Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Learning Math Home
Measurement Session 10, Grades 6-8: Classroom Case Studies
Session 10 Session 10 6-8 Part A Part B Part C Homework
measurement Site Map
Session 10 Materials:

A B C 


Solutions for Session 10, Part C

See solutions for Problems: C1 | C2 | C3

Problem C1

With each rotation, the tire covers the distance of its circumference. So, the circumference of one 26-inch tire rotation = 26 = approx. 81.68 inches = approx. 7 feet. 5,280 feet per mile 7 feet per tire rotation = approx. 754 pedal turns per mile.

 Answers to Questions:


The measurement ideas addressed by this problem include determining circumference; solving problems involving ratio and proportion; converting units; and applying appropriate techniques, tools, and formulas to determine measurements.


Working on this problem, students are likely to extend their understanding of circumference as well as ratio and proportion through problem solving. The problem prepares students for rate problems and other algebraic concepts like solving equations.


This problem draws upon the work that was done on exploring circles, in particular the relationship of diameter to circumference. This problem poses a situation where such prior knowledge is applied to solve a practical problem.


You can play with the proportional aspect of this problem by giving students different information, such as distance traveled and number of pedal turns, and asking students to determine tire size. You might also ask students to consider what tire size would be needed in order to reduce the number of pedal turns needed to ride 1 mile by 10% or 25%.


You can make this lesson concrete for students by having them ride a stationary bike, determine the number of tire rotations per pedal and length of rotation, and then answer the question about number of pedal turns per mile. This problem could also serve as a springboard for scale problems involving area of circles.

<< back to Problem C1


Problem C2


Scale Factor


Surface Area



1 by 1 by 1

6 square units (un2)

1 cubic unit (un3)


2 by 2 by 2

22 • 6 = 24 un2

23 = 8 un3


3 by 3 by 3

32 • 6 = 54 un2

33 = 27 un3


4 by 4 by 4

42 • 6 = 96 un2

43 = 64 un3


5 by 5 by 5

52 • 6 = 150 un2

53 = 125 un3


10 by 10 by 10

102 • 6 = 600 un2

103 = 1,000 un3


25 by 25 by 25

252 • 6 = 3,750 un2

253 = 15,625 un3

The surface area of a cube is increased by the scale factor squared. The volume is increased by the scale factor cubed.

 Answers to Questions:


Students working on this problem are likely to deepen their understanding of measurable attributes of a cube (dimension length, surface area, and volume) as well as the relationship between the attributes. The big idea for students to consider is how the surface area and volume grow when a cube is increased (or decreased) by a certain scale factor.


The problem extends and deepens understanding of surface area and volume of cubes, perfect squares, cubed numbers, and patterns. It prepares students for continued work with exponents.


This problem also draws on the work that was done in Session 8. In that session, the participants explored how each of the dimensions change based on the scale factor. As a result, this will affect how surface area and volume change for scaled cubes.


Have students use manipulatives to build the cube and determine the dimensions, surface area, and volume. Explore how the volume also changes. If you know the volume of a cube, can you determine the dimensions and surface area? Why?


Students can begin to examine and compare the surface area-to-volume ratio for both rectangular prisms and cubes. They can also make comparisons with other shapes or objects, and explore which of them will have a maximum or minimum surface area-to-volume ratio.

<< back to Problem C2


Problem C3

Since Dave and Charlene are the same height, the 28-degree angle measures exactly the height of the cliff. Visualize closing off the angle to form a triangle and then sliding that triangle down to water level. The side opposite the 28-degree angle would align exactly with the cliff. Use similar triangles to determine the height of the cliff. Using a protractor, draw a right triangle with a 28-degree angle opposite the vertical leg forming the right angle. Measure the length of the two legs of the right triangle with a ruler. Then set up a proportion between those two sides and the height of the cliff and 50 m length in the surfing triangle. The solution should give you a cliff height of roughly 26.5 m.

 Answers to Questions:


The measurement ideas in this problem include using appropriate techniques to accurately measure angles and lengths and using ratio and proportion to determine length. Students will use similar triangles. Note: Some reasoning skills are necessary to understand that knowing that Charlene and Dave's heights are equal allows you to find the height of the cliff.


This problem builds on prior work with protractors as well as proportional reasoning. The problem provides an opportunity to apply knowledge of similar triangles.


In the course, teachers explored similar triangles to solve similar problems.


Students could be given the height of the cliff and then asked to determine the distance Charlene is from the cliff. They could also be asked to think about how changing the angle formed by the horizontal side and Charlene's line of sight affects the height of the cliff.


This problem can be followed with a series of "shadow" problems in which students have to determine the height of trees, buildings, or flagpoles by measuring the shadow of a person looking at the top of the structure.

<< back to Problem C3


Learning Math Home | Measurement Home | Glossary | Map | ©

Session 10, Grades 6-8 | Notes | Solutions | Video


© Annenberg Foundation 2017. All rights reserved. Legal Policy