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Faces Edges and Vertices

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When you look into a photo, the objects inside it appear flat. More precisely, they have no depth. In contrast, everything we see with our eyes right now in this world is in three dimensions. This means that nothing is entirely flat! Just by looking at your surroundings, you can tell that the objects around you consist of unique shapes. These shapes are governed by their faces, edges and vertices that make them seem so three-dimensional!

In this article, we shall be introduced to such concepts in three-dimensional figures and observe how they play a role in common solids we may come across in everyday life. Following that, we will look into the idea of surfaces, areas and volumes to aid our visualisation of these mentioned solids.

Unlike an object in two dimensions,

A **three-dimensional figure** is one defined by a space that considers three dimensions: length, width and depth.

Let us look at the example below.

Here we have a square. Figure 1 below shows an example of a two-dimensional figure.

Fig. 1 - Example of a square

Notice how the square above only consists of two dimensions: its length and its width. Now, let's try to look at this as a three-dimensional figure. This is known as a cube.

Fig. 2 - Example of a cube

In this case, we have an additional dimension that is considered, namely the depth. Figure 2 above shows an example of a three-dimensional figure.

Three-dimensional figures are called **solids** and consist of three basic elements: a face, edge and vertex. The next section will introduce these terms explicitly.

To begin our main discussion today, we shall become acquainted with three new terms, as defined below.

The **face **refers to a flat surface on a solid. This is indicated by the blue shaded region in Figure 3 below.

Fig. 3 - The face of a solid

An **edge **is a line segment in which two faces meet. This is indicated by the blue line (side) in Figure 4 below.

Fig. 4 - The edge of a solid

A **vertex **(or corner) is a point in which two edges meet. This is indicated by the circled region in Figure 5 below.

Fig. 5 - The vertex of a solid

A solid's face doesn't necessarily have to spread in one direction, like the ones you see on the solid above. A face can wrap around the space. Can you imagine this?

When this happens, we say we have a **curved face**, which describes a curved surface. Refer to the arrow in Figure 6 below.

Fig. 6 - The curved face of a solid

In this section, we shall familiarise ourselves with the number of faces, edges and vertices of several solids we may come across throughout this syllabus. The table below illustrates this.

Solid | Diagram | Number of Faces | Number of Edges | Number of Vertices | Number of Curved Faces |

Sphere | Sphere | 0 | 0 | 0 | 1 |

Ellipsoid | Ellipsoid | 0 | 0 | 0 | 1 |

Cone | Cone | 1 | 1 | 1 | 1 |

Cylinder | Cylinder | 2 | 2 | 0 | 1 |

Tetrahedron | Tetrahedron | 4 | 6 | 4 | 0 |

Square Pyramid | Square pyramid | 5 | 8 | 5 | 0 |

Triangular Prism | Triangular prism | 5 | 9 | 6 | 0 |

Cube | Cube | 6 | 12 | 8 | 0 |

Cuboid | Cuboid | 6 | 12 | 8 | 0 |

Octahedron | Octahedron | 8 | 12 | 6 | 0 |

Pentagonal Prism | Pentagonal prism | 7 | 15 | 10 | 0 |

Hexagonal Prism | Hexagonal Prism | 8 | 18 | 12 | 0 |

All solids have surface area and volume. Let's begin by defining these terms.

The** surface area** is the total area covered by all the faces of a given solid.

The **volume **is the amount of space occupied within a given solid.

To picture the volume of a solid, imagine filling up a cylindrical bottle of water, for example. The amount of water this bottle can hold is the volume of this cylindrical shape.

The surface area can be found by adding up all the areas of each face of a given three-dimensional figure. However, rather than executing such a cumbersome action, there is a shortcut to this!

Every common solid has a particular formula to find its surface area. The same goes for its volume. The table below exhibits the formula of the surface area and volume of several notable solids.

Solid | Diagram | Surface Area | Volume | |

Sphere | Sphere | \[A=4×π×r^2\] | \[V=4×π×\frac{r^3}{3}\] | \(r\) = radius |

Hemisphere | Hemisphere | \[A=3×π×r^2\] | \[V=2×π×\frac{r^3}{3}\] | \(r\) = radius |

Cone | Cone | \[A=π×r×(s+r)\] | \[V=π×r^2×\frac{h}{3}\] | \(r\) = radius \(s\) = slant height \(h\) = height |

Cylinder |
Cylinder | \[A=2×π×r×(r+h)\] | \[V=π×r^2×h\] | \(r\) = radius \(h\) = height |

Pyramid | Pyramid | \[A=bl+2bs\] | \[V=l×b×\frac{h}{3}\] | \(l\) = length \(b\) = base \(h\) = height \(s\) = slant height |

Cube | Cube | \[A=6×l^2\] | \[V=l^3\] | \(l\) = length |

Cuboid | Cuboid | \[A=2×(lb+bn+lh)\] | \[V=l×b×h\] | \(l\) = length \(b\) = base \(h\) = height |

Triangular Prism | Triangular prism | \[A=bh+lb+2ls\] | \[V=l×b×\frac{h}{2}\] | \(l\) = length \(b\) = base \(h\) = height \(s\) = slant height |

Trapezoidal Prism | Trapezoidal prism | \[A=(a+b)h+bl+al+2ls\] | \[V=(a+b)×h×\frac{l}{2}\] | \(l\) = length \(b\) = base \(h\) = height \(s\) = slant height \(a\) = top length |

Let's look at two examples.

Calculate the surface area and volume of a cone whose radius is 5 units, slant height is 8 units and perpendicular height is 6 units.

**Solution **

The dimensions of this cone are given by \(r=5\), \(s=8\) and \(h=6\). Now applying the formula for the surface area of a cone, we obtain

\[A=π×5×(8+5)=65\pi\approx 641.52\]

Thus, this cone has a surface area of 641.52 units^{2} correct to 2 decimal places. Next, we shall use the formula for the volume of a cone.

\[V=π×5^2×\frac{6}{3}=50\pi\approx 157.08\]

Hence, this cone has a volume of 157.08 units^{3} correct to 2 decimal places.

Here is a final example for this section before we move on to Euler's formula.

What is the radius of a sphere whose surface area is 67 units^{2}? Use this result to determine its volume as well.

**Solution**

Here, we are given the surface area of this sphere as \(A=67\). In order to determine its radius, we need to rearrange the formula for the surface area of a sphere so that \(r\) becomes the subject.

\[A=4×π×r^2\implies r^2=\frac{A}{4\times \pi}\]

Now take the square root on both sides,

\[r=\sqrt{\frac{A}{4\times \pi}}\]

Note that we only need to consider the positive root in this case since we are dealing with dimensions and measures, which are always taken as positive. We can now substitute our given values to determine the radius.

\[r=\sqrt{\frac{67}{4\times \pi}}=\approx 5.33\]

Thus, the radius is 5.33 units correct to two decimal places. We can now use this result to determine the volume of this sphere. To ensure accuracy, we will not take the approximate form of the radius, but the fractional form, that is, \(r=\sqrt{\frac{67}{4\times \pi}}\).

\[V=4×π×\frac{(\sqrt{\frac{67}{4\times \pi}})^3}{3}\approx 634.87\]

Therefore, the volume of this sphere is 634.87 units^{3}.

Euler's Formula states that for any polyhedron that does not intersect itself or have any holes, the number of faces plus the number of vertices minus the number of edges always equals two. This can be written by the expression below.

\[F+V-E=2\]

where

F = number of faces;

V = number of vertices;

E = number of edges.

Recall that a **polyhedron **is a solid with only flat faces and a **non-polyhedron** is a solid with at least one curved face.

In this section, we shall look at several examples that apply Euler's formula. Refer back to the tables above to help you verify the answers to the examples here.

A **square-based pyramid** is a type of pyramid with a square base. It contains 5 sides comprising the square base and 4 congruent triangular lateral faces. Figure 7 shows an illustration of a square-based pyramid.

Fig. 7 - Square-based pyramid

Let's look at an example.

Verify that Euler's Formula is satisfied for a square-based pyramid.

**Solution**

From our table above, a square-based pyramid has the following features:

Number of Faces: 5

Number of Vertices: 5

Number of Edges: 8

Now, applying Euler's Formula, we obtain

\[F+V-E=5+5-8=2\]

Thus, Euler's Formula holds true for a square-based pyramid.

A **cuboid **is a three-dimensional figure bounded by six rectangular faces. Another name for a cuboid is a hexahedron. Figure 8 shows an illustration of a cuboid.

Fig. 8 - Cuboid

There are two special cases for cuboids to consider.

- The square cuboid: this type of cuboid has two (or more) opposite square faces.
- The cube: this type of cuboid has only square faces.

Here is an example.

Verify that Euler's Formula is satisfied for a cuboid.

**Solution**

From our table above, a cuboid has the following features:

Number of Faces: 6

Number of Vertices: 8

Number of Edges: 12

Now, applying Euler's Formula, we obtain

\[F+V-E=6+8-12=2\]

Thus, Euler's Formula holds true for a cuboid.

A **triangular prism** is a type of prism made up of two triangular bases and three rectangular sides. Figure 9 shows an image of a triangular prism.

Fig. 9 - Triangular prism

Let us now look at an example.

Verify that Euler's Formula is satisfied for a triangular prism.

**Solution**

From our table above, a triangular prism has the following features:

Number of Faces: 5

Number of Vertices: 6

Number of Edges: 9

Now, applying Euler's Formula, we obtain

\[F+V-E=5+6-9=2\]

Thus, Euler's Formula holds true for a triangular prism.

A** hexagonal prism** is a type of prism made of** **two hexagonal bases and six rectangular sides. It is sometimes referred to as an octahedron. Figure 10 illustrates a hexagonal prism.

Fig. 10 - Hexagonal prism

Recall that a hexagon has six sides.

Here is an example.

Verify that Euler's Formula is satisfied for a hexagonal prism.

**Solution**

From our table above, a hexagonal prism has the following features:

Number of Faces: 8

Number of Vertices: 12

Number of Edges: 18

Now, applying Euler's Formula, we obtain

\[F+V-E=8+12-18=2\]

Thus, Euler's Formula holds true for a hexagonal prism.

A **cylinder **is a type of prism made of** **two circular bases and one curved face. It is considered a curvilinear geometric shape. Figure 11 shows a cylinder.

Fig. 11 - Cylinder

Now, notice that a** **cylinder has one *curved *face. Therefore, it is not a polyhedron. This means that Euler's formula does not apply here.

Let's check to see if this is true.

Does Euler's Formula satisfy a cylinder?

**Solution**

From our table above, a cylinder has the following features:

Number of Flat Faces: 2

Number of Curved Faces: 1

Number of Vertices: 0

Number of Edges: 2

First, notice that we cannot insert the number of curved faces into Euler's formula as there is no variable that indicates it. Even if we did not consider this curved face and only took into account the flat faces and edges of the cylinder, we would get

\[F+V-E=2+0-2=0\]

which certainly fails Euler's formula.

- The face refers to a flat surface on a solid.
- An edge is a line segment in which two faces meet.
- A vertex (or corner) is a point in which two edges meet.
- The surface area is the total area covered by all the faces of a given solid.
- The volume is the amount of space occupied within a given solid.
- Euler's formula is given by \(F+V-E=2\)

A cylinder has 2 faces, 2 edges and no vertices.

A cone has 1 face, 1 edge and 1 vertex.

A cube has 6 faces, 12 edges and 8 vertices.

A sphere has no face, no edges and no vertices.

More about Faces Edges and Vertices

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