Upon completion, have students discuss their solutions to Questions 24–26.
- Explain how you found the diameter when you knew the circumference of a bicycle wheel (Question 24). How did you know it was a reasonable answer? (Possible response: One turn of the bike wheel is the same as its circumference. The diameter is about 1/3 of that. 75 inches ÷ π = 23.9 inches. That makes sense because 24 × 3 = 72, which is close to 75.)
- Explain how you estimated the circumference of the sequoia tree (Question 25). Why is it only an estimate? (Possible response: It is only an estimate because fifth-graders have different arm spans. I multiplied the average arm span, 140 cm × 30 = 4200 cm, to estimate the distance around the tree.)
- Once you estimated the circumference, how did you estimate the diameter of the tree? (Possible response: I divided 4200 cm ÷ π on my calculator. The diameter is about 1337 cm.)
- What if someone got an answer of 13,194.7 cm? What do you think might of happened? (They might have multiplied 4200 ÷ π instead of divided it by π.
- How did you solve the multistep problem in Question 26? (Possible response: I knew one turn of the wheel would be the same as the circumference, so first I found the circumference by multiplying the diameter 64 × π = 201.1 cm. Then I made 201.1 cm into 2.011 meters so I could divide meters by meters. 3000 meters ÷ 2.011 meters = 1491.8 turns, about 1492 turns.)
Use students' responses to Check-In: Questions 20–26 in the Explore Circumference and Diameter pages the Student Guide and the corresponding Feedback Box to assess their abilities to representing the relationship between variables as ratios [E2]; find equivalent fractions and ratios using a variety of strategies [E3]; use ratios and proportions to solve problems [E4]; use patterns in tables and line graphs to make predictions and solve problems [E12]; know the problem [MPE1]; and use labels [MPE6].
