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` .