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# 1.11: Surface Area and Volume

Difficulty Level: At Grade Created by: CK-12

Pacing

Day 1 Day 2 Day 3 Day 4 Day 5
Exploring Solids Surface Area of Prisms & Cylinders Finish Surface Area of Prisms & Cylinders

Quiz 1

Start Surface Area of Pyramids and Cones

Finish Surface Area of Pyramids and Cones
Day 6 Day 7 Day 8 Day 9 Day 10
Volume of Prisms & Cylinders

Quiz 2

Start Volume of Pyramids & Cones

Finish Volume of Pyramids & Cones

Investigation 11-1

Surface Area and Volume of Spheres

Quiz 3

\begin{align*}*\end{align*}Start Extension: Exploring Similar Solids

Day 11 Day 12 Day 13 Day 14 Day 15
\begin{align*}*\end{align*}Finish Extension: Exploring Similar Solids \begin{align*}*\end{align*}Extension Quiz Start Review of Chapter 11 Review of Chapter 11 Chapter 11 Test Start Chapter 12

## Exploring Solids

Goal

The purpose of this lesson is to introduce students to three-dimensional figures. Polyhedral figures are presented in this lesson and common terms such as edge, vertex, and face are explained, as well as how to name polyhedra.

Relevant Review

Students should know the definition of a regular polygon and how to find the area of various triangles and quadrilaterals.

Teaching Strategies

Like with previous chapters, it might be helpful for students to make flash cards of the vocabulary and theorems learned.

Discuss with students where they would see polyhedra, prisms, pyramids, cylinders, and cones in real life. Tell students to visualize themselves in a grocery store; there are several examples of these solids there. Examples could be: soup can (cylinder), Toblerone chocolate bar (triangular prism), oranges (sphere), or waffle cones (cone). Write down the items students come up with and draw a representation.

After going over Example 1 and putting the faces, vertices, and edges into a table, see if students can come up with Euler’s Theorem on their own. Students might wonder why they need to know Euler’s Theorem, when they can just count the number of vertices, edges, and faces. One application could be when the prism has so many sides that it becomes difficult to count the edges (E\begin{align*}E\end{align*} is always the largest of F,V,\begin{align*}F, V,\end{align*} and E\begin{align*}E\end{align*} in the formula). Other applications are Examples 3 and 4, when you are not given a picture in the problem.

One way to “view” a three-dimensional solid is to use cross-sections. You might need to use physical models to help students understand this concept. For example, you could mold Play-doh into a cylinder or cube and “cut” them in different ways to show the different cross-sections. A cylinder can have a circle, oval (or portion of an oval), or rectangle as a cross-section.

Another way to “view a three-dimensional solid is to use a net. A net takes all the faces of a solid and lies them out flat and adjoining. Students need to be careful with nets. There can be several answers for one solid and depending on how the faces are laid out, the net might not even work. Be sure to show students nets that do not work. The activity in the link, http://illuminations.nctm.org/activitydetail.aspx?ID=84, gives students 20 possible nets of a cube and they need to find the 11 that work.

## Surface Area of Prisms and Cylinders

Goal

Students will learn how to find the surface area of prisms and cylinders.

Relevant Review

Students should know how to find a net of a three-dimensional solid. They will also need to be able to find the area of triangles and quadrilaterals.

Teaching Strategies

The surface area can be difficult for students to visualize. Make sure they have a firm grasp on how to find nets of a three-dimension figure. Students need to find all the lengths and heights of the faces of a prism before finding the total surface area. For example, in Example 2, students need to find the length of the hypotenuse of the base because it is also the length of a rectangle.

Additional Example: Find the surface area of the trapezoidal prism. The bases are isosceles trapezoids.

Solution: First find the length of the legs of the isosceles trapezoid bases.

Here is the net for this prism.

Surface Area:

Trapezoids A=12(35+19)15=405Rectangles A=1740=680A=3540=1400A=1940=760 Total=405+405+680+680+1400+760  =4330 units2\begin{align*}& \text{Trapezoids} \ A= \frac{1}{2}(35+19)15=405\\ & \text{Rectangles} \ A=17 \cdot 40=680\\ & \qquad \qquad \quad A=35 \cdot 40=1400\\ & \qquad \qquad \quad A=19 \cdot 40=760\\ & \qquad \quad \ \text{Total} = 405 + 405 + 680 + 680 + 1400 + 760\\ & \qquad \qquad \quad \ \ = 4330 \ units^2\end{align*}

The surface area of a cylinder can be difficult for students to visualize. As the FlexBook suggests, take the label off of a soup can so that it opens up to a rectangle. Then, students will see that the width of the rectangle is the circumference of the circular base.

## Surface Area of Pyramids and Cones

Goal

Students will learn the formulas for the surface area of a regular pyramid and cone.

Relevant Review

Students should review nets and the formula for the area of triangles and quadrilaterals. Students should also be able to apply the Pythagorean Theorem, Pythagorean triples, and special right triangle ratios.

Teaching Strategies

The slant height is a new term that applies to the lateral faces. It is only used to find the surface area of pyramids and cones. Students may need to use the actual height to find the slant height through the Pythagorean Theorem. Help students develop a formula for finding the slant height for different types of pyramids and cones. In this lesson, we will find the surface area of equilateral triangle based pyramids, square based pyramids, and right cones.

The most difficult of the three is the equilateral based pyramid. For the review questions in this section, the slant height for these pyramids is given. However, students will still need to find the altitude of the base in order to find the area. Or, if students remember, they can use the formula, A=s234\begin{align*}A=\frac{s^2\sqrt{3}}{4}\end{align*} for the area of any equilateral triangle, where s\begin{align*}s\end{align*} is the length of the sides.

Since students only need to worry about two types of pyramids (for surface area), then you could generate more specific surface area formulas for each one. They are:

Equilateral triangle based pyramid:Square based pyramid:SA=B+12nbl=s234+12(3)sl=s234+32slSA=B+12nbl=s2+12(4)sl=s2+2sl\begin{align*}& Equilateral \ triangle \ based \ pyramid: && SA=B+\frac{1}{2}nbl=\frac{s^2\sqrt{3}}{4}+\frac{1}{2}(3)sl=\frac{s^2\sqrt{3}}{4}+\frac{3}{2}sl\\ & Square \ based \ pyramid: && SA=B+\frac{1}{2}nbl=s^2+\frac{1}{2}(4) sl=s^2+2sl\end{align*}

Where s\begin{align*}s\end{align*} is the edge length and l\begin{align*}l\end{align*} is the slant height. Once students know the basic surface area, derive these with the class for n=3\begin{align*}n = 3\end{align*} and n=4\begin{align*}n = 4\end{align*}.

Compared to the pyramid, the surface area of a cone is relatively simple. Students will always be given two of the three pieces of information needed; h,r\begin{align*}h, r\end{align*}, or l\begin{align*}l\end{align*} (in the picture above). If the problem does not give l\begin{align*}l\end{align*}, the slant height, then students will have to use the Pythagorean Theorem to solve for it.

## Volume of Prisms and Cylinders

Goal

Students will learn how to find the volume of prisms and cylinders. They will also discover that the volume of an oblique prism or cylinder is the same as a right prism or cylinder with the same height.

Relevant Review

Students will still need to know how to find the area of triangles and quadrilaterals so they can properly apply the volume formulas in this lesson. Also, review with students how to “cube” and “cube root” a number. Lastly, review the definitions of prisms and cylinders, so students are clear about which faces are bases. This will be very helpful when using the volume formulas.

Teaching Strategies

Begin with a discussion of volume. Students might get volume, mass, and density confused. Volume is simply the amount of physical space a three-dimension takes up. Think of two different objects that have the same shape; a 12-pound bowling ball and a plastic empty sphere of the same size. Both of these solids have the same volume, but very different masses or densities. Therefore, an empty solid has the same volume as a full solid of the same size.

Volume is measured in cubic units, or units3\begin{align*}\text{units}^3\end{align*}. This is because the formula for volume always comes back to length×width×height\begin{align*}length \times width \times height\end{align*}. Each of these have a unit associated with them. So, the answer would be unit×unit×unit=units3\begin{align*}unit \times unit \times unit = units^3\end{align*}.

Example 1 explores the concept of placing cubes (of whatever unit) into a prism to find its volume. Students can count all these cubes and find that there are 60 within the prism. Another option would be to show students what the area of each face is and then multiply that by the relative height. This will demonstrate that the order of multiplication does not matter.

V=(3×5)×4=15×4=60V=(3×4)×5=12×5=60V=(5×4)×3=20×3=60\begin{align*}& V=(3 \times 5) \times 4=15 \times 4=60\\ & V=(3 \times 4) \times 5=12 \times 5=60\\ & V=(5 \times 4) \times 3=20 \times 3=60\end{align*}

Regardless of the order, the formula for the volume of a rectangular prism will be length×width×height\begin{align*}length \times width \times height\end{align*}. It does not matter which face is the base for a rectangular prism.

Students will take the words “base” and “height” literally. However, in the formula for volume, the “height” might not always be the vertical length. As in Example 3, the apparent height of the solid is only the height of the base. The “height” in the formula is actually 7 ft, which looks like the length of the base. Students will think that the base of this tent is the rectangle on the bottom. However, we know that the bases are actually the triangles at the front and back of the tent. Be very careful when discussing the formula for volume and the definition of a prism. Review with students that the “bases” are the two congruent parallel faces. Before starting on a volume problem, students should examine the solid so they know which faces are the bases.

As with the case of an oblique prism, the sides are all parallelograms, however the bases will still be parallel. Students should know that the height of an oblique prism is not going to be an edge (like it is in a right prism), but a vertical length that is outside the solid. Review with students that the base of a cylinder is a circle.

## Volume of Pyramids and Cones

Goal

In this lesson, students will discover that the volume of a prism is one-third the volume of a prism with the same base. This property also applies to cones. Then, they will find the volume of composite solids.

Relevant Review

Make sure students are comfortable finding the volume of prisms and cylinders from the previous lesson.

Teaching Strategies

Investigation 11-1 should be a teacher-led activity. You should pre-make the open nets of a cube and pyramid and then demonstrate that filling the pyramid three times will completely fill the cube.

If you decide to allow students to do the investigation, be prepared for it to take 20-30 minutes. You should make the nets with students in case they have questions. Have students complete this activity in pairs and each student in the pair can make one net.

In this lesson, students should be able to find the volume of any triangle or quadrilateral based pyramid and any type of cone. If students are ever given the slant height, they will need to solve for the overall height of the pyramid or cone. If a cone is not a right cone, it will not have a slant height and students will need to be given the height in order to find the volume.

Like with surface area of pyramids, you can generate more specific formulas for the volume of an equilateral triangle based pyramid and a square based pyramid. They would be:

Equilateral triangle based pyramid:Square based pyramid:V=13Bh=13(s234)h=s2h312V=13Bh=13s2h\begin{align*}&Equilateral \ triangle \ based \ pyramid: && V = \frac{1}{3} Bh = \frac{1}{3} \left (\frac{s^2\sqrt{3}}{4} \right ) h = \frac{s^2h\sqrt{3}}{12}\\ &Square \ based \ pyramid: && V = \frac{1}{3} Bh = \frac{1}{3} s^2 h\end{align*}

s\begin{align*}s\end{align*} and h\begin{align*}h\end{align*} are the base edge length and the height. You can also generate formulas for the volume of a rectangular based pyramid and a right triangle based pyramid.

Problems 3 and 6 will be quite difficult for students to complete (in the Review Questions). This is because the bases are equilateral triangles and they are given the slant height. If you desire, you can change the values given as the slant height to be the height of the pyramids. If you do this, the answers will be 25633\begin{align*}\frac{256\sqrt{3}}{3}\end{align*} and 2403\begin{align*}240\sqrt{3}\end{align*}, respectively. Students will still need to find the area of the equilateral triangle bases (s234)\begin{align*}\left (\frac{s^2\sqrt{3}}{4} \right )\end{align*}. If you do not alter the problems, students will have to find the height of the pyramids, using the slant height. This will be a difficult process, because students will have to recall the properties of the centroid (the point where the vertical height hits the base). Therefore, the distance from the bottom of the slant height to the centroid will be one-third the length of the entire altitude. (See the picture in the teaching tips for Surface Area of Pyramids and Cones.)

## Surface Area and Volume of Spheres

Goal

Students will learn how to find the surface area and volume of spheres and hemispheres.

Relevant Review

Make sure students are comfortable with the formulas for circumference and area of a circle. Also, students will apply the formulas for the surface area and volume of cylinders and cones to composite solids.

Teaching Strategies

Ask students where they have seen spheres and hemispheres in real life. Discuss the parts of a sphere that are the same as a circle and those parts that are different. Be sure to show students the animated derivations of the surface area and volume by Russell Knightley.

When finding the surface area or volume of composite solids discuss with students what the parts are before starting each problem This will make it easier for students to complete the problem. Also, make sure they understand when to not include the top or bottom of cylinders or hemispheres in the total surface area (see Example 5).

## Extension: Exploring Similar Solids

Goal

Students will understand the relationship between similar solids, their surface areas, and their volumes.

Relevant Review

The areas of similar polygons should be reviewed before starting this extension. Students should know that the square of the scale factor is the ratio of the areas of two similar shapes.

Teaching Strategies

Help students make the connection between the ratios of the areas of two similar shapes is the same as the ratio of the surface areas of two similar solids. Both are area, so both ratios will be the square of the scale factor. Following this pattern, ask students if they have an idea as to what the ratio of the volumes of two similar solids would be.

Additional Example: The two square based pyramids below are similar. Find the surface area and volume of both solids.

Solution: The scale factor is 812=23\begin{align*}\frac{8}{12} = \frac{2}{3}\end{align*}. The slant height of the smaller pyramid is l=42+82=16+64=80=45\begin{align*}l = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\end{align*}. Using the scale factor, the slant height of the larger pyramid is 3245=65\begin{align*}\frac{3}{2} \cdot 4\sqrt{5} = 6 \sqrt{5}\end{align*}.

SAsmaller=82+4(12845)=64+645SAlarger=122+4(121265)=144+1445Vsmaller=13(82)8=5123Vsmaller=13(122)12=576\begin{align*}& SA_{smaller} = 8^2+4 \left (\frac{1}{2} \cdot 8 \cdot 4\sqrt{5} \right ) = 64+64 \sqrt{5} && V_{smaller}= \frac{1}{3} (8^2) 8 = \frac{512}{3}\\ & SA_{larger}=12^2 + 4 \left (\frac{1}{2} \cdot 12 \cdot 6 \sqrt{5} \right ) = 144 + 144 \sqrt{5} && V_{smaller} = \frac{1}{3} (12^2) 12 = 576\end{align*}

Encourage students to find the ratios of the surface areas and volumes above to reinforce what was learned in this lesson.

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