 cm^2`

Total surface area `= [n*S^2 cot (pi/n)]/2 + S*H*n `

`= [6*5^2*cot (pi /6)]/2 + 6*5*10`

`= (48*1.732) + 300`

`= 129.93 + 300`

`= 429.93 cm^2`

Practice problems on these form the bases of a prism:

1. Calculate area of the base and the total area of the pentagonal prism with the height 10 cm and side length of the base 6cm.

Answer: Total surface area = 423.9 cm2 and Area of a base= 61.94 cm2.

2. Calculate area of the base and the total area of the octagonal prism with the height 9cm and side length of the base 3cm.

Answer: Total surface area = 302.9 cm2 and Area of a base = 43.46 cm2.

Surface Area of Part of a Sphere

Introduction to Sphere:

Sphere is the one of the type of the geometrical figure.  Sphere is the three dimensional figure of the circle.  Surface area is the area covered by the sphere.  Radius of the sphere is the distance between centers to the circumference of the sphere.  Here we have to discuss about the surface area of sphere with example problems.

Brief Explanation Surface Area of Part of a Sphere

Sphere:

Sphere is a solid 3D shape figure.  It is look like a ball.  The picture of the sphere is look like in the following diagram

Here A is the area of the sphere and r is the radius of the sphere. And o is the center of the sphere.

Surface Area:

The surface area of the part of the sphere is mentioned by the following formulas,

Total surface area of the sphere =4 p r2 Square units
Curved surface area of the sphere=3 p r2 Square units
In the above formulas we have to substitute the radius of the sphere and the value of p is always equal to 3.14 or `(22)/(7)` . We can get the surface area of the sphere.  Curved surface area of the sphere is also called as lateral area of the sphere.


Example and Practice Problems

Find the Surface area of part of a sphere if the radius of the sphere is 105 cm.

Solution:

Given radius of the sphere r=105 cm.

Curved Surface Area:

A= 3 p r2 Square units

Here we have to substitute the values of r and p then we can get,

A= 3 x `(22)/(7)` x 105 Square centimeter

Divided by 7 we can get,

A=3 x 22 x15 Square centimeter

Multiplying this we can get,

A=990 Square centimeter

Total Surface Area:

A= 4 p r2 Square units

Here we have to substitute the values of r and p then we can get,

A= 4 x`(22)/(7)` x 105 Square centimeter

Divided by 7 we can get,

A=4 x 22 x15 Square centimeter

Multiplying this we can get,

A=1320 Square centimeter

Practice problems:

Find the surface area of the  sphere if the radius is 0.25 meter,        ans = 0.785 squrae meter
Find the surface area of the  sphere if the radius is 82 inches           ans = 84453.44 square inches.

Rectangular Pyramid Vertices

Introduction to rectangular pyramid vertices

The pyramid is the solid form objects through a polygon for a bottom.  Each faces joint on one point. The bottom of a rectangular pyramid is for all time a rectangle.  The rectangular pyramid includes the five vertices. It normally contains five sides.  Let us see about the rectangular pyramid vertices.

Rectangular Pyramid Vertices

A rectangular pyramid contains five vertices.

Rectangular pyramid is geometrical form within math. Usually pyramids are objects which contain a pyramid like formation through a triangular or else rectangular or else square or else pentagonal base etc. The bottom is which categorized the kind of pyramids. Pyramid through a rectangular bottom is identified as rectangular pyramid.

The rectangular pyramid contains the five vertices and five sides, eight edges.

In the above rectangular pyramid contains five vertices.

The volume of right rectangular pyramid = `1/3`  x base x height

Example

Given the length = 12 cm, width = 15 cm, height = 18 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (12 x 15) x 18

= `1/3`  x 180 x 18

= `1/3`  x 3240

= 1080

Therefore the volume of right rectangular pyramid is 1080 cm3

Between, if you have problem on these topics geometric probability formula, please browse expert math related websites for more help on math word problem solver online.

Examples for Rectangular Pyramid

Example 1

Compute the volume of rectangular pyramid if length = 8 cm, width = 10 cm, height = 14 cm.                             

Solution

Given the length = 8 cm, width = 10 cm, height = 14 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (8 x 10) x 14

= `1/3` * 80 x 14

= `1/3` * 1120

= 373.3

Therefore the volume of right rectangular pyramid is 373.3 cm3

Example 2

Compute the volume of rectangular pyramid if length = 6 cm, width = 12 cm, height = 20 cm.                             

Solution

Given the length = 6 cm, width = 12 cm, height = 20 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (6 x 12) x 20

= `1/3`  x 72 x 20

= `1/3`  x 1440

= 480

Therefore the volume of right rectangular pyramid is 480 cm3

Polar Equation Cartesian

Introduction on polar equation Cartesian :

Polar Equations:

The polar equation system is a two-dimensional equation system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction.

Cartesian Coordinates:

A Cartesian coordinate system specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances from the point to two fixed perpendicular directed lines, measured in the same unit of length.

Complex Form of Polar Equation Cartesian:
A complex number is of the form x + iy, where x, y belongs to R and 'i' is called imaginary unit (i = `sqrt(-1)` )

Let z = x+ iy,

Real part of z = Re(z) = x, and imaginary part of z = Im(z) = y.

The point (2, 8) written as 2 + 8i. Cartesian form are used to solve non linear shallow -water equations on the sphere.

Let, z1 = a + ib; z2 = c + id

z1 = z2; a = c; b = d.

Sinz = Sin (a+ib)

= Sina Cos hb + iCosa Sin hb

Cosz = Cos (a+ib)

= Cosa Cos hb  - iSina Sin hb

Sinz = Cosz

Sina Cos hb = Cosa Cos hb

Cosa Sin hb = Sina Sin hb

I am planning to write more post on geometric probability formula, combination probability formula. Keep checking my blog.

Steps for Solving Polar Equation Cartesian:

The following steps are used to solve from Cartesian form to polar form,

Find the value of mod(z), | z | = `sqrt(x^2 + y^2)`

Find the `theta` value by using `tan theta = (y)/(x)`

Find q in radians.

By writing the equation in polar form, `z = r(cos theta + i sin theta)`.

Example for Solving Polar Equation Cartesian:

Question: Solve z = 1 + i in polar form.

Solution:  1) r = | z | = `sqrt(1^2 + 1^2) = sqrt(1 + 1) = sqrt2`

2) `tan theta = ` `1 / 1`  => `theta = ` 45°

3) `theta = (pi)/(4)`

4) The polar form is, z = r (`cos theta + i sin theta` )

=> z = `sqrt(2)(cos (pi /4) + sin(pi/4))`

The Polar form of z = 1 + i is, z = `sqrt(2)(cos (pi /4) + sin (pi/4))`

Semicircle Learning

Introduction of semicircle learning :

Semicircle is defined as half of a circle. That is, the angle is 180 degree arc of a circle. A triangle decorated in a semicircle is always called a right triangle.

If two curves or arcs are equal, then both the segments and sectors are similar. This each part of term is called as semicircle region.

Formulas of Semicircle Learning

A semicircle is the area enclosed by a diameter and an arc of the circle joining its two ends. The length of the resulting segment is called the geometric mean, which can be proved using the concept of Pythagorean Theorem.

Formulas:

Area of semicircle (A) =circle /2

A = (pr2)/2

Circumference of semicircle(C) = (2pr)/2

C = pr

A circumference of a semicircle is calculated for the circumference of circle divided by 2.we get,

C = 2pr ==> C/2 = pr

p = 3.14 ( approximately )

A perimeter of a semicircle is the sum of circumference and diameter of a semicircle. We get,

P = pr + 2r = r (p+2)

P = 5.14 r

Between, if you have problem on these topics math equation solver free, please browse expert math related websites for more help on how to solve a math word problem.

Examples of Semicircle Learning:

Semicircle learning Ex 1!:

Find the area of semicircle with radius of 12.5 cm.

Semicircle learning sol :

We can find the area of semicircle by using the following formula,

Area = (pr2)/2

Substitute the values of p and the radius into the above formula. Then we get,

= (3.14*(12.5)2)/2

Squaring the values of radius and multiplying with 3.14 then dividing by the value of 2.

= (3.14*156.25)/2

= (490.625)/2

Then we get the final answer.

=245.3 cm2

Answer: 245.3 cm2

Semicircle learning Ex 2:

Find the perimeter of semicircle with the radius 10 cm.

Semicircle learning sol :

We can find the perimeter of semicircle by using the following formula,

Perimeter = 5.14*r

Substitute the value of r into the above formula,

=5.14*10

=51.4 cm

Answer: 51.4 cm

Semicircle learning Ex 3:

Find the circumference of semicircle with the radius of 7.5 cm.

Semicircle learning sol :

We can find the circumference of semicircle by using the following formula,

Circumference C = (2pr)/2

Circumference C = pr

Substitute the value of p and the radius.

C = 3.14*7.5

C = 23.55 cm2

Answer: C = 23.55 cm2