Volume of Equilateral Triangle

Introduction

In geometry, an equilateral triangle is a triangle in which all three sides are equal. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. They are regular polygons, and can therefore also be referred to as regular triangles.As, equilateral triangle is two dimensional, it is impossible to determine its volume.So, lets know how to determine volume of the prism with equilateral triangular base.

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Formula for Volume of Equilateral Triangle Base Prism:

The volume of equilateral triangle, ` v = 1/3 B xx H`

Example for Volume of Equilateral Triangle Base :

Example 1. Find the volume of equilateral triangle base prism of side 10 cm and height `17/2 cm`

Solution:

Let, H = `17/2 cm` , B = 10cm

Formula used:  ` V = 1/3 B H`

Therefore, volume of equilateral triangle base prism ,

`V = 1/3 xx 10 xx 17/2` = `85/3 cm3`

Example2. Find the volume of equilateral triangle base prism of side 100 m and height 85 m

Solution:

Let, `H = 17/2 m` , B = 10 m

Formula used: ` V = 1/3 B H`

Therefore, volume of equilateral triangle base prism,

`V = 1/3 xx 10 xx 17/2`

= 2833.33 m3

Example3. Find the volume of equilateral triangle base prism of side 9 cm and height `15/2` cm

Solution:

Let, `H = 15/2 cm` , B = 9 cm

Formula used:   `V = 1/3 B H`

Therefore, volume of equilateral triangle base prism ,

`V = 1/3 xx 9 xx 15/2`

= `45/2` cm3

Example4. Find the volume of equilateral triangle base prism of side 90 cm and height 75 cm

Solution:

Let, H = 75cm, B = 90 cm

Formula used:      `V = 1/3 B H`

Therefore, volume of equilateral triangle base prism ,

`V = 1/3 xx 90 xx 75`

= 2250 cm3



Practice Problems Volume of Equilateral Triangle Base Prism

1. Find the volume of equilateral triangle base prism of side 180 m and height 156 m

2. Find the volume of equilateral triangle base prism of side 450 cm and height 390 cm

Solution for practice problems of volume of equilateral triangle base prism :

1. 9360 m3

2. 58500cm3

Understanding the Concepts of Set Theory Using Venn Diagram

Sets can be termed as a collection of objects. The objects must be similar in nature. In mathematics, the objects must be related to the mathematical world. The set theory can be understood with the help of diagrammatically explanation. The Venn diagram example can be used for this purpose. The Venn diagram questions relate to the set theory and can help one to understand the set theory better. So, basically set theory can be best understood with the help of these. The definition of Venn diagram states that they are diagrams depicting sets and the operations on them. The operation on sets can be easily explained with the help of these. Once the definition has been understood the same concept can be used for the solving of the problems. The problems can be easily solved once the concept is clear.

The picture of a Venn diagram represents sets pictorially and helps in solving the operations on it. The various operations on sets include Union and intersection. The Venn diagram symbols have to be learnt first and then the whole thing can be implemented. There can problems on how to find the union or intersection of two sets. The union of sets will result in a set which will contain all the elements present in both the sets and the intersection will contain only those elements which are common to both the sets. The diagram will show the common area between the sets. This common area will denote the intersection of the sets. There can be also special type of sets in set theory. Some of them are empty set or unit set. When the intersection of two sets does not have a common area it means that there are no common elements between and the resultant set is an empty set. In the unit set there will be only one element.Please express your views of this topic how many faces does a triangular prism have by commenting on blog.

There can sets to include the various elements and can be named by the elements present in them. There are natural number sets or whole number sets. There are also sets containing the real numbers and the rational numbers. Complex numbers can also have sets in their name which will contain the collection of complex numbers. The square root of a negative number can be an example of a complex number. The number of elements in a set is denoted by the term called cardinality.

Plug in and Solve Parabolas

Introduction to Plug in and Solve Parabolas:

A curve which is formed by the intersection of a right circular cone and half of the circle is called as parabola. Directrix of a parabola is a set of all points located at same distance from a fixed line. The fixed point is called as focus and not on the directrix. The midpoint between focus of a parabola and vertex of a parabola is called as vertex. A line passing through vertex and focus of a parabola is called as axis of symmetry. Finding the above four criteria by solving parabola equation. Let us see about plug in and solve parabolas in this article.

Worked Examples to Plug in and Solve Parabolas

The general form the parabolic curve is` y = ax^(2) + bx + c` or `y^(2) = 4ax` . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.

Example 1 for Plug in and Solve Parabolas – Vertex:

Find the vertex of a parabola equation `y = x^ (2) + 4x + 3` .

Solution:

Given parabola equation is `y = x^ (2) + 4x + 3` .

To find the vertices of a given parabola, we have to plug `y = 0` in the above equation, we get,

` 0 = x^ (2) + 4x + 3`

Now we have to factor the above equation, we get,

So `x^ (2) + x + 3x + 3 = 0`

`x(x + 1) + 3 (x + 1) = 0`

`(x + 1) (x + 3) = 0`

From this `x + 1 = 0` and `x + 3 = 0`

Then `x = - 1` and `x = - 3`

So, the vertices of given parabola equation is` (-1, 0)` and `(-3, 0)` .

Example 2 for Plug in and Solve Parabolas – Focus:

What is the focus of the following parabola equation `y^ (2) = 8x` ?

Solution:

Given parabola equation is `y^ (2) = 8x` is of the form y`^ (2) = 4ax`

We know that the formula for focus, `p = 1 / (4a)`

Now compare the given equation y2 = 8x with the general equation `y^ (2) = 4ax` . So, that `4a = 8`

From this,` p = 1 / (4a) = 1 / 8`

So, the focus of a parabola equation is `(0, 1/8)` .

Other Example Problems to Plug in and Solve Parabolas

Example 3 for Plug in and Solve Parabolas – Axis of Symmetry:

What is the axis of symmetry for parabola equation `y = 5x^ (2) + 15x + 12` `?`

Solution:

Given parabola equation is `y = 5x^ (2) + 15x + 12`

From the above equation, plug `a = 5` and `b = 15` in the axis formula.

So the axis of the symmetry of the given parabola is `-b/ (2a) = - 15/ (2 xx 5) = -15/10 = - 3/2`

Therefore, the axis of symmetry for a given parabolic curve equation is` -3/2` .

Example 4 for Plug in and Solve Parabolas – Latus Rectum:

Find the latus rectum of the given parabola equation `y^ (2) = 12x` .

Solution:

The given parabola equation is `y^ (2) = 12x`

To find the latus rectum, we have to find the value of` p` .

The parabola equation is of the form `y^ (2) = 4ax`

Here `4a = 12`

So, `p = 1/ (4a) = 1/12`

The formula for latus rectum is `4p` .

Plug the value for `p =1/12` in the latus rectum formula.

From this, the latus rectum of the parabola is `= 4p = 4 (1/12) = 4/12 = 1/3`

Therefore, the latus rectum for the parabola equation is `1/3` .

Distinct Points

Introduction for distinct point:

The distance between any two different points (x1, y1) and (x2, y2).  The distance between two different points is basic concept in geometry. We now give an algebraic expression for the same.  Let P1 (x1, y1) and P2(x2, y2) be two distinct points in the Cartesian plane and denote the distance between P1 and P2 by d(P1, P2) or by P1P2. Draw the line segment P1P2. There are three cases are following.

Cases for Distinct Point

Case (i):

The segment `bar (P_(1)P_(2))` is parallel to the x-axis.  Then y1 = y2. Illustrate P1L and P2M, perpendicular in the direction of the y-axis. Then d(P1, P2) is equal to the distance between L and M.  But L is (x1, 0) and M is (x2, 0). So the length LM = |x1 – x2|.  Hence d(P1, P2) = |x1 – x2|.

Case (ii):

The segment `bar (P_(1)P_(2))` is parallel to the y-axis.  Then x1 = x2. Illustrate P1L and P2M, perpendicular in the direction of the y-axis. Then d(P1, P2) is equal to the distance between L and M.  But L is (0, y1) and M is (0, y2). So the length LM = |y1 – y2|.  Hence d(P1, P2) = |y1 – y2|.

Case (iii):

The line segment `bar (P_(1)P_(2))` is neither parallel to the x-axis nor parallel to the y-axis. Draw a line through P1 parallel to x-axis and a line through P2  parallel to y-axis. Let these lines intersect at the point P3. Then P3 (x2, y1). The length of the line segment P1P3 is |x1-x2| and the length of the segment P3P2 is |y1-y2|. We observe that the triangle ΔP1P3P2 is a right triangle.

Formula for distinct point:

`sqrt((x_(2) - x_(1)^(2)) + (y_(2) - y_(1))^(2))`

Problems for Distinct Points:

Let us some problems of distinct points:

Problem 1:

Find the distance between the points A(10, 5) and B(4, 8).

Solution:

Let d is the distance between the two points A and B.

Formula for distinct point:

`sqrt((x_(2) - x_(1))^(2)) + (y_(2) - y_(1))^(2))`

` = sqrt((4 - 10^(2)) + (8 - 5)^(2))`

`= sqrt( ((-6)^(2)) + (3)^(2))`

`= sqrt (36 + 9)`

` =sqrt ( 45)`

` = 3sqrt ( 5)`

So, the dietance is `3sqrt(5)`

Problem 2:

Find the distance between the points A(7, 11) and B(20, 10).

Solution:

Let d is the distance between the two points A and B.

Formula for distinct point:

`sqrt((x_(2) - x_(1)^(2)) + (y_(2) - y_(1))^(2))`

`= sqrt((11 - 7^(2)) + (20 - 10)^(2))`

`= sqrt( ((4)^(2)) + (10)^(2))`

`= sqrt ( 160)`

` = 4sqrt ( 10)`

`These are problems of distinct points.`

Least-squares Line

Introduction to least squares line:

There are many methods avilable for curve fitting. The most popular method of curve fitting is the principle of least squares line method. Curve fitting is a process of finding a functional relationship betweent the variables. It is useful in the study of correlation and regression.

Definition of least Squares Line :

Let (xi, yi) be the observed set of values of the variables (x, y), where i = 1, 2, 3,…,n. Let y = f(x) be  a functional relationship between x and y. Then di = yi - f(xi) which is the difference between the observed value of y and the value of y is determined by the functional relation is called the residuals. The priniciple of least squares states that the parameters involved in f(x) should be chosen in such a way that `sum`  di2 is minimum.

Fitting a Straight Line Using least Squares Method

Consider the fitting of the straight line y = ax + b to the data (xi, yi), i = 1, 2, 3, …, n. The residual

di is given by di = yi - (axi + b).

Therefore, `sum` di2 =`sum`  (yi - axi - b)2 = R (say).

Since we are using the principle of least squares, we have to determine the value of a and b so that R is minimum.

Determine the Parameters of a and B Using Leats Squares Line Method:

Since R is minimum, `(del R)/(dela)`   = 0 `=>` - 2 `sum` (yi - axi - b)xi = 0

`=>`  `sum` (xiyi - axi2 - bxi) = 0.

Therefore, a`sum`xi2 + b `sum`xi =  `sum`xiyi  ————(1)

`(del R)/(del b)`   = 0 `=>` - 2 `sum` (yi - axi - b) = 0

Therefore, a`sum`xi + nb =  `sum`yi  ————(2)

Equations (1) and (2)  are called normal equations from which a and b can be found.

Note:  If the given data is not in linear form it can be brought to linear form by some suitable transformation of variables. Then using the priniciple of least squares the curve of best fit can be achieved.

Polar Coordinates R

Introduction :

The polar coordinates R system is an option for rectangular system. In polar coordinate system, instead of a using (x, y) coordinates, a point is represented by (r, θ). Where r delineate the length of a straight line from the point to the origin and θ delineate the angle that straight line makes with the horizontal axis. The θ as the angular coordinate and r is generally referred to as the radial coordinate. From the origin the distance of a point P is consider by a point r (an arbitrary fixed point provided by the symbol Q).

Equations for Polar Coordinates R:

Consider θ =angle between the radial line from point P to Q and the given line “θ = 0”, a kind of positive axis for polar coordinates r system. Polar coordinates r are referred in terms of ordinary Cartesian coordinates through the transformations

x = r cos θ
y = r sin θ

Where r ≥0 0≤ θ < 2π.

From these relation we can see that the polar coordinates r of point P delineates the Relation x2 + y2 = r2 (cos2 θ + sin2 θ) ⇒ x2 + y2 = r2 (so that, as we indicated, P(x, y) point is on a circle of radius r centered at Q), other hand, we can find θ by calculating the equation

tan θ = y/x =⇒ θ = arctan (y/x),

for θ in the interval 0 ≤ θ < 2π.

Examples of Polar Coordinates R:

1) The following are typical “slices” in polar coordinates r (see the margin):

Radial slice = {(r, θ): θ = π/4, 1 ≤ r ≤ 2}

Radial slice = {(r, θ): θ = 3π/2, 0.5 ≤ r ≤ 0.8}

Circular slice = {(r, θ): r = 1.2, π/4≤ θ ≤ π/2}

Circular slice = {(r, θ): r = 3, 3π/4≤ θ ≤ π}

Now we can start describing regions using slices.

2) The ideas in Example 6 show that the circumference, C, of the circle x2 + y2 = R2 can be described by both in polar coordinates r.

C = {(r, θ): r = R, and 0 ≤ θ < 2π},

Along with the Cartesian description

C = {(x, y): |y| = R2 − x2, and − R ≤ x ≤ R}.

Example of Congruence

Introduction to congruence:

Two objects are congruent if they consist of the similar shape with size. The given two triangles are congruent if their equivalent sides are equal within length also their equivalent angles are the same in size. Assume the triangles DEF and RST is congruent. These can be written as ? DEF ? ? RST. In geometry two congruent triangles contains the equal corresponding angles.

Example of Congruence

The followings are the important congruence test

ASA congruence

The two angles with the integrated faces of one triangle are equal to the corresponding two angles with the integrated faces of another triangle.

SAS congruence

The two faces with the included angle of triangle are the same to two faces with the included angle of another triangle.

AAS congruence

The two angles along throughout a non integrated side of one triangle are congruent to the identical measurements of a different triangle.

SSS congruence

Three sides of one triangle are identical to corresponding three sides of another triangle.

Example

Consider the following two triangles.




The triangle IJK is congruent to the triangle LMN

Angle I = Angle L

Angle J = Angle M

Angle K = Angle N

Length IJ = Length LM

Length JK = Length MN

Length KI = Length LN

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Examples for Congruence

Example 1 for congruence

In the following triangles are congruent then find the length of sides a, b, c.



Solution

The given triangles are congruent. Therefore the lengths of the sides of the triangles are equal.

Length EG = 52

Therefore the length VT = a = 52

Length FG = 48

Therefore the length UT = b = 48

Length EF = 50

Therefore the length UV = c = 50

Thus the a = 52, b = 48, c = 50.

Example 2 for congruence

Prove that triangle LMN is congruent to triangle PQR.



Solution

Given figure the angle L and angle P are the same.

Angle L = Angle P = 75 degree

Given figure the angle N and angle Q are the same.

Angle N = Angle Q = 65 degree.

Line segment LM is equal to the line segment PR.

Line LM = Line PR = 40 cm.

Therefore ? LMN and ? PQR are congruent through AAS congruence.