Topics
5. Modeling in Two & Three Dimensions
topic badge
United States of AmericaTennessee
Integrated Math 3

5.03 Prisms and cylinders

Lesson
Practice
Lesson

Concept summary

Many common solids fall under the category of prisms, or the related category of cylinders:

Prism

A figure with two congruent, parallel, polygonal bases that are connected by rectangular faces.

A pair of three dimensional solids. One has a rectangular base and rectangular sides, the other has a triangular base and rectangular sides.
Triangular prism

A prism with triangular bases.

A three dimensional solid with two congruent triangular bases and three congruent rectangular faces.
Rectangular prism

A prism with rectangular bases. Cubes form a sub-class.

A three dimensional solid with a pair of congruent rectangular bases and four congruent rectangular sides.
Cube

A rectangular prism with six congruent square faces.

A three dimensional solid with six congruent square faces.
Right prism

A prism in which the bases are connected by rectangular faces that are perpendicular to each base.

A rectangular prism. The angles between the base and the faces are marked as right angles.
Oblique prism

A prism in which the bases are connected by rectangular faces that are not perpendicular to each base.

A rectangular prism. The angles between the base and the faces are not right angles.
Cylinder

A figure containing two congruent, parallel, circular bases whose edges are connected by a curved surface.

A three dimensional solid with a pair of congruent circular bases, connected by a curved surface.
Right cylinder

A cylinder in which the bases are connected by a curved surface that is perpendicular to each base.

A cylinder with the height drawn from the center of the top circle to the center of the bottom circle. The angle between the height and the bottom circle is marked as a right angle
Oblique cylinder

A cylinder in which the bases are connected by a curved surface that is not perpendicular to each base.

A cylinder with the height drawn from the center of the top circle to the center of the bottom circle. The angle between the height and the bottom circle is not a right angle.

Prisms are formed by a pair of congruent polygons joined by rectangles. Cylinders are similar, but the bases are circles instead of polygons, and so they are joined by a curved surface instead of rectangles.

Each solid has a number of properties that are associated with them.

Net

A diagram of the faces of a three-dimensional figure arranged in such a way that the diagram can be folded to form the three-dimensional figure.

A square pyramid is shown on the left. On the right is a two dimensional shape as if the pyramid were unfolded. The shape consists of a square in the center, with a congruent triangle attached to each side of the square.
A cube and its net. The net is made up of 6 identical squares: a column of 4 squares, and 2 squares adjacent to the left and right of the third square from the top.
A cube and its net
A cylinder and its net. The net is made up of a rectangle, and 2 identical circles. The circles are tangent to the lengths of the rectangle.
A right cylinder and its net

Note that even some solids with curved faces, such as cylinders, have nets consisting of flat, two-dimensional shapes. Additionally, a solid can be represented with multiple different, equivalent nets.

Volume

The amount of space enclosed by a soild.

The volume of a prism can be calculated using the formula

\displaystyle V = Bh
\bm{B}
area of the base
\bm{h}
perpendicular height between bases

The volume of a cylinder can be calculated identically; by multiplying the area of the circular base by the perpendicular height between bases.

A cube is a particular sub-class of prism, which has six congruent square faces. In particular, if the side length of the cube is s, then the area of the base is B = s^2 and the perpendicular height is h = s. So the volume of a cube can be expressed as

\displaystyle V = s^3
\bm{s}
side length of the cube
Surface area

The total area of the outer surfaces of a solid.

The surface area of a solid is equivalent to the total area of a net of that solid.

The surface area of a prism can be calculated using the formula

\displaystyle SA = 2B + Ph
\bm{B}
area of the base
\bm{P}
perimeter of the base
\bm{h}
perpendicular height between bases

The surface area of a cylinder can be calculated identically; by adding the area of two circular bases to the product of the circumference abd the perpendicular height between bases.

Once again, we can form a simpler expression for the surface area of a cube with side length s. Since a cube consists of 6 congruent square faces, the surface area can be expressed as

\displaystyle SA = 6s^2
\bm{s}
side length of the cube
Density

The compactness of a substance. Usually measured in weight per unit of volume.\text{Density} = \dfrac{\text{Mass}}{\text{Volume}}

Cavalieri's Principle

If two three-dimensional figures have the same height and the same cross-sectional area at every level, then they have the same volume.

The following three figures each reach the same perpendicular height h and have the same base area B, so by Cavalieri's principle we know that the they have the same volume:

A diagram showing a rectangular prism, a triangular prism, and a cylinder with their bases lying on a same plane. All figures have a perpendicular height of h. A plane parallel to the bases of the 3 figures is drawn. For each figure, the cross section created has an area of B.

Notice that it is the perpendicular height that is important, rather than the slant height, even if the solids are not right prisms/cylinders.

Worked examples

Example 1

Find the density of a cube with side length 4 \text{ ft} and weight 300 \text{ lb}. Round your answer to two decimal places.

Approach

We can first find the volume of the cube, using the formula V = s^3, where s is the side length.

Then we can calculate the density by dividing the given mass by the volume.

Solution

Finding the volume of the cube, we have:

\displaystyle V\displaystyle =\displaystyle s^3Volume of a cube
\displaystyle =\displaystyle 4^3Substitute side length
\displaystyle =\displaystyle 64Simplify

So the volume of the cube is 64 \text{ ft}^3.

We can now use this to calculate the density:

\displaystyle \text{Density}\displaystyle =\displaystyle \dfrac{\text{Mass}}{\text{Volume}}Density formula
\displaystyle =\displaystyle \dfrac{300}{64}Substitute known values
\displaystyle =\displaystyle 4.6875Simplify

Therefore the density of the cube is 4.69 \, \text{lb}/\text{ft}^3.

Reflection

Note that the density is calculated by dividing the mass (in pounds) by the volume (in cubic feet), and so the density has units of mass per volume - in this case, pounds per cubic feet.

Example 2

Find the surface area of a right cylinder with radius 2 \text{ in} and height 5\text{ in}.

Approach

We know that the surface area of a cylinder is given bySA = 2B+Phwhere B is the area of the base, P is the perimeter of the base, and h is the perpendicular height.

We are given the height in the question, and it is a right cylinder, so this is perpendicular height.

We are also given the radius, which we can use to calculate both the area of the circular base and its perimeter (i.e. circumference).

Solution

Let's start by calculating the area of the base:

\displaystyle B\displaystyle =\displaystyle \pi r^2Area of a circle
\displaystyle =\displaystyle \pi \cdot 2^2Substitute radius measure
\displaystyle =\displaystyle 4\piSimplify

So the area of the base is 4 \pi.

Next, we find the perimeter of the base:

\displaystyle P\displaystyle =\displaystyle 2 \pi rCircumference formula
\displaystyle =\displaystyle 2\pi \cdot 2Substitute radius measure
\displaystyle =\displaystyle 4 \piSimplify

So the perimeter of the base is 4 \pi.

Putting this together with the known perpendicular height of 5 inches, we can calculate the surface area:

\displaystyle SA\displaystyle =\displaystyle 2B+PhSurface area of a cylinder
\displaystyle =\displaystyle 2\cdot 4\pi+ 4 \pi \cdot 5 Substitute known values
\displaystyle =\displaystyle 28\piSimplify

So the surface area of the cylinder is 28 \pi \text{ in}^2.

Outcomes

M3.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.

M3.N.Q.A.1.C

Define and justify appropriate quantities within a context for the purpose of modeling.

M3.G.MG.A.1

Use geometric shapes, their measures, and their properties to model objects found in a real-world context for the purpose of approximating solutions to problems.*

M3.G.GMD.A.1

Understand and explain the formulas for the volume and surface area of a cylinder, cone, prism, and pyramid.

M3.G.GMD.A.2

Use volume and surface area formulas for cylinders, cones, prisms, pyramids, and spheres to solve problems in a real-world context.*

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP7

Look for and make use of structure.

M3.MP8

Look for and express regularity in repeated reasoning.

What is Mathspace

About Mathspace