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

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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

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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

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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

Measuring Irregular Triangles

Introduction for measuring irregular triangles:

The irregular triangle is nothing but the triangle where the three sides are not equal and the angles present in it also different during its measurement. The only irregular triangle is the scalene triangle. Now we are going to see about the measuring of irregular triangle with some example problems.


About Measuring of Irregular Triangle:
Now we are going to see about the irregular triangles and its measurement. The irregular triangle is nothing but the scalene triangle where the sides are unequal in its length and the angles in it also unequal.

When the two sides and an angle are given and if we want to find the third side of an irregular triangle, then use the formula which is given below,

c2 = a2 + b2 - 2ac * cos ?

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Problems for Measuring Irregular Triangle:

Example 1:

The sides of the triangle are 5cm and 10 cm and the angle measuring is 40. Determine the third side of the irregular triangle.

Solution:

The third side of the triangle can be calculated by using the formula,

c2 = a2 + b2 - 2ac * cos ?

Now substitute the values in the formula we get,

c2 = 52 + 102  2(5)(10) * cos 40

Now square the values which are substituted in the formula as follows,

= 25 + 100  100 cos 40

c2 = 48.39

c = `sqrt 48.39`

c = 6.95

The value can be rounded as 7cm.

Example 2:

Find the third side of the triangle whose measurements of the triangle are 7cm, 8cm and angle measuring is about 50 degree.

Solution:

The third side of the triangle can be calculated by using the formula,

c2 = a2 + b2 - 2ac * cos ?

Now substitute the values in the formula we get,

c2 = 72 + 82  2(7)(8) * cos 40

= 49 + 64  112 cos 50

c2 = 41

c = ` sqrt 41`

c = 6.4

The value can be rounded as 6cm.

Non Coplanar Definition

Introduction to non-coplanar points:
The points which do not lie in the same plane or geometrical plane are called as non-coplanar points. Any 3 points can be enclosed by one plane or geometrical plane but four or more points cannot be enclosed by one. The points belong to the same plane are called as coplanar points. In this article we shall be discussing the non-coplanar points. Now we know what non-coplanar point is and we shall see some examples of the non-coplanar points and solve it for the same.

Example for Non Co-planar Points
1)      From the below shown figure the points are non coplanar points as they do not lie on the same plane it lies in different planes.

2)      We can see four planes with the help of four non co-planar points.

3)      Plane is the two dimensional geometrical object.

Non co-planar points

Non co-planar points

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Solved Examples for Non Co-planar Points:

Ex 1:  Check whether the following lines are co-planar or not.

3x+6y+9 = 0 and 4x+4y+11 = 0

Sol :  The given equations are  3x+6y+9 = 0 and 4x+4y+11 = 0

The slope intercept form can be given as y = mx+b

Where m indicates slope.

Comparing the above equation with the given equation, we get:

6y = -2x-9

Dividing by 6 on both sides we get:

We get,

--- (1)

The slope intercept form can be given as y = nx+b

Where n indicates slope.

Comparing the above equation with the given equation, we get:

4y = -4x-11

Dividing by 4 on both sides we get:

y = -1

We get n = -1--------- (2)

Equation (1) (2), that is m n

That is the slopes of the two equations are not equal and therefore the points lie on the two lines are non co-planar points.

Ex 2:   Check whether the following lines are co-planar or not.

x+5y+9 = 0 and 2x+10y+11 = 0

Sol :  The given equations are  x+5y+9 = 0 and 2x+10y+11 = 0

The slope intercept form can be given as y = mx+b

Where m indicates slope.

Comparing the above equation with the given equation, we get:

5y = -x-9

Dividing by 5 on both sides we get:

We get,

------- (1)

The slope intercept form can be given as y = nx+b

Where n indicates slope.

Comparing the above equation with the given equation, we get:

10y = -2x-11

Dividing by 10 on both sides we get:

We get

--------- (2)

Equation (1) =(2), that is m = n

That is the slopes of the two equations are equal and therefore the points lie on the two lines are co-planar points.