Math problem set Grade 10 Math

Brindlewick Center-Pivot Irrigation: Area, Volume, and Flow Rate

A grade 10 problem set using circle area and density to size a fictional farm's center-pivot irrigation, from inches of applied water to gallons and flow rate.

Grade: Grade 10
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSG-MG.A.1 Common Core Math

Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).

How this material aligns: Problems 1 and 9 model the irrigated field as a circle and use the pivot radius in the area formula, the real-object geometric modeling this standard describes.

All materials for HSG-MG.A.1 →

HSG-MG.A.2 Common Core Math

Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).

How this material aligns: Problem 7 applies a density of gallons of water applied per acre, using area and volume together, exactly the area-and-volume density reasoning this standard targets.

All materials for HSG-MG.A.2 →

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 3 through 6 carry inches of applied water through feet, cubic feet, gallons, and gallons per minute, choosing consistent units to size the pump flow.

All materials for HSN-Q.A.1 →

Download Word file

This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate the area a center-pivot system irrigates using the circle-area formula.
Students will convert a depth of applied water into a total volume in gallons.
Students will find the flow rate a pivot must deliver to finish a pass in a set time.

Vocabulary

Flow rate: Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: center pivot, acre, application depth, density

Teacher overview

A farm near the town of Brindlewick, a fictional setting used for this problem set, irrigates a field with a center-pivot system that sweeps a long arm in a circle. The problems build one scenario: students find the circular area the pivot waters, convert a depth of applied water into a total volume, and find the flow rate the system must deliver to finish a pass on schedule. A common misconception in this work is using the pivot's full length as a diameter instead of a radius in the area formula; another is multiplying by a depth in inches without first converting it to feet.

Brindlewick Center-Pivot Irrigation: Area, Volume, and Flow Rate — Grade 10 Math

Name: ____________________________ Date: ______________

Problems

A farm near the town of Brindlewick runs a center-pivot irrigation system. The pivot arm is 1,320 feet long, so it waters a circle with a radius of 1,320 feet. Use these facts for all problems, and use pi ≈ 3.14159:

Circle area = pi × radius².
1 acre = 43,560 square feet.
1 cubic foot holds 7.48 gallons.
A depth of water applied over an area gives a volume: area (square feet) × depth (feet) = cubic feet.

Find the area the pivot waters, in square feet, using a radius of 1,320 feet. Round to the nearest square foot.
Convert the pivot's area to acres. Round to the nearest hundredth.
The farmer applies 1.5 inches of water in one pass. Convert 1.5 inches to feet.
Find the volume of water applied in one pass, in cubic feet, by multiplying the area from item 1 by the depth from item 3. Round to the nearest cubic foot.
Convert the volume from item 4 to gallons. Round to the nearest gallon.
The pivot completes one full pass in 24 hours. Find the flow rate, in gallons per minute, the system must deliver to apply that volume in 24 hours. Round to the nearest tenth.
Find the water applied per acre in one pass, in gallons per acre, by dividing the total gallons from item 5 by the acres from item 2. Round to the nearest ten gallons.
The farm's well can supply 3,000 gallons per minute. Is that enough to run the pivot at the rate from item 6? If not, by how many gallons per minute does it fall short?
Explain, using the circle-area formula, how the watered area would change if the pivot arm were lengthened from 1,320 feet to 2,640 feet.
A center-pivot system waters a circle, leaving the corners of a square field dry. Explain one practical way a farmer can still make use of those dry corners.

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

Answer Key

Answer: 5,473,906 sq ft.
Area = pi × radius² = 3.14159 × 1,320² = 3.14159 × 1,742,400 = 5,473,906.4, which rounds to 5,473,906 square feet.
Answer: 125.66 acres.
5,473,906 square feet ÷ 43,560 square feet per acre = 125.663, which rounds to 125.66 acres.
Answer: 0.125 ft.
1.5 inches ÷ 12 inches per foot = 0.125 feet.
Answer: 684,238 cu ft.
Volume = area × depth = 5,473,906 × 0.125 = 684,238.25, which rounds to 684,238 cubic feet.
Answer: 5,118,100 gal.
684,238 cubic feet × 7.48 gallons per cubic foot = 5,118,100.24, which rounds to about 5,118,100 gallons.
Answer: 3,554.2 gpm.
One pass lasts 24 hours × 60 minutes = 1,440 minutes. 5,118,100 gallons ÷ 1,440 minutes = 3,554.24, which rounds to 3,554.2 gallons per minute.
Answer: 40,730 gal per acre.
5,118,100 gallons ÷ 125.66 acres = 40,729.7, which rounds to about 40,730 gallons per acre.
Answer: No, short by 554.2 gpm.
The pivot needs 3,554.2 gallons per minute (item 6), but the well supplies only 3,000. The shortfall is 3,554.2 − 3,000 = 554.2 gallons per minute, so the well cannot run the pivot at the 24-hour pace.
Answer: The area becomes four times as large.
Area depends on the square of the radius. Doubling the arm from 1,320 to 2,640 feet doubles the radius, and 2² = 4, so the watered area grows by a factor of 4 rather than 2.
Answer: Plant the corners in dryland or rain-fed crops.
Because the pivot cannot reach the square field's corners, a farmer often plants those corners with a crop that does not need irrigation, uses them for equipment access or storage, or leaves them as unplanted buffer strips.

Differentiation

Support

Provide the pivot's area (item 1) already calculated so students can begin at item 2, and give a printed reference card with the circle-area formula and the conversions for acres and gallons. Remind students that the pivot length is the radius, not the diameter.

Extension

Ask students to find how long, in hours, the well at 3,000 gallons per minute would need to apply the full one-pass volume from item 5, then compare that to the 24-hour schedule and explain what the farmer would have to change to stay on time.

ELL supports

Provide a top-view diagram of a square field with the pivot's circle drawn inside it, labeling the radius and shading the dry corners. Pre-teach "radius," "depth," and "flow rate" with the diagram, and give a sentence frame for item 10: "The farmer can use the dry corners to ___."