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2.11: Surface Area and Volume

Difficulty Level: At Grade Created by: CK-12
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Exploring Solids

Polygon or Polyhedron - A polyhedron is defined using polygons, so in the beginning students will understand the difference. After some time has passed though, students tend to get these similar sounding words confused. Remind them that polygons are two-dimensional and polyhedrons are three-dimensional. The extra letters in polyhedron represents it spreading out into three-dimensions.

The Limitations of Two Dimensions - It is difficult for students to see the two-dimensional representations of three-dimensional figures provided in books and on computer screens. A set of geometric solids is easily obtained through teacher supply companies, and are extremely helpful for students as they familiarize themselves with three-dimensional figures. When first counting faces, edges, and vertices most students need to hold the solid in their hands, turn it around, and see how it is put together. After they have some experience with these objects, students will be better able to read the figures drawn in the text to represent three-dimensional objects.

Assemble Solids - A valuable exercise for students as they learn about polyhedrons is to make their own. Students can cut out polygons from light cardboard and assemble them into polyhedrons. Patterns are readily available. This hands-on experience with how three-dimensional shapes are put together will help them develop the visualization skills required to count faces, edges, and vertices of polyhedrons described to them.

Computer Representations - When shopping on-line it is possible to “grab” and turn merchandise so that they can be seen from different perspectives. The same can be done with polyhedrons. With a little poking around students can find sites that will let them virtually manipulate a three-dimensional shape. This is another possible option to develop the students’ ability to visualize the solids they will be working with for the remainder of this chapter.

Using the Contrapositive - If students have already learned about conditional statements, point out to them that Example 4 in this section makes use of the contrapositive. Euler’s formula states that if a solid is a polyhedron, then \begin{align*}V+F=E+2\end{align*}V+F=E+2. The contrapositive is that if \begin{align*}V+F \neq E+2\end{align*}V+FE+2, then the solid is not a polyhedron. Students need periodic review of important concepts in order to transfer them to their long-term memory. For more review of conditional statements, see the second chapter of this text.

Each Representation Has Its Use - Each of the methods for making two-dimensional representations of three-dimensional figures was developed for a specific reason and different representations are most appropriate depending on what aspect of the geometric solid is of interest.

Perspective – used in art, and when one wants to make the representation look realistic

Isometric View – used when finding volume

Orthographic View – used when finding surface area

Cross Section – used when finding volume and the study of conic sections (circles, ellipses, parabolas, and hyperbolas) is based on the cross sections of a cone

Nets – used when finding surface area or assembling solids

Ask the students to think of other uses for these representations.

When students know and fully understand the options, they will be able to choose the best tool for each task they undertake.

Isometric Dot Paper - If students are having trouble making isometric drawings, they might benefit from the use of isometric dot paper. The spacing of the dots allows students to make consistent lengths and angles on their polyhedrons. After some practice with the dot paper, they should be able to make decent drawings on any paper. A good drawing will be helpful when calculating volumes and surface areas.

Practice - Most students will need to make quite a few drawings before the result is good enough to be helpful when making calculations. The process of making these representations provides the student with an opportunity to contemplate three-dimensional polyhedrons. The better their concept of these solids, the easier it will be for them to calculate surface areas and volumes in the sections to come.

Additional Exercises:

1. In the next week look around you for polyhedrons. Some example may be a cereal box or a door stop. Make a two-dimensional representation of the object. Choose four objects and use a different method of representation for each.

These can make nice decorations for classroom walls and the assignment makes students look for way to apply what they will learn in this chapter about surface area and volume.

Surface Area of Prisms and Cylinders

The Proper Units - Students will frequently use volume units when reporting a surface area. Because the number describes a three-dimensional figure, the use of cubic units seems appropriate. This shows a lack of understanding of what exactly it is that they are calculating. Provide students with some familiar applications of surface area like wrapping a present or painting a room, to improve their understanding of the concept. Insist on the use of correct units so the student will have to consider what exactly is being calculated in each exercise.

Review Area Formulas - Calculating the surface area and volume of polyhedrons requires the students to find the areas of different polygons. Before starting the new material, take some time to review the area formulas for the polygons that will be used in the lesson. When students are comfortable with the basic area calculations, they can focus their attention on the new skill of working with three-dimensional solids.

A Prism Does Will Not Always Be Sitting On Its Base - When identifying prisms, calculating volumes, or using the perimeter method for calculating surface area, it is necessary to locate the bases. Students sometimes have trouble with this when the polyhedron in question is not sitting on its base. Remind students that the mathematical definition of the bases of a prism is two parallel congruent polygons, not the common language definition of a base, which is something an object sits on. Once students think they have identified the bases, they can check that any cross-section taken parallel to the bases is congruent to the bases. Thinking about the cross-sections will also help them understand why the volume formula works later in this chapter.

Understand the Formula - Many times students think it is enough to remember and know how to apply a formula. They do not see why it is necessary to understand how and why it works. The benefit of fully understanding what the formula is doing is versatility. Substituting and simplifying works wonderfully for standard cylinders, but what if the surface area of a composite solid needs to be found?

Example 1: Find the surface area of the composite solid shown below.

Answer: \begin{align*}426 \ u^2\end{align*}426 u2

There are a number of ways to “divide” up this figure into numerous rectagular pieces to find the surface area. One way is to find the areas of the rectangular “steps” and add those to the rectangles formed on the sides of the figure by drawing vertical lines.

\begin{align*}2(4 \times 12)+2(3 \times 12)+2(4 \times 12)+2 \left \{(4 \times 11)+(3 \times 7)+(4 \times 4) \right \}=426 \ u^2\end{align*}2(4×12)+2(3×12)+2(4×12)+2{(4×11)+(3×7)+(4×4)}=426 u2

Make and Take Apart a Cylinder - Students have a difficult time understanding that the length of the rectangle that composes the lateral area of a cylinder has length equal to the circumference of the circular base. First, review the definition of circumference with the students. A good way to help students visualize this is a soup can label. When the label is removed and placed flat on a table, it is a rectangle. Another way to describe the circumference is to talk about an ant walking around the circle. Next, let them play with some paper cylinders. Have them cut out circular bases, and then fit a rectangle to the circles to make the lateral surface. After some time spent trying to tape the rectangle to the circle, they will understand that the length of the rectangle matches up with the outside of the circle, and therefore, must be the same as the circumference of the circle.

Surface Area of Pyramids and Cones

Prism or Pyramid - Some students have trouble deciding if a solid is a prism or a pyramid. Most try to make the determination by looking for the bases. This is especially tricky if the figure is not sitting on its base. Another method for differentiating between these solids is to look at the lateral faces. If there are a large number of parallelograms, the figure is probably a prism. If there are more triangles, the figure is most likely a pyramid. Once the student has located the lateral faces, then they can make a more detailed inspection of the base or bases.

Height, Slant Height, or Edge - A pyramid contains a number of segments with endpoints at the vertex of the pyramid. There is the altitude which is located inside right pyramids, the slant height of the pyramid is the height of the triangular lateral faces, and there are lateral edges, where two lateral faces intersect. Students frequently get these segments confused. To improve their understanding, give them the opportunity to explore with three-dimensional pyramids. Have the students build pyramids out of paper or light cardboard. The slant height of the pyramid should be highlighted along each lateral face in one color, and the edges where the lateral faces come together in another color. A string can be hung from the vertex to represent the altitude of the pyramid. The lengths of all of these segments should be carefully measured and compared. They should make detailed observations before and after the pyramid is assembled. Once the students have gained some familiarity with pyramids and these different segments, it will make intuitive sense to them to use the height when calculating volume, and the slant height for surface area.

Example 1: A square pyramid is placed on top of a cube. The cube has side length 3 cm. The slant height of the triangular lateral faces of the pyramid is 2 cm. What is the surface area of this composite solid?

Answer: \begin{align*}57 \ cm^2\end{align*}57 cm2

Five faces of the cubic base are visible and therefore part of the surface area. Their total area is: \begin{align*}5 \times 3 \times 3 = 45 \ cm^2\end{align*}5×3×3=45 cm2. The surface area of the pyramid is just the lateral surface area because the base is not visible. This lateral surface area is calculated by multiplying the perimeter of the base by the slant height of the pyramid and then multiplying by one half as shown: \begin{align*}\frac{1}{2} \times 4 \times 3 \times 2 = 12 \ cm^2\end{align*}12×4×3×2=12 cm2. Adding these two areas together yields the answer shown above.

Example 2: Calculate the surface area of the composite figure shown below.

Answer: \begin{align*}226.19 \ u^2\end{align*}226.19 u2

The surface area of this figure is the surface area of the cylinder minus the area of the “top” base plus the lateral surface area of the cone. The area of the “top” of the base and the “base” of the cone are not visible and therefore not part of the total surface area of the composite solid. Students are prone to just find the surface areas of the two figures and add them together. This will result in an answer that includes the areas of these two circular bases.

\begin{align*}\pi 3^2 + 2(3) \pi (8)+ \pi (3)(5)\approx 226.19 \ u^2\end{align*}π32+2(3)π(8)+π(3)(5)226.19 u2

Volume of Prisms and Cylinders

The Volume Base - In the past, when students used formulas, they just needed to identify the correct number to substitute in for each variable. Calculating a volume requires more steps. To find the correct value to substitute into the \begin{align*}B\end{align*}B in the formula \begin{align*}V=Bh\end{align*}V=Bh, usually requires an additional calculation with an area formula. Students will often forget this step, and use the length of the base of the polygon that is the base of the prism for the \begin{align*}B\end{align*}B. Emphasize the difference between \begin{align*}b\end{align*}b, the linear measurement of the length of a side of a polygon, and \begin{align*}B\end{align*}B, the area of the two-dimensional polygon that is the base of the prism. Students can use dimensional analysis to check their work. Volume is measured in cubic units, so three linear measurements, or a linear unit and a squared unit must be fed into the formula.

Example 1: The volume of a 4 in tall coffee cup is approximately \begin{align*}50 \ in^3\end{align*}50 in3. What is the radius of the base of the cup?

Answer: The cup has a radius of approximately 2 inches.

Since we know the volume we can use the volume formula with the given height and solve for the radius as shown below:

\begin{align*}&50 = \pi r^2 (4)\\ &3.97887 \ldots \approx r^2\\ &r \approx 2\end{align*}50=πr2(4)3.97887r2r2

Example 2: A prism has a base with area \begin{align*}15 \ cm^2\end{align*}15 cm2 and a height of 10 cm. What is the volume of the prism?

Answer: \begin{align*}V=15 \times 10 = 150 \ cm^3\end{align*}V=15×10=150 cm3

Example 3: A triangular prism has a height of 7 cm. Its base is an equilateral triangle with side length 4 cm. What is the volume of the prism?

Answer: \begin{align*}V = Bh = \frac{1}{2} (4)(2 \sqrt{3})(7) = 28 \sqrt{3} \approx 48.5 \ cm^3\end{align*}V=Bh=12(4)(23)(7)=28348.5 cm3

Given that the sides of the equilateral triangle are 4, then the altitude (height) of the triangle is \begin{align*}2\sqrt{3}\end{align*}23. So, the area of the base is \begin{align*}\frac{1}{2} (4)(2 \sqrt{3})\end{align*}12(4)(23). Then we can multiply this by the height, 7.

Example 4: The volume of a cube is \begin{align*}27 \ cm^3\end{align*}27 cm3. What is the cube’s surface area?

Answer: \begin{align*}36 \ cm^2\end{align*}36 cm2

This is a two step problem. First, students should find the length of one edge of the cube by finding the cubed root of 27. This is 3. Next, find the area of one square face \begin{align*}(3^2 = 9)\end{align*}(32=9) and multiply it by 6 because there are 6 faces.

Volume of Pyramids and Cones

Don’t Forget the \begin{align*}\frac{1}{3}\end{align*}13 – The most common mistake students make when calculating the volume of a pyramid is to forget to divide by three. They also might mistakenly divide by three when trying to find the volume of a prism. The first step students should take when beginning a volume calculation, is to make the decision if the solid is a prism or a pyramid. Once they have chosen, they should immediately write down the correct volume formula.

Prism or Pyramid - Some students have trouble deciding if a solid is a prism or a pyramid. Most try to make the determination by looking for the bases. This is especially tricky if the figure is not sitting on its base. Another method for differentiating between these solids is to look at the lateral faces. If there are a large number of parallelograms, the figure is probably a prism. If there are more triangles, the figure is most likely a pyramid. Once the student has located the lateral faces, then they can make a more detailed inspection of the base or bases.

Mix’em Up - Students have just learned to calculate the surface area and volume of prisms, cylinders, and cones. Most students do quite well when focused on one type of solid. They remember the formulas and how to apply them. It is a bit more difficult when students have to choose between the formulas for all four solids. Take a review day here. Have the students work in small groups during class on a worksheet or group quiz that has a mixture of volume and surface area exercises for these four solids. The extra day will greatly help to solidify the material learned in the last few lessons.

Example 1: A square pyramid is placed on top of a cube. The cube has side length 4 cm. The height of the pyramid is 6 cm. What is the volume of this composite solid?

Answer: \begin{align*}96 \ cm^3\end{align*}96 cm3

The volumes of the cube and square pyramid should be calculated separately and then added together to get the total volume of the composite solid.

Example 2: Calculate the volume of the composite figure shown below.

Answer: \begin{align*}263.89 \ u^2\end{align*}263.89 u2

The volume of this composite solid can be calculated by finding the volumes of the cylinder and the cone and adding them together. The Pythagorean theorem (or recognition of a Pythagorean Triple) will be used to determine the height (4) of the cone.

\begin{align*}\pi 3^2 (8) + \frac{1}{3} \pi (3^2 )(4) \approx 263.89 \ u^2\end{align*}π32(8)+13π(32)(4)263.89 u2

Surface Area and Volume of Spheres

Expand on Circles - Students learned about circles earlier in the course. Review and expand on this knowledge as they learn about spheres. Ask the students what they know about circles. Being able to demonstrate their knowledge will build their confidence and activate their minds. Now, modify the definitions that the students have provided to fit the three-dimensional sphere. Students will learn the new material quickly and will remember it because it is now integrated with their knowledge of circles.

Explore Cross-Sections - One of the goals of this chapter is to develop the students’ ability to think about three-dimensional objects. Most students will need a significant amount of practice before becoming competent at this skill. Take some time and ask the students to think about what the cross-sections of a sphere and a plane will look like. Explore trends. What happens to the cross-section as the plane moves farther away from the center of the circle? A cross-section that passes through the center of the sphere makes the largest possible circle, or the great circle of the sphere.

Cylinder to Sphere - It would be a good exercise for students to take the formula for the surface area of a cylinder and derive the formula for the surface area of a sphere. It is just a matter of switching a few variables, but it would be a good exercise for them. During the lesson, ask them to do it in their notes, wait a few minutes and then do it on the board or ask one of them to put their work on the board. It should look something like this:

\begin{align*}A_{cylinder} & = bases + lateral \ area\\ A_{cylinder} & = 2 \pi r^2 + 2 \pi rh\end{align*}AcylinderAcylinder=bases+lateral area=2πr2+2πrh

Now, replace \begin{align*}h\end{align*}h with \begin{align*}r\end{align*}r to get:

\begin{align*}A_{sphere} &= 2 \pi r^2 + 2 \pi r(r)\\ A_{sphere} & = 2 \pi r^2 + 2 \pi r^2\\ A_{sphere} & = 4 \pi r^2\end{align*}AsphereAsphereAsphere=2πr2+2πr(r)=2πr2+2πr2=4πr2

Point out to students that in the last line the terms could be combined because they both had \begin{align*}\pi r^2\end{align*}πr2 and are therefore like terms. The coefficients could have been different, but to combine terms using the distributive property they must have the exact same variable combination. Here the \begin{align*}\pi\end{align*}π is being treated as a variable even though it represents a number. This is a more complex application of like terms than students are used to seeing.

Limits - Another way to derive the volume of the sphere is to consider a limit. Essentially, the idea is to sum the volume of an infinite number of pyramids. The base of each pyramid is a regular polygon on the surface of the sphere and its height is the sphere’s radius.

\begin{align*}V_{pyramid} = \frac{1}{3} Bh = \frac{1}{3} Br\end{align*}Vpyramid=13Bh=13Br

If we let the areas of each of the infinite number of bases be \begin{align*}B_1,B_2,B_3,\ldots\end{align*}B1,B2,B3, we get a volume formula for the sphere of:

\begin{align*}V_{sphere} = \frac{1}{3} B_1 r + \frac{1}{3} B_2 r + \frac{1}{3} B_3 r+ \ldots\end{align*}Vsphere=13B1r+13B2r+13B3r+

Now factor out the \begin{align*}\frac{1}{3}\end{align*}13 and the \begin{align*}r\end{align*}r to get:

\begin{align*}V_{sphere} = \frac{1}{3} r(B_1 + B_2 + B_3, \ldots )\end{align*}Vsphere=13r(B1+B2+B3,)

Now, we can replace the sum of the infinite base areas with the surface area of the sphere to get:

\begin{align*}V_{sphere} &= \frac{1}{3} r(4 \pi r^2)\\ V_{sphere} &=\frac{4}{3} \pi r^3\end{align*}VsphereVsphere=13r(4πr2)=43πr3

You can do this in reverse to go from the formula for the volume of a sphere to the surface area of the sphere:

\begin{align*}V_{sphere}&=\frac{4}{3} \pi r^3 = \frac{1}{3} r(4 \pi r^2 )\\ \frac{4}{3} \pi r^3 &= \frac{1}{3} r(B_1+B_2+B_3, \ldots )\end{align*}Vsphere43πr3=43πr3=13r(4πr2)=13r(B1+B2+B3,)

Now, dividing both sides by \begin{align*}\frac{1}{3} r\end{align*}13r we get:

\begin{align*}4 \pi r^2 = (B_1+B_2+B_3,\ldots )\end{align*}4πr2=(B1+B2+B3,)

Initially, the logic might seem fuzzy to them. The limit is a fundamental concept to all of calculus. It is worthwhile to give it some attention here and some students are more interested in a formula and mathematics in general when they understand where it comes from.

Extension: Exploring Similar Solids

Surface Area is Squared - Surface area is a two-dimensional measurement taken of a three-dimensional object. Students are often distracted by the solid and use cubed units when calculating surface area or mistakenly cube the ratio of linear measurements of similar solids when trying to find the ratio of the surface areas. Remind them, and give them many opportunities to practice with exercises where surface area and volume are both used.

Don’t Forget to Adjust the Ratio - There are three distinct ratios that describe the relationship between similar solids. When the different ratios and their uses are the subject of the lesson, students usually remember to use the correct ratio for the given situation. In a few weeks when it comes to the chapter test or on the final at the end of the year, students will frequently forget that the area ratio is different from the volume ratio and the linear ratio. They enjoy writing proportions and when they recognize that a proportion will be used, they get right to it without analyzing the ratios. One way to remind them is to have them use units when writing proportions. The units on both sides of the equal sign have to match before they can cross-multiply. Give them opportunities to consider the relationship between the different ratios with questions like the one below.

Example 1: If a fully reduced ratio is raised to a power, will the resulting ratio be fully reduced? Explain your reasoning.

Answer: Yes, two numbers make a fully reduced ratio if they have no common factors. Raising a number to a power increases the exponent of each factor already present, but does not introduce new factors. Therefore, the resulting two numbers will still not have any common factors.

These concepts frequently appear on the SAT. It will serve the students well to practice them from time to time to keep the knowledge fresh.

Example 2: The ratio of the surface areas of two cubes is 25:49. What is the ratio of their volumes?

Answer: In order to find the ratio of the volumes, we need to square root the ratio of the areas and then cube the resulting ratio:

\begin{align*}\left (\sqrt{25} \right )^3 & : \left (\sqrt{49} \right )^3\\ 5^3 & :7^3\\ 125 & :343\end{align*}(25)353125:(49)3:73:343

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