Getting into Solids: Pyramids - ETA hand2mind

I would like to dedicate this book to my brothers and sisters: Mary Lilja, Virginia 

Bellrichard, Thomas Errthum, Christine Breuer, Charles Errthum, and Joseph 

Errthum. A special thank you to Charles, for encouraging me to patent the original 

models. It is impossible for me to express my true appreciation for the guidance 

and support all of you have provided me over the years. You are wonderful brothers 

and sisters. 

I would like to also thank Vincent Kotnik III and his wife, Cathy, who have been 

like a brother and sister to me. Vincent, you have been a wonderful friend and a 

great colleague. Thank you for mentoring me my first three years of teaching; you 

guided me well. Cathy, your positive attitude is an inspiration not only for me, but 

for anyone who meets you. I cherish the fact that I have both of you as friends. 

I have great respect and admiration for each of you. You are truly very special people. 

Please know that you are all near and dear to my heart. 

Emily Errthum 

ETA 75545 

Copyright ©2005 by Classroom Products. All rights reserved. 

Except as permitted under the United States Copyright Act and where expressly permitted, no part of this publication 

may be reproduced or distributed in any form by any means, or stored in a database or retrieval system, 

without prior written permission from the publisher. However, the publisher grants permission to classroom teachers 

to reproduce activity pages for one classroom only. The reproduction of any part for an entire school or system 

is strictly prohibited. 

Send all inquiries to: 

ETA/Cuisenaire ® 

500 Greenview Court 

Vernon Hills, IL 60061

INTRODUCTION 

Prior to Getting Into Solids, the solids were just that — solid. The lateral faces were made of plastic 

that tightly enclosed each solid. It was impossible to get inside the model to identify the height, apothem, 

radius, central angle measure, and the right triangles needed to find missing lengths. 

Leeanne Branham, from Fresno Pacific University, best identified the problem with solids in the March 

1998 issue of CMC ComMuniCator. In the article “Getting Into Pyramids” Leeanne wrote, (referring to 

the solid lateral face models) “but one drawback that these models all share was that I couldn’t get 

inside them.” Her article was designed, with worksheets, to assist students in solving surface area and 

volume problems by redrawing the parts of a figure on one-dimensional paper. I identified with her 

frustration as well as observed the frustration level of my own students. This inspired me to design the 

three-dimensional open models, Getting Into Solids — Pyramids and Getting Into Solids — Prisms. 

These new models literally allow the learner to “get inside” the solid. 

The Getting Into Solids — Prisms models allow students to: 

A. Identify the different parts of the regular prisms and right cylinders by colors. 

1. Brass rod – height 

2. Green string – lateral edge, height 

3. Etched green line on the base – radius 

4. Etched red line on the base – apothem 

5. Etched blue line on the base – base edge 

B. Attach numbers to the model for ease of visualization. 

1. Write on the plexiglass base with overhead markers 

2. Attach stick-on notes — with unit measure — on the height 

C. Employ all learning styles. 

1. Visual 

2. Auditory (when guided by the teacher or students teaching others in groups) 

3. Kinesthetic/Tactical 

1

Section One of this book contains reproducible worksheets for finding the surface area and volume of regular 

prisms and right cylinders. Each model has three worksheets. The first worksheet includes dimensions. 

The second worksheet serves as an answer key for the first worksheet. Finally, the third worksheet 

is blank so that the instructor or student can assign dimensions to any illustration — specific from 

a classroom textbook. The worksheets in this section are intended to guide students through a frequently 

long and difficult process. Helping students to organize their mathematical steps provides students 

with immediate success and, as a result, promotes a positive experience for finding surface area 

and volume. 

Section Two of this book (beginning on page 22) also contains reproducible worksheets for finding the 

surface area and volume of regular prisms and right cylinders. This section is less structured than 

section one. It is designed for students to take on the challenge of organizing their thought process in 

a logical order. Students will need to show multiple mathematical steps to yield the desired result. With 

prior work in section one, students are easily able to make this transition. Students may find it more 

difficult, but many will show extreme growth in their organizational skills. Section two utilizes concepts 

from the book Getting Into Solids – Pyramids. Students will need to know how to calculate the volume 

of a pyramid and the volume of a cone. Brass rods are provided to build and display each of the six 

prism models at one time. A shorter brass rod is included to allow the construction of a 16 cm x 16 

cm x 16 cm cube, or to change the height of any of the other models. 

Easy to assemble instructions: (see diagram below) 

Green String 

1. Screw the brass rod into polygon center (representing height). 

2. Screw a congruent polygon on the opposite side of the brass rod. 

3. Slide the end of the green string (representing lateral edge, height) 

into the notch of each base. The string should be perpendicular to 

the bases. 

4. Repeat Step 3 for all green strings, alternating from side to side. 

5. Stick-on notes can be attached to the green string, with unit 

measure (representing lateral edge, height). 

6. Turning one of the prism bases 180º will create symetrical 

pyramid models. Use twist ties, string, or tape to retain the 

desired shape. 

Brass Rod 

Notches 

Twist to Create Pyramids 

Polygon 

Base 

2

Vocabulary for Regular Prisms and Right Cylinders 

L.A. = lateral area P = perimeter of base r = radius 

S.A. = surface area B = area of base h = height 

V = volume 

π = pi 

Formulas for Regular Prisms and Right Cylinders 

Regular Prisms 

Right Cylinders 

L.A. = P h 

L.A. = 2 π r h 

S.A. = L.A. + 2 B 

S.A. = L.A. + 2 B 

S.A. = P h + 2 B S.A. = 2 π r h + 2 π r 2 

V = B h 

V = B h 

V = π r 2 h 

TABLE OF CONTENTS 

Section One: Finding Surface Area and Volume of Regular Solids 

Cylinder. ..................................................4 

Triangular Prism. ............................................7 

Cube. ...................................................10 

Rectangular Prism ..........................................13 

Pentagonal Prism ...........................................16 

Hexagonal Prism ...........................................19 

Section Two: Challenge Problems 

Finding Surface Area and Volume of Regular Solids 

Two Attached Rectangular Prisms ...............................22 

Rectangular Prism and Regular Square Pyramid ......................24 

Two Cylinders; Maximizing Lateral Area ............................26 

Cylinder with Two Cones Inserted ................................28 

Rectangular Prism with a Rectangular Hole .........................30 

Rectangular Prism with a Cylindrical Hole. ..........................32 

3

Name __________________________________________ 

Find Surface Area and Volume 

of the regular solid. 

Fill in the blanks and show all your work. 

Name of the solid: _______________________________ 

GETTING INTO SOLIDS — PRISMS 

Worksheet 1 

diameter = 7 ft 

height = 20 ft 

Equation for formula: S.A. = _____________________________ 

Equation for formula: V = ________________________________ 

First, I will solve for the area of the Base, which is a _____________________________. 

Describe your process through a series of clearly shown mathematical steps. 

B = ____________________ 

Next, I will solve for the Lateral Area, which is a _____________________________. 


L.A. = ____________________ 

Write the equations. S.A. = __________________________ V = __________________________ 

Replace variables with 

their value. S.A. = __________________________ V = __________________________ 

Solve the equations. 

(Show all algebraic steps.) S.A. = __________________________ V = __________________________ 

4

Name __________________________________________ 




Right Cylinder 



Worksheet 1: Answer Key (difficulty — medium) 

diameter = 7 ft 


L.A. + 2 B 


B h 


circle 

First, I will solve for the area of the Base, which is a ____________________________. 


First, I will find the radius, r. 

Now I can find the area of the base. 

r = 1 / 2 diameter B = π r 2 

r = 1 / 2 (7) B = π (3.5) 2 

r = 3.5 B = 12.25π 

12.25 π ft 2 , rectangle 

B = ____________________ 

Next, I will solve for the Lateral Area which is a _____________________________. 


L.A. = 2 π r h 

L.A. = 2 π (3.5) (20) 

L.A. = 140 π 

140 π ft 2 L.A. + 2 B B h 

L.A. = ____________________ 




140 π + 2 (12.25 π) 12.25 π (20) 

S.A. = 140 π + 24.5 π 


(Show all algebraic steps.) S.A. = __________________________ 164.5 π ft 2 V = __________________________ 

245 π ft 3 

S.A. = 516.79 ft 2 V = 769.69 ft 3 

5

Name __________________________________________ 




Worksheet 2 







B = ____________________ 

Next, I will solve for the Lateral Area which is a _____________________________. 


L.A. = ____________________ 






6

Name __________________________________________ 






Worksheet 3 

4 in. 

18 in. 





B = ____________________ 



L.A. = ____________________ 






7

Name __________________________________________ 




Triangular Prism 



Worksheet 3: Answer Key (difficulty - medium) 

4 in. 

18 in. 





B = S 2 3 

4 

B = 4 2 3 

4 

B = 4 3 

L.A. + 2 B 

B h 

triangle 

B = ____________________ 

4 3 in. 2 rectangle 



First I will find the 

Now I can find the 

perimeter of the base, P. 

lateral area, L.A. 

P = 3 S 

L.A. = P h 

P = 3 (4) L.A. = 12 (18) 

P = 12 L.A. = 216 

L.A. = ____________________ 






8 

216 in. 2 L.A. + 2 B B h 

216 + 2 (4 3 ) 4 3 (18) 

216 + 8 3 in. 2 72 3 in. 3 

S.A. = 229.86 in. 2 V = 124.71 in. 3

Name __________________________________________ 




Worksheet 4 







B = ____________________ 



L.A. = ____________________ 






9

Name __________________________________________ 




Worksheet 5 



5 cm 





B = ____________________ 



L.A. = ____________________ 






10

Name __________________________________________ 




Worksheet 5: Answer Key (difficulty - easy) 


Cube 


5 cm 

L.A. + 2 B 


B h 


square 



B = S 2 

B = 5 2 

B = 25 

25 cm 2 square 

B = ____________________ 







P = 4 S 

L.A. = P h 

P = 4 (5) L.A. = 20 (5) 

P = 20 L.A. = 100 

100 cm 2 L.A. + 2 B B h 

L.A. = ____________________ 





100 + 2 (25) 25 (5) 

S.A. = 100 + 50 

150 cm 2 125 cm 3 


11

Name __________________________________________ 




Worksheet 6 







B = ____________________ 



L.A. = ____________________ 






12

Name __________________________________________ 




Worksheet 7 



3 m 



4 m 

15 m 



B = ____________________ 



L.A. = ____________________ 






13

Name __________________________________________ 




Worksheet 7: Answer Key (difficulty - easy) 


Rectangular Prism 


L.A. + 2 B 


B h 


4 m 

15 m 

3 m 

rectangle 



B = l w 

B = 15(4) 

B = 60 

Note: Area of the base, B, could also be 12 m or 45 m. 

Teachers will need to specify which base to select 

in a rectangular prism. The surface area and 

volume results will not be affected. 

60 m 2 rectangle 

B = ____________________ 







P = 2 l+ 2 w 

L.A. = P h 

P = 2 (15) + 2 (4) L.A. = 38 (3) 

P = 30 + 8 L.A. = 114 

P = 38 

114 m 2 L.A. + 2 B B h 

L.A. = ____________________ 





(Show all algebraic steps.) S.A. = __________________________ 234 m 2 V = __________________________ 

180 m 3 

14 

114 + 2 (60) 60 (3) 

S.A. = 114 + 120

Name __________________________________________ 




Worksheet 8 







B = ____________________ 



L.A. = ____________________ 






15

Name __________________________________________ 







Worksheet 9 

regular pentagon 

base edge = 2 cm 

apothem = 1.376 cm 

13 cm 




B = ____________________ 



L.A. = ____________________ 






16

Name __________________________________________ 




Pentagonal Prism 


L.A. + 2 B 



Worksheet 9 Answer Key: (difficulty - hard) 

regular pentagon 

base edge = 2 cm 

apothem = 1.376 cm 

13 cm 

B h 


pentagon 





perimeter of the base, P. area of the base, B. 

P = 5 S 

B = 1 / 2 Pa 

P = 5 (2) B = 1 / 2 (10)(1.376) 

P = 10 B = 6.88 

6.88 cm 2 rectangle 

B = ____________________ 



L.A. = P h 

L.A. = 10(13) 

L.A. = 130 

130 cm 2 L.A. + 2 B B h 

L.A. = ____________________ 




130 + 2 (6.88) 6.88 (13) 

S.A. = 130 + 13.76 


(Show all algebraic steps.) S.A. = __________________________ 143.76 cm 2 V = __________________________ 

89.44 cm 3 

17

Name __________________________________________ 




Worksheet 10 







B = ____________________ 



L.A. = ____________________ 






18

Name __________________________________________ 





Worksheet 11 

regular hexagon 

base edge = 10 ft 







B = ____________________ 



L.A. = ____________________ 






19

Name __________________________________________ 




Hexagonal Prism 



Worksheet 11: Answer Key (difficulty - hard) 

regular hexagon 

base edge = 10 ft 






B = ____________________ 

L.A. + 2 B 

B h 

hexagon 

First, I will find the Second, I will find the Now I will find the 

perimeter of the base, P. apothem of the base. area of the base B. 

P = 6 S 1 /2 the base edge of 10 B = 1 / 2 Pa 

P = 6 (10) is 5, which is the short B = 1 / 2 (60)(5 3 ) 

P = 60 leg of a special 30˚-60˚-90˚ B = 150 3 

triangle. So the apothem, 

long leg, is 5 3. 



L.A. = P h 

L.A. = 60(40) 

L.A. = 2,400 

5 3 

150 3 ft 2 60 

5 

rectangle 

30 

2,400 ft 2 L.A. + 2 B B h 

L.A. = ____________________ 





(Show all algebraic steps.) S.A. = __________________________ 2,400 + 300 3 ft 2 V = __________________________ 

6,000 3 ft 3 

20 

2,400 + 2 (150 3 ) 150 3 (40) 

S.A. = 2,919.62 ft 2 V = 10, 392.30 ft 3

Name __________________________________________ 




Worksheet 12 







B = ____________________ 



L.A. = ____________________ 






21

Name __________________________________________ 

6 ft 

2 ft 

14 ft 


Worksheet 13 

Surface Area 

Challenge 

Find the surface area of the right solid. Describe 

your process through a series of clearly shown 

mathematical steps. 

9 ft 

10 ft 

Solution: ______________________________ 

22

Name __________________________________________ 

2 ft 


Worksheet 13: Answer Key 

Surface Area 

Challenge 

6 ft 

14 ft Find the surface area of the right solid. Describe 



9 ft 

10 ft 

Step 1: I will find the surface area of the large rectangular prism less the surface area of the 2 x 9 area. 

First, I will find the perimeter Second, I will find the Third, I will find the 

of the base, P, using the formula lateral area, L.A. area of the base, B, 

P = 2 l + 2 w L.A. = P h using the formula 

P = 2 (10) + 2 (9) L.A. = 38 (6) B = l w 

P = 20 + 18 L.A. = 228 B = 10 (9) 

P = 38 B = 90 

Fourth, total surface area of Finally, I will subtract 18 

the large rectangular prism is (the area of the 2 x 9 area) 

S.A. = L.A. + 2 B 

because it is hidden by the 

S.A. = 228 + 2 (90) 

top rectangular prism. 

S.A. = 228 + 180 S.A. = 408 - 18 

S.A. = 408 S.A. = 390 

Step 2: I will find the surface area of the small rectangular prism less the surface area of the 2 x 9 area. 

I will find the perimeter of the Second, I will find the Third, I will find the 

base, P, using the formula lateral area, L.A. area of the base, B, 

P = 2 l + 2 w L.A. = P h using the formula 

P = 2 (2) + 2 (9) L.A. = 22 (8) B = l w 

P = 4 + 18 L.A. = 176 B = 2 (9) 

P = 22 B = 18 

Fourth, total surface area of Finally, I will subtract 18 (the area of the 2 x 9 area) 

the small rectangular prism is because it is hidden by the bottom rectangular prism. 

S.A. = L.A. + 2 B S.A. = 212 - 18 

S.A. = 176 + 2 (18) S.A. = 194 

S.A. = 176 + 36 

S.A. = 212 

Step 3: The surface area of large rectangular prism + surface area of small rectangular prism will 

equal the total surface area. 

S.A. = 390 + 194 

S.A. = 584 ft 2 

Solution: ______________________________ 

584 ft 2 

23

Name __________________________________________ 

8 in. 8 in. 

9 in. 

9 in. 

10 in. 

12 in. 


Worksheet 14 

Volume 

Challenge 




24 

Solution: ______________________________

Name __________________________________________ 

8 in. 8 in. 

9 in. 

9 in. 

10 in. 

12 in. 



Volume 

Challenge 




Step 1: I will find the volume of the square pyramid. You will need to reference the first book entitled, 

Getting Into Solids — Pyramids. 

First, I will find the area of Second, I will use the formula 

the base, B. 

for the volume of a regular pyramid. 

B = S 2 

V = 1 / 3 B h 

B = 9 2 V = 1 / 3 (81)(10) 

B = 81 V = 270 

Step 2: I will find the volume of the square prism. 

Note: Since the base of the Now I will find the volume of the 

square pyramid is exactly square prism using the formula 

the same as the base of the V = B h 

square prism, the area of the V = 81 (12) 

bases will be the same. V = 972 

Therefore, B = 81 

Step 3: I will now add the volume of the square pyramid to the volume of the rectangular prism to 

find the total volume. 

V = 270 + 972 

V = 1,242 in. 3 V = 1,242 in. 3 

Solution: ______________________________ 

25

Name __________________________________________ 


Worksheet 15 

11.2 cm 

3.5 cm 

4 cm 

8.4 cm 

Surface Area 

Challenge 

Your company, Premium Sauce, is going to purchase one 

of the two 16-ounce cans shown and fill them with your 

secret tomato sauce products. Your task is to determine 

which can will minimize the paper that will go 

around the can to advertise the secret sauce. Write a 

memo to the advertising department informing them of 

the dimensions. Describe your process through a series 

of clearly shown mathematical steps. 

Solution: ______________________________ 

26

Name __________________________________________ 



11.2 cm 

3.5 cm 

4 cm 

8.4 cm 

Surface Area 

Challenge 

Your company, Premium Sauce, is going to purchase one 

of the two 16 ounce cans shown and fill them with your 

secret tomato sauce products. Your task is to determine 

which can will minimize the paper that will go 

around the can to advertise the secret sauce. Write a 

memo to the advertising department informing them of 

the dimensions. Describe your process through a series 

of clearly shown mathematical steps. 

Step 1: I will find the lateral area of the tall right cylinder. 

L.A. = 2 π r h 

L.A. = 2 π (3.5)(11.2) 

L.A. = 78.4 π 

L.A. = 246.301 cm 2 

Step 2: I will find the lateral area of the short right cylinder. 

L.A. = 2 π r h 

L.A. = 2 π (4)(8.4) 

L.A. = 67.2 π 

L.A. = 211.115 cm 2 

Step 3: The lateral area that yields the minimum paper is the short right cylinder. There will be a 

savings of 35.186 cm 2 of paper for each can. 

Step 4: I will now find the dimensions of the lateral area, which is a rectangle. The circumference of 

the circle will be the length. I will find circumference using the formula 

C = 2 π r 

C = 2 π (4) 

C = 8 π 

C = 25.133 

Step 5: The dimensions of the rectangle will be circumference x height. The dimensions are 

25.1 cm x 8.4 cm 

(The memo written to the advertising department will vary. However, the dimensions should be 

the same.) 

25.1 cm x 8.4 cm 

Solution: ______________________________ 

27

Name __________________________________________ 

21 in. 

8 in. 

9 in. 

9 in. 


Worksheet 16 

Volume 

Challenge 

Find the volume of the cylinder if the two cones do not 

contain any substance. Describe your process through a 

series of clearly shown mathematical steps. 

28 

Solution: ______________________________

Name __________________________________________ 

21 in. 

8 in. 

9 in. 

9 in. 



Volume 

Challenge 

Find the volume of the cylinder if the two cones do not 

contain any substance. Describe your process through a 


Step 1: I will find the volume of the two right cones. You will need to reference the first book entitled, 

Getting Into Solids — Pyramids. 

First, I will find the area of the 

base, B, of the cone, which is 

a circle. 

B = π r 2 

B = π (8) 2 

B = 64 π 

Second, I will find the volume of 

the right cone using the formula 

V = 1 / 3 B h 

V = 1 / 3 (64 π)(9) 

V = 192 π 

Third, since the two right cones are the same, I will double the volume to find the total volume of 

both cones. 

V of both cones = 192 π(2) 

V of both cones = 384 π 

Step 2: I will find the volume of the right cylinder. 

Note: Since the base of the right Now I will find the volume of the 

cylinder is exactly the same as the right cylinder using the formula, 

base of the right cones, the area V = B h 

of the bases will be the same. V = 64π (21) 

Therefore, B = 64 π 

V = 1,344 π 

Step 3: I will find the volume of the solid by subtracting the volume of the two right cones from the 

volume of the right cylinder. 

V = volume of the cylinder - volume of the 2 cones 

V = 1,344 π - 384 π 

V = 960 π in. 3 3,015.93 in. 3 

960 π in. 3 (or) 

Solution: ______________________________ 

29

Name __________________________________________ 


Worksheet 17 

20 cm 

20 cm 

10 cm 

Surface Area 

Challenge 

Find the surface area of the rectangular prism with a 

10 cm x 12 cm hole through the center. Describe 



Solution: ______________________________ 

30

Name __________________________________________ 

20 cm 

20 cm 

10 cm 



Surface Area 

Challenge 

Find the surface area of the rectangular prism with a 

10 cm x 12 cm hole through the center. Describe 



Step 1: I will find the surface area of the large rectangular prism and subtract the area of the two 

10 cm x 12 cm regions. 

First, I will find the Second, I will find the L.A. Third, I will find the area 

perimeter of the base, P. L.A. = P h of the base, B. 

P = 2 w + 2 l L.A. = 80(10) B = l w 

P = 2(20) + 2(20) L.A. = 800 B = 20(20) 

P = 80 B = 400 

Fourth, I will calculate Fifth, I will find the area Finally, the surface area of 

the surface area of the of the two open regions that the outside region of the large 

rectangular prism using 10 cm x 12 cm. rectangular prism is the S.A. less 

the S.A. formula. A = 2(lw) the area of the two open regions. 

S.A. = L.A. + 2 B A = 2(10)(12) S.A. = 1,600 - 240 

S.A. = 800 + 2(400) A = 240 S.A. = 1,360 

S.A. = 1,600 

Step 2: I will calculate the lateral area of the inside rectangular prism with 10 cm x 12 cm dimensions. 

First, I will find the perimeter of Second, I will find the lateral area, L.A. 

the base, P, of the 12 x 10 region. L.A. = P h 

P = 2 l + 2 w L.A. = 44(10) 

P = 2(12) + 2(10) L.A. = 440 

P = 44 

Step 3: Iʼll combine the S.A. of the outside region plus the L.A. of the inside region. 

S.A. = Surface area of the outside region + inside lateral area 

S.A. = 1,360 + 440 

S.A. = 1,800 cm 2 1,800 cm 2 

Solution: ______________________________ 

31

Name __________________________________________ 


Worksheet 18 

32 in. 

Volume 

Challenge 

Find the volume of the rectangular prism with a 

cylindrical hole through the prism. The diameter of the 

cylinder is 10 inches. Describe your process through a 


25 in. 

29 in. 

Solution: ______________________________ 

32

Name __________________________________________ 

32 in. 



Volume 

Challenge 

Find the volume of the rectangular prism with a 

cylindrical hole through the prism. The diameter of the 

cylinder is 10 inches. Describe your process through a 


25 in. 

29 in. 

Step 1: I will find the volume of the rectangular prism. 

First, I will find the area Second, I will find the volume 

of the base, B. 

of the rectangular prism. 

B = l w 

V = B h 

B = 25(29) V = 725 (32) 

B = 725 V = 23,200 

Step 2: I will find the volume of the cylinder with a diameter of 10. 

First, I will find the radius Second, I will find the volume 

of the circle, r. 

of the right cylinder. 

d = 2 r 

V = π r 2 h 

10 = 2 r V = π (5) 2 (32) 

5 = r V = 800 π (or 2,513.274) 

Step 3: I will subtract the volume of the right cylinder from the volume of the rectangular prism. 

V = Volume of rectangular prism - volume of the right cylinder 

V = 23,200 - 800 π in. 3 

V = 20,686.726 in. 3 

V = 20,686.726 in. 3 

Solution: ______________________________ 

33

Getting into Solids: Pyramids - ETA hand2mind

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