| Vertex (Vertices plural) | A point where two or more line segments meet (i.e. a corner). |
| Edge | A line segment joining one vertex/corner to another or a line segment where two faces meet. |
| Face | Any individual flat surface of a 3D figure. |
| Surface Area | Total area of the surfaces of a 3D figure. |
| Lateral Area | Total area of the figure's surfaces, excluding its top and base if they exist. |
| Volume | The amount of 3D space a figure takes up. |
Prisms are solids with two identical faces (called bases) joined by lateral edges. The shape of the base decides the type of prism. A triangular prism has a triangle as the base, a rectangular prism has a rectangle as the base, etc.
Regardless of the shape of a prism, we always calculate the volume in the same way:
Volume = Area of base x height
As a formula, this is:
\[ V = A_b h \]
Note:
The height, also called altitude, is not always the same as the lateral length. If the prism is skewed so that the bases are not directly above each other, the height will be shorter than the lateral length.
Example 1
A rectangular-based cardboard box has 3 m by 4 m base and a lateral length of 7 m. The box gets squashed so that its altitude becomes 6.5 m. What is the volume of the squashed box?
Solution
We calculate the area of the base first. Since it is a rectangle, the area is \(3 \times 4 = 12 m^2\). Now we use the altitude to calculate the box's volume. \[V = 12m^2 \times 6.5 m = 78m^3 \]
Cylinders are very similar to prisms in many ways. They also have two identical bases, but because the bases are circles, a cylinder is smooth and has no lateral edges.
We calculate the volume similarly to a prism:
\[ V = A_b h \]
The area of the base depends on its radius, and can be calculated by the formula: \[A_b = \pi r^2 \]
So we could also write the volume formula as \( V = \pi r^2 h \).
Example 2
A cylindrical pop can is 12 cm tall and has a radius of 2.5 cm. You pour the pop into a glass and measure it as 65 mL. How much of the pop can is empty space?
Solution
Using the formula for volume of a cylinder to find the total volume the can could hold, we have
\[V = \pi (2.5)^2(12) = 75 cm^3 = 75 mL\]
Calculating the difference, we have \(75 - 65 = 10 mL\). So 10 mL of the pop can is empty.
Like a prism, a pyramid has a base that can be any polygon - triangle, rectangle, pentagon, etc. However, instead of lateral edges, a pyramid has lateral sides that are triangular shaped. The triangles meet at a point called the vertex.
The height of a pyramid is the length of a perpendicular line from the vertex to the base. To calculate the pyramid's volume, we use the formula:
\[V = \frac{1}{3} A_b h \]
This is similar to the formula for the volume of a prism, but it gets scaled down by the \(\frac{1}{3}\) to account for the pinched top.
Example 3
A pyramid has the same height as a hexagonal prism, and the pyramid's base is a hexagon with the same dimensions as the base of the prism. If the prism has a volume of \(450 cm^3\), what is the volume of the pyramid?
Solution
A pyramid has \(\frac{1}{3}\) of the volume of a prism with the same base and height, so \(V = \frac{1}{3} \cdot 450 = 150 m^3\).
If pyramids are pinched prisms, then cones are like pinched cylinders. While pyramids have multiple lateral faces shaped like triangles, cones are made up of one triangle wrapped around a circular base. Just as pyramids have \(\frac{1}{3}\) the area of corresponding prisms, cones have \(\frac{1}{3}\) the area of a cylinder with the same base.
\[V = \frac{1}{3} A_b h \]
Example 4
Calculate the volume of a cone that has a base with diameter \(3 cm^2\) and a height of \(6 cm\).
Solution
We first need to calculate the base area. Using the formula for the area of a circle,
\[ A_b = \pi (d \div 2)^2 = \pi (3 \div 2)^2 = 7.069 cm^2 \]
Now we can plug this into the formula:
\[ V = 7.069 \cdot 6 = 42.41 cm^3 \]
Surface area is the total area of all the faces of a solid. Lateral area is the surface area without the area of the base(s). The calculations for surface and lateral area depend on the shape of each face.
Example 5
What is the lateral area of a cone with a diameter of \(7 cm\) and a slant height of \(12 cm\)?
Solution
The lateral surface of the cone is a triangle with a height of \(12 cm\) and a base equal to the circumference of the cone's base. To find the circumference of the cone's base, we use the formula \(c = \pi d\), where \(d\) is the diameter.
\[c = 7 \pi = 21.99 \]
The area of a triangle is \(\frac{1}{2}\) base \(\times\) height. So we calculate the lateral area to be
\[ \frac{1}{2} (21.99)(12) = 131.94 \]
Thus the lateral area is \(131.94 cm^2\).
Example 6
What is the surface area of a 3 by 6 by 8 rectangular prism?
Solution
The prism has 6 faces total. Two of the faces are 3 by 6, two of the faces are 3 by 8, and two of the faces are 6 by 8. So we will calculate the surface area by adding:
\(SA\) = 2(area of 3 by 6 rectangle) + 2(area of 3 by 8 rectangle) + 2(area of 6 by 8 rectangle) = \(2(3 \cdot 6) + 2(3 \cdot 8) + 2(6 \cdot 8) \)
This gives:
\[SA = 36 + 48 + 96 = 180 cm^2 \]
