Faces Edges and Vertices

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.

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 Units² 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 units³ correct to 2 decimal places.

What is the radius of a sphere whose surface area is 67 units²? 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, \(r=\sqrt{\frac{67}{4\times \pi}}\).

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

Therefore, the volume of this sphere is 634.87 units³.

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.

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.

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.

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.

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.