Definition 4.2.1. Prism.
A prism is a solid consisting of two identical polygons connected by parallelograms.
Section 4.2 Geometric Reasoning 3D
Subsection 4.2.1 Properties
The surface area of a 3D shape is the cummulative area of all the 2D areas of the shape.
The volume of a 3D shape is a measure of what it takes to fill a 3D shape.
Subsection 4.2.2 3D Shapes
These look like a polygon has been extruded. If the sides are rectangles, then it is called a right prism. Prisms are named for their base shape. For example, there are triangular prisms and pentagonal prisms.
Shape |
Surface Area |
Volume |
 |
sum of area of all sides |
\(V = B h\) |
Example 4.2.3.
(a)
What is the surface area of the right triangular prism in Figure 4.2.4?
Solution.
The surface area consists of the areas of the three rectangles and the two right triangles. The rectangle areas are \(2 \cdot 10 = 20\text{,}\) \(3 \cdot 10 = 30\text{,}\) and \(\sqrt{13} \cdot 10 \approx 36\text{.}\) Because the triangles are right triangles, they both have area \(\frac{1}{2} \cdot 2 \cdot 3 = 3\text{.}\) The total area is \(20+30+36+6 \approx 92\text{.}\)
(b)
What is the volume of the triangular prism in Figure 4.2.4?
Solution.
The volume is the area of the triangle, 3, times the height of the prism, 10. Thus the area is \(3 \cdot 10 = 30\text{.}\)
Definition 4.2.5. Cylinder.
A cylinder is a circular prism.
Shape |
Lateral Surface Area |
Volume |
 |
\(A=2\pi r \cdot h\) | \(V = \pi r^2 h\) |
Note the surface area of the side a cylinder can be imagined to be the result of peeling off the surface which results in a rectangle.
Shape |
Surface Area |
Volume |
 |
\(A=4\pi r^2\) | \(V = \frac{4}{3}\pi r^3 \) |
Shape |
Surface Area |
Volume |
 |
sum of surfaces |
\(V = \frac{1}{3} B h \) |
Checkpoint 4.2.9.
What is the relationship between the volume of a pyramid to the volume of a prism with the same base?
Example 4.2.10.
(a)
What is the surface area of the pyramid in Figure 4.2.11
Solution.
This pyramid consists of four triangular sides and a square base. The area of a triangle can be found using Heron’s formula.
\begin{align*}
s & = \frac{1}{2}(\sqrt{10}+\sqrt{10}+2).\\
A & = \sqrt{s(s-\sqrt{10})(s-\sqrt{10})(s-2)}\\
& = \sqrt{(1+\sqrt{10})(1)(1)(\sqrt{10}-1)}\\
& = 3.
\end{align*}
The area of the base is 4. Thus the surface area is \(4+4(3)=16\text{.}\)
(b)
What is the volume of the pyramid in Figure 4.2.11
Solution.
The volume of a pyramid is a third of base times height, so \(V=\frac{1}{3}(2^2)(2\sqrt{2}) \approx 3.77\text{.}\)
Shape |
Lateral Surface Area |
Volume |
 |
\(A=\pi r \sqrt{h^2+r^2}\) or \(A=\pi r s\) | \(V = \frac{\pi}{3} r^2 h \) |
The latin word frustrum means “cut off”. This is the etymology of frustrated which refers to a cut off hope. For these formulas the bases must be parallel. Note \(P_i\) below refers to the perimeter of the bases.
Shape |
Lateral Surface Area |
Volume |
 |
\(\frac{1}{2}s(P_1+P_2)\) | \(V = \frac{1}{3} h (B_1+B_2+\sqrt{B_1 B_2}) \) |
Example 4.2.14.
(a)
What is the surface area of the frustrum of a pyramid in Figure 4.2.15?
Solution.
The perimeter of the lower base is 9 and the perimeter of the upper base is 3. The area of the three sides is therefore \(\frac{1}{2}2(9+3)=12\text{.}\) To calculate the area of the two bases we will need Heron’s formula.
\begin{align*}
s & = \frac{9}{2}.\\
A_2 & = \sqrt{\frac{9}{2}\left(\frac{3}{2}\right)^3}\\
& = \frac{9\sqrt{3}}{4}.\\
A_1 & = \sqrt{\frac{3}{2}\left(\frac{1}{2}\right)^3}\\
& = \frac{\sqrt{3}}{4}.
\end{align*}
Thus the surface area is \(12+\frac{9\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \approx 16.33\text{.}\)
(b)
What is the volume of the frustrum of a pyramid in Figure 4.2.15?
Solution.
Using the values provided
\begin{align*}
V & = \frac{1}{3} \cdot \frac{4}{3}\left(\frac{9\sqrt{3}}{4}+\frac{\sqrt{3}}{4}+\sqrt{\frac{9\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4}}\right)\\
& = \frac{13\sqrt{3}}{12}\\
& \approx 1.88
\end{align*}
Shape |
Lateral Surface Area |
Volume |
 |
\(\pi s(r_1+r_2)\) | \(V = \frac{1}{3} h (B_1+B_2+\sqrt{B_1 B_2}) \) |
Checkpoint 4.2.17.
For the surface area of a frustrum of a cone we use the slant height: the distance from the edge of the bottom base to the edge of the top base. Here we consider limitations on that length.
For these questions suppose the bottom base has radius 5. The top base will vary depending on the slant height.
(a)
If the height is 3, can the slant height be 2?
(b)
What is the bottom of the range of possible slant heights for this frustrum?
(c)
If the height is 3 and the top base has radius 1, what is the slant height?
(d)
If the height is 3 and the top base has radius 0.5, what is the slant height?
(e)
What is the top of the range of possible slant heights for this frustrum? Note this is for bottom base radius 5, height 3, and top base radius unrestricted (but smaller than bottom base).
Subsection 4.2.3 Applying Geometry
Example 4.2.18.
(a)
An ice cream cone has the dimensions shown in Figure 4.2.19. What is the volume of the ice cream?
Solution.
This is a right circular cone with a half sphere on top. The volume of the cone is \(V_c = \frac{\pi}{3} 2^2 \cdot 10 \approx 42\text{.}\) The volume of the half sphere is \(\frac{1}{2} \cdot \frac{4}{3} \pi 2^3 \approx 17\text{.}\) Thus the total volume of the ice cream is 59.
(b)
An ice cream cone has the dimensions shown in Figure 4.2.19. What is the surface area of the ice cream? Note the greater the area the faster it melts.
Solution.
This is a right circular cone with a half sphere on top. The surface area of the cone is \(A_c = 2\pi (2+\sqrt{10^2+2^2}) \approx 77\text{.}\) The surface area of the half sphere is \(\frac{1}{2} \cdot 4 \pi 2^2 \approx 25\text{.}\) Thus the total surface area of the ice cream is 102.
