Instructional Week 6: February 8-12
Grade 7 ISTEP+
Focus Topic: Cylinders, Three-Dimensional Objects, and Nets
Paced Standards:
7.GM.6: Solve real-world and other mathematical problems involving volume of cylinders and three-dimensional
objects composed of right rectangular prisms. +
7.GM.7: Construct nets for right rectangular prisms and cylinders and use the nets to compute the surface area;
apply this technique to solve real-world and other mathematical problems.
PS: 1, 2, 3, 4, 5, 6, 7, and 8 +
Key Vocabulary
Cylinders-A solid object with: • two identical flat ends that are circular
Right Regular Prism- a geometric solid that has a regular polygon as its base and vertical sides perpendicular to the
base. The base and top surface are the same shape and size.
Nets- a flattened three-dimensional figure which can be turned into the sold by folding it.
Connection to other 7th grade Standards
None
Prerequisite/Foundational Standards
6.GM.3: Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find
the length of a side joining points with the same first coordinate or the same second coordinate; apply
these techniques to solve real-world and other mathematical problems.
6.GM.4: Find the area of complex shapes composed of polygons by composing or decomposing into
simple shapes; apply this technique to solve real-world and other mathematical problems.
6.GM.6: Construct right rectangular prisms from nets and use the nets to compute the surface area of
prisms; apply this technique to solve real-world and other mathematical problems.
Teacher Background: Samples Problems for 7.GM.6
1. The 7th graders at Sunview Middle School were helping to renovate a playground for the kindergartners at a
nearby elementary school. City regulations require that the sand underneath the swings be at least 15 inches
deep. The sand under both swing sets was only 12 inches deep when they started.
The rectangular area under the small swing set measures 9 feet by 12 feet and required 40 bags of sand to
increase the depth by 3 inches. How many bags of sand will the students need to cover the rectangular area
under the large swing set if it is 1.5 times as long and 1.5 times as wide as the area under the small swing
set? (Illustrative Mathematics)
2. Leo's recipe for banana bread won't fit in his favorite pan. The batter fills the 8.5 inch by 11 inch by 1.75 inch
pan to the very top, but when it bakes it spills over the side. He has another pan that is 9 inches by 9 inches
by 3 inches, and from past experience he thinks he needs about an inch between the top of the batter and
the rim of the pan. Should he use this pan?( Illustrative Mathematics)
3. Rolled oats (dry oatmeal) come in cylindrical containers with a diameter of 5 inches and a height of
9 1/2 inches. These containers are shipped to grocery stores in boxes. Each shipping box contains six rolled
oats containers. The shipping company is trying to figure out the dimensions of the box for shipping the rolled
oats containers that will use the least amount of cardboard. They are only considering boxes that are
rectangular prisms so that they are easy to stack.
a. What is the surface area of the box needed to ship these containers to the grocery store that uses the least
amount of cardboard?
b. What is the volume of this box?
Teacher Background Notes:
1.
Solution: Finding the scale factor the hard way
3 inches is 1/4=0.25 foot, so the volume of sand that was used is 0.25×9×12=27 cubic feet. The amount of sand
needed for an area that is 1.5 times as long and 1.5 times as wide would be 0.25×(1.5⋅9)×(1.5⋅×12)=60.75 cubic feet.
We know that 40 bags covers 27 cubic feet. Since the amount of sand for the large swing set is 60.75÷27=2.25
times as large, they will need 2.25 times as many bags. Since 2.25×40=90, they will need 90 bags of sand for the
large swing set.
Solution: Finding the scale factor the easy way
Since we have to multiply both the length and the width by 1.5, the area that needs to be covered is 1.52=2.25
times as large. Since the depth of sand is the same, the amount of sand needed for the large swing set is 2.25 times
as much as is needed for the small swing set, and they will need 2.25 times as many bags. Since 2.25×40=90, they
will need 90 bags of sand for the large swing set.
Solution: Using a unit rate
The area they cover under the small swing set is 9×12=108 square feet. Since the depth is the same everywhere,
and we know that 40 bags covers 108 square feet, they can cover 108÷40=2.7 square feet per bag.
The area they need to cover under the large swing set is 1.52=2.25 times as big as the area under the small swing
set, which is 2.25×108=243 square feet. If we divide the number of square feet we need to cover by the area
covered per bag, we will get the total number of bags we need: 243÷2.7=90. So they will need 90 bags of sand for
the large swing set.
Solution: The other unit rate
The area they cover under the small swing set is 9×12=108 square feet. Since the depth is the same everywhere,
and we know that 40 bags covers 108 square feet, they can cover 40÷108=10/27 bags per square foot.
The area they need to cover under the large swing set is (3/2)2=9/4 times as big as the area under the small swing
set, which is 9/4×108=243 square feet. If we multiply the number of square feet they need to cover by the number
of bags needed per square foot, we will get the total number of bags we need: 243×10/27=90
So they will need 90 bags of sand for the large swing set.
2.
In order to find out how high the batter will be in the second pan, we must first find out the total volume of
the batter that the recipe makes. We know that the recipe fills a pan that is 8.5 inch by 11 inch by 1.75 inches. We
can calculate the volume of the batter multiplying the length, the width, and the height:
V=8.5 in×11 in×1.75 in
We know that the batter will have the same volume when we pour it into the new pan. When the batter is poured
into the new pan, we know that the volume will be 9×9×h where h is the height of the batter in the pan. We already
know that
V=163.625 in3, so:
V=lxwxh
163.625 in3 = 9 in x 9 in x h
163.625 in3= 81 in2x h
163.625 in3 = h
81 in2
2.02 in ≈h
Therefore, the batter will fill the second pan about 2 inches high. Since the pan is 3 inches high, there is nearly an
inch between the top of the batter and the rim of the pan, so it will probably work for the banana bread (assuming
that Leo is right that that an inch of space is enough).
3.
a
i. At first students may find the sum of all the areas of the faces.
Sum of the areas of the 6 faces: SA=47.5+47.5+150+150+285+285=965 in2
Using the surface area formula:
SA=2wl+2lh+2wh
SA=2(5)(30)+2(30)(9.5)+2(5)(9.5)
SA=965 in2
Students may continue to use the sum of all the areas of the faces to find the surface area which gives the
same value as using the surface area formula which is shown below.
Using the surface area formula:
SA=2wl+2lh+2wh
SA=2(5)(30)+2(30)(9.5)+2(5)(9.5)
SA=965 in2
Students may continue to use the sum of all the areas of the faces to find the surface area which gives the
same value as using the surface area formula which is shown below.
ii.
SA=2wl+2lh+2wh
SA=2(10)(15)+2(15)(9.5)+2(10)(9.5)
SA=775 in2
iii.
iv.
v.
b.
The best solution is a setup of 2 cylinders wide by 1 cylinder high by 3 cylinders long. This is the same as 3
cylinders wide by 1 cylinder high by 2 cylinders long.
Resources for 7.GM.6
 This provides web-based examples and guidance:
http://www.onlinemathlearning.com/volume-of-a-cylinder.html
 This provides a tutorial. (Volume is needed for this standard and not surface area):
https://www.khanacademy.org/math/basic-geo/basic-geo-volume-surface-area/basic-geo-volumes/v/cylindervolume-and-surface-area
 This provides a tutorial:
https://learnzillion.com/lessons/3540-find-the-volume-of-a-cylinder

This provides specific guidance and activities pertaining to objects composed of right rectangular prisms.
Click on the link at the top of the following page:
http://webcache.googleusercontent.com/search?q=cache:VNDFHPRajTkJ:https://www.engageny.org/file/64066/
download/math-g7-m6-topic-e-lesson-26teacher.pdf%3Ftoken%3DPj4Zdev6mEJOIszd6AFGuW_D5aHyCUV5WFGJRWcy04U+engageny+volume+obje
cts+composed+of+right+rectangular+prisms&cd=3&hl=en&ct=clnk&gl=us&client=firefox-a
Teacher Background: Samples Problems for 7.GM.7
1. Samuel wants to make a pencil holder for his grandfather out of a can. He needs to cover the sides and
bottom of the can with fabric, but leave the top uncovered. The can has a diameter of 3 inches and a height
of 5 inches. How much fabric should Samuel buy? Round up to the nearest inch.
2. What is the value of t if the surface area of this right rectangular prism is 352 in2? Draw a net to represent
this problem
8 in
Teacher Background Notes:
1. Use the net to calculate the surface area of the can minus the top which is not covered in fabriic.
S.A. = 1.52𝜋+ 3𝜋(5) = 54.19 in2 therefore 55 in2
2. Use the net of the rectangular prism to write an equation for the surface area.
2(8)(8) + 4(8)(t) + = 352 in2
128+32t = 352
32t = 224
t = 7 in
there are other possibilities
Resources for 7.GM.7
 This tutorial provides guidance for using nets to find the surface area of a prism:
https://www.khanacademy.org/math/basic-geo/basic-geo-volume-surface-area/basic-geo-surfacearea/e/surface-area
 This link provides a tutorial for constructing nets of geometric figures:
http://www.onlinemathlearning.com/geometry-nets.html
 This link provides a tutorial for using the net to find the surface area of a cylinder:
http://www.virtualnerd.com/geometry/surface-area-volume-solid/prisms-cylinders-area/calculate-surfacearea-cylinder-net
 This link provides a tutorial for using the net to find the surface area of a rectangular prism:
http://virtualnerd.com/geometry/surface-area-volume-solid/prisms-cylinders-area/calculate-surface-arearectangular-prism-net

This provides an opportunity for students to practice applying the relationship between surface area and
nets:
http://cdn.kutasoftware.com/Worksheets/Geo/10-Surface%20Area%20of%20Prisms%20and%20Cylinders.pdf
Process Standards to Emphasize with Instruction
PS.1: Make sense of problems and persevere in solving them.
PS.2: Reason abstractly and quantitatively.
PS.3: Construct a viable argument and critique the reasoning of others.
PS.4: Model with mathematics.
PS.5: Use appropriate tools strategically.
PS.6: Attend to precision.
PS.7: Look for and make use of structure.
PS.8: Look for and express regularity in repeated reasoning.
Instructional Week 6 Assessment
ISTEP+ Grade 7
Name______________________________
(7.GM.7) 1. Lydia needs to bring a homemade musical instrument to class. She can only find an empty paint can in the
basement to use for this project. She needs to cover the entire can including the top and bottom with
construction paper, so it will look like a drum. How much construction paper will Lydia need for this project?
Show all work. Round up to the nearest centimeter.
(7.GM.6) 2. A shipping company wants to ship its boxes in a trailer that has a height of 108 inches, a width of 8 feet, and a
length of 20 feet long. The boxes have dimensions of 2 feet by 5 feet by 3 feet.
a. What is the maximum number of boxes that can fit in the trailer if they have to be stacked so that the bottom
of each box measures 3 ft x 5 ft?
b. What is the volume of space left “over” after the trailer has been loaded?
(7.GM.7) 3. If the surface area of the box below is 556 cm2, what must the value of x be?
x
Instructional Week 6 Assessment Key
ISTEP+ Grade 7
(7.GM.7) 1. 2 points
S.A. = 2990.79; therefore 2, 221 cm2 of construction paper is needed.
(7.GM.6) 2. 4 points total; 2 each. Methods may vary
a. 32 boxes
b. 480 ft2
(7.GM. 7) 3. 2 points
10 cm
All constructed response problems will be graded using the ISTEP+ Content Rubric.