PARALLELEPIPED: CHARACTERISTICS, TYPES, AREA, VOLUME - MATHS - 2025

A parallelepiped is a geometric body made up of six faces, the main characteristic of which is that all of its faces are parallelograms and also that its opposite faces are parallel to each other. It is a common polyhedron in our daily lives, since we can find it in shoe boxes, the shape of a brick, the shape of a microwave, etc.

Being a polyhedron, the parallelepiped encloses a finite volume and all its faces are flat. It is part of the group of prisms, which are those polyhedra in which all its vertices are contained in two parallel planes.

Elements of the Parallelepiped

Faces

They are each of the regions formed by parallelograms that limit the parallelepiped. A parallelepiped has six faces, where each face has four adjacent faces and one opposite. Also, each face is parallel with its opposite.

Edges

They are the common side of two faces. In total, a parallelepiped has twelve edges.

Vertex

It is the common point of three faces that are adjacent to each other two by two. A parallelepiped has eight vertices.

Diagonal

Given two faces of a parallelepiped opposite each other, we can draw a line segment that goes from the vertex of one face to the opposite vertex of the other.

This segment is known as the diagonal of the parallelepiped. Each parallelepiped has four diagonals.

Center

It is the point at which all the diagonals intersect.

Characteristics of the Parallelepiped

As we already mentioned, this geometric body has twelve edges, six faces, and eight vertices.

In a parallelepiped, three sets formed by four edges can be identified, which are parallel to each other. Furthermore, the edges of said sets also have the property of having the same length.

Another property that parallelepipeds possess is that they are convex, that is, if we take any pair of points belonging to the interior of the parallelepiped, the segment determined by said pair of points will also be within the parallelepiped.

In addition, parallelepipeds, being convex polyhedra, comply with Euler's theorem for polyhedra, which gives us a relationship between the number of faces, the number of edges, and the number of vertices. This relationship is given in the form of the following equation:

C + V = A + 2

This characteristic is known as the Euler characteristic.

Where C is the number of faces, V the number of vertices and A the number of edges.

Types

We can classify parallelepipeds based on their faces, into the following types:

Orthohedron

They are the parallelepipeds where their faces are formed by six rectangles. Each rectangle is perpendicular to those that share an edge. They are the most common in our daily lives, this being the usual form of shoe boxes and bricks.

Regular cube or hexahedron

This is a particular case of the previous one, where each of the faces is a square.

The cube is also part of the geometric bodies called Platonic solids. A Platonic solid is a convex polyhedron, so that both its faces and its internal angles are equal to each other.

Rhombohedron

It is a parallelepiped with rhombuses for its face. These rhombuses are all equal to each other, since they share edges.

Rhombohedron

Its six faces are rhomboids. Recall that a rhomboid is a polygon with four sides and four angles that are equal two to two. Rhomboids are parallelograms that are neither squares, nor rectangles, nor rhombuses.

On the other hand, Oblique Parallelepipeds are those in which at least one height does not agree with their edge. In this classification we can include rhombohedra and rhombohedra.

Diagonals calculation

To calculate the diagonal of an orthohedron we can use the Pythagorean theorem for R ³.

Recall that an ortohedron has the characteristic that each side is perpendicular to the sides that share an edge. From this fact we can deduce that each edge is perpendicular to those that share a vertex.

To calculate the length of a diagonal of an orthohedron we proceed as follows:

1. We calculate the diagonal of one of the faces, which we will put as a base. For this we use the Pythagorean theorem. Let's name this diagonal d _b.

2. Then with d _b we can form a new right triangle, such that the hypotenuse of said triangle is the diagonal D we are looking for.

3. We use the Pythagorean theorem again and we have that the length of this diagonal is:

Another way to calculate diagonals in a more graphic way is with the addition of free vectors.

Recall that two free vectors A and B are added by placing the tail of vector B with the tip of vector A.

The vector (A + B) is the one that begins at the tail of A and ends at the tip of B.

Let us consider a parallelepiped for which we wish to calculate a diagonal.

We identify the edges with conveniently oriented vectors.

Then we add these vectors and the resulting vector will be the diagonal of the parallelepiped.

Area

The area of ​​a parallelepiped is given by the sum of each of the areas of its faces.

If we determine one of the sides as the base, A _L + 2A _B = Total Area

Where A _L is equal to the sum of the areas of all the sides adjacent to the base, called the lateral area and A _B is the area of ​​the base.

Depending on the type of parallelepiped we are working with, we can rewrite this formula.

Area of ​​an ortohedron

It is given by the formula

A = 2 (ab + bc + ca).

Example 1

Given the following orthohedron, with sides a = 6 cm, b = 8 cm and c = 10 cm, calculate the area of ​​the parallelepiped and the length of its diagonal.

Using the formula for the area of ​​an ortohedron we have that

A = 2 = 2 = 2 = 376 cm ².

Notice that since it is an orthohedron the length of any of its four diagonals is the same.

Using the Pythagorean theorem for space we have that

D = (6 ² + 8 ² + 10 ²) ^1/2 = (36 + 64 + 100) ^1/2 = (200) ^1/2

Area of ​​a cube

Since each edge has the same length, we have that a = b and a = c. Substituting in the previous formula we have

A = 2 (aa + aa + aa) = 2 (3a ²) = 6a ²

A = 6a ²

Example 2

The box of a game console is shaped like a cube. If we want to wrap this box with gift wrap, how much paper would we spend knowing that the length of the edges of the cube is 45 cm?

Using the formula for the area of ​​the cube we obtain that

A = 6 (45 cm) ² = 6 (2025 cm ²) = 12150 cm ²

Area of ​​a rhombohedron

Since all their faces are the same, just calculate the area of ​​one of them and multiply it by six.

We have that the area of ​​a rhombus can be calculated through its diagonals with the following formula

A _R = (Dd) / 2

Using this formula it follows that the total area of ​​the rhombohedron is

A _T = 6 (Dd) / 2 = 3Dd.

Example 3

The faces of the following rhombohedron are formed by a rhombus whose diagonals are D = 7 cm and d = 4 cm. Your area will be

A = 3 (7cm) (4cm) = 84cm ².

Area of ​​a rhombohedron

To calculate the area of ​​a rhombohedron we must calculate the area of ​​the rhomboids that compose it. Since parallelepipeds fulfill the property that opposite sides have the same area, we can associate the sides in three pairs.

In this way we have that your area will be

A _T = 2b ₁ h ₁ + 2b ₂ h ₂ + 2b ₃ h ₃

Where the b _i are the bases associated with the sides and the h _i their relative height corresponding to these bases.

Example 4

Consider the following parallelepiped,

where side A and side A '(its opposite side) have a base b = 10 and a height h = 6. The marked area will have a value of

A ₁ = 2 (10) (6) = 120

The B and B 'have b = 4 and h = 6, so

A ₂ = 2 (4) (6) = 48

YC and C 'have b = 10 and h = 5, thus

A ₃ = 2 (10) (5) = 100

Finally the area of ​​the rhombohedron is

A = 120 + 48 + 100 = 268.

Volume of a parallelepiped

The formula that gives us the volume of a parallelepiped is the product of the area of ​​one of its faces by the height corresponding to that face.

V = A _C h _C

Depending on the type of parallelepiped, this formula can be simplified.

Thus we have for example that the volume of an orthohedron would be given by

V = abc.

Where a, b and c represent the length of the edges of the ortohedron.

And in the particular case of the cube is

V = a ³

Example 1

There are three different models for cookie boxes and you want to know in which of these models you can store more cookies, that is, which of the boxes has the largest volume.

The first is a cube whose edge has a length of a = 10 cm

Its volume will be V = 1000 cm ³

The second has edges b = 17 cm, c = 5 cm, d = 9 cm

And therefore its volume is V = 765 cm ³

And the third has e = 9 cm, f = 9 cm and g = 13 cm

And its volume is V = 1053 cm ³

Therefore, the box with the largest volume is the third.

Another method to obtain the volume of a parallelepiped is to use vector algebra. In particular, the triple dot product.

One of the geometric interpretations that the triple scalar product has is that of the volume of the parallelepiped, whose edges are three vectors that share the same vertex as a starting point.

In this way, if we have a parallelepiped and we want to know what its volume is, it is enough to represent it in a coordinate system in R ^{3 by} making one of its vertices coincide with the origin.

Then we represent the edges that coincide at the origin with vectors as shown in the figure.

And in this way we have that the volume of said parallelepiped is given by

V = - AxB ∙ C-

Or, equivalently, the volume is the determinant of the 3 × 3 matrix, formed by the components of the edge vectors.

Example 2

When representing the following parallelepiped in R ³ we can see that the vectors that determine it are the following

u = (-1, -3,0), v = (5, 0, 0) and w = (-0.25, -4, 4)

Using the triple scalar product we have

V = - (uxv) ∙ w-

uxv = (-1, -3,0) x (5, 0, 0) = (0,0, - 15)

(uxv) ∙ w = (0,0, - 15) ∙ (-0.25, -4, 4) = 0 + 0 + 4 (- 15) = - 60

From this we conclude that V = 60

Let us now consider the following parallelepiped in R3 whose edges are determined by the vectors

A = (2, 5, 0), B = (6, 1, 0) and C = (3, 4, 4)

Using determinants gives us that

Thus we have that the volume of said parallelepiped is 112.

Both are equivalent ways of calculating volume.

Perfect parallelepiped

An orthohedron is known as a Euler brick (or Euler's block) that fulfills the property that both the length of its edges and the length of the diagonals of each of its faces are whole numbers.

Although Euler was not the first scientist to study the ortohedra that fulfill this property, he did find interesting results about them.

The smallest Euler brick was discovered by Paul Halcke and the lengths of its edges are a = 44, b = 117 and c = 240.

An open problem in number theory is as follows

Are there perfect ortohedra?

At present, this question has not been answered, since it has not been possible to prove that such bodies do not exist, but neither has any been found.

What has been shown so far is that perfect parallelepipeds do exist. The first to be discovered has the length of its edges the values ​​103, 106 and 271.

Bibliography

1. Guy, R. (1981). Unsolved problems in number theory. Springer.
2. Landaverde, F. d. (1997). Geometry. Progress.
3. Leithold, L. (1992). The calculation with analytic geometry. HARLA, SA
4. Rendon, A. (2004). Technical drawing: Activity book 3 2nd Bachillerato. Tebar.
5. Resnick, R., Halliday, D., & Krane, K. (2001). Physics Vol. 1. Mexico: Continental.