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Math problem set Grade 8 Math

Bramblewood Middle School Cistern: Cylinder Volume and Roof Runoff

Nine problems sizing a fictional school's cylindrical stormwater cistern against roof runoff, using the real cylinder-volume and roof-runoff formulas.

Objectives

  1. Students will calculate the volume of a cylindrical cistern in cubic feet and convert it to gallons.
  2. Students will apply the roof-runoff formula to calculate stormwater volume from a roof area and rainfall amount.
  3. Students will compare calculated volumes to determine whether a storage structure is adequately sized for a given storm.

Vocabulary

Cistern
A large tank that stores rainwater collected from a roof so it can be used later, usually for watering plants.

Also: cylinder, runoff, cubic foot

Teacher overview

Bramblewood Middle School (a fictional school used for this problem set) installed a cylindrical cistern to capture roof runoff instead of letting it wash straight into the storm drain. This set walks students through finding the cistern's volume, calculating how much runoff a storm produces, and comparing the two to see whether the cistern can hold an entire storm's worth of water. A common misconception in this math is calculating the cistern's volume in cubic feet and then comparing it directly to a runoff total already given in gallons, skipping the 7.48-gallons-per-cubic-foot conversion that both quantities need to share before they can be compared.

Bramblewood Middle School Cistern: Cylinder Volume and Roof Runoff — Grade 8 Math
Name: ____________________________ Date: ______________

Problems

Bramblewood's cistern is a cylinder 10 feet in diameter and 8 feet deep, sitting beneath a downspout from the school's 4,000-square-foot roof.

  1. Calculate the cistern's volume in cubic feet, using the formula Volume (cu ft) = 0.785 × Diameter² × depth.
  2. Convert your answer from problem 1 to gallons, using the conversion that one cubic foot holds 7.48 gallons.
  3. Using the roof-runoff formula (roof area in square feet × rainfall in inches × 0.623 = gallons of runoff), calculate how many gallons of runoff a 1.5-inch storm produces from the school's 4,000-square-foot roof.
  4. Compare your answers from problems 2 and 3. Will the empty cistern hold all of this storm's runoff? By how many gallons of spare capacity, or shortfall, does it miss or clear the total?
  5. Suppose the cistern starts this storm already 20% full from an earlier, smaller storm. How many gallons of empty capacity does it have available at the start of this new storm?
  6. Using your answer from problem 5, will the cistern hold all of this storm's runoff? By about how many gallons does it clear or fall short?
  7. The school is considering a second, identical cistern to catch any overflow. How many total gallons of storage would two cisterns the same size as problem 2's provide together?
  8. A contractor proposes a taller cistern with the same 10-foot diameter but a depth of 12 feet instead of 8 feet. Calculate its volume in gallons.
  9. What percent increase in storage capacity does the 12-foot-deep cistern (problem 8) provide compared to the original 8-foot-deep cistern (problem 2)? Round to the nearest whole percent.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 628 cu ft.
    Volume = 0.785 × 10² × 8 = 0.785 × 100 × 8 = 628 cubic feet.
  2. Answer: 4,697 gal.
    628 cu ft × 7.48 gal/cu ft = 4,697.44, which rounds to 4,697 gallons.
  3. Answer: 3,738 gal.
    4,000 sq ft × 1.5 in × 0.623 = 3,738 gallons of runoff. This is the rain that falls on the roof, the most that could reach the cistern; a real system collects somewhat less.
  4. Answer: clears it by about 959 gal.
    Comparing the cistern's 4,697 gallons of capacity to the storm's 3,738 gallons of runoff, the cistern has 4,697 − 3,738 = 959 gallons of spare capacity and does not overflow.
  5. Answer: 3,758 gal.
    If the cistern is 20% full, 80% of its capacity is empty: 0.80 × 4,697.44 = 3,757.95, which rounds to 3,758 gallons available.
  6. Answer: clears it by about 20 gal.
    Comparing the available 3,758 gallons to the storm's 3,738 gallons of runoff, the cistern still holds the full storm, but with only about 3,758 − 3,738 = 20 gallons of empty capacity left afterward, a much tighter margin than in problem 4.
  7. Answer: 9,395 gal.
    Two cisterns provide 2 × 4,697.44 = 9,394.88, which rounds to 9,395 gallons of combined storage.
  8. Answer: 7,046 gal.
    Volume = 0.785 × 10² × 12 = 0.785 × 100 × 12 = 942 cubic feet. Converting: 942 × 7.48 = 7,046.16, which rounds to 7,046 gallons.
  9. Answer: 50 percent.
    Comparing the exact volumes, (7,046.16 − 4,697.44) ÷ 4,697.44 × 100 = 2,348.72 ÷ 4,697.44 × 100 ≈ 50%. This matches the fact that increasing depth from 8 to 12 feet (a factor of 1.5) with the same diameter increases volume by exactly the same factor, since volume scales directly with depth when diameter stays fixed.

Differentiation

Support (below grade level)

Provide the cylinder volume formula pre-filled with the diameter and depth substituted in for problem 1, so students only need to complete the multiplication. Give a printed conversion card listing "1 cu ft = 7.48 gal" and the roof-runoff formula for reference throughout.

Extension (above grade level)

Ask students to find what depth (keeping the 10-foot diameter) the cistern would need to exactly match, not exceed, the 3,738-gallon runoff total from problem 3, working backward from the volume and conversion formulas.

ELL supports

Provide a labeled diagram of the cylinder showing diameter and depth as distinct measurements, along with a simple arrow diagram showing "roof → runoff → cistern" before problem 1. Allow students to show their work using the diagram and numbers rather than full sentences for problems 1 through 3.