Geometry Modeling and Multistep Measurement Problems

14.6: Geometry Modeling and Multistep Measurement Problems. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 6 of 63 parts Practice Flashcards

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Material needed for a region

Objective: Combine area with a per-area coverage rate.

Prerequisite skills: measurement formulas and equations. A coverage rate such as liters per square meter multiplies area to give material quantity. A stated number of coats multiplies the coverage again. If a package count is requested, round up only at the final stage. Common error: multiplying a coverage rate by perimeter. Restrictions and interpretation: Keep unit dimensions consistent and apply any count, packing, or feasibility restrictions after modeling.

Formulas

$material=area×rate×coats$

Worked examples

Try itA 26-by-4-meter wall needs two coats of paint. Each coat uses 0.5 liter per square meter. How many liters are needed?Show answer

Area=26×4=104 m². Each coat uses 104×0.5=52 liters. Two coats use 104 liters. Therefore the requested answer is 104.

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Material needed for a region

Objective: Combine area with a per-area coverage rate.

Formulas

$material=area×rate×coats$

Worked examples

Try itA 26-by-4-meter wall needs two coats of paint. Each coat uses 0.5 liter per square meter. How many liters are needed?Show answer

Area=26×4=104 m². Each coat uses 104×0.5=52 liters. Two coats use 104 liters. Therefore the requested answer is 104.

Volume difference

Objective: Model an added layer as a difference of volumes.

An increase in a container’s fill level adds base area times the height change when the base cross section is constant. Use the change rather than the final height if the question asks only about added material. Common error: using the final full volume as the added volume. Restrictions and interpretation: Keep unit dimensions consistent and apply any count, packing, or feasibility restrictions after modeling.

Formulas

ΔV=B·Δh for a constant-base prism

Worked examples

Try itA rectangular tank has base 26 by 3 meters. Water depth rises from 2 meters to 26 meters. What volume of water was added in cubic meters?Show answer

Base area=26×3=78 m². Depth change=26−2=24 m. Added volume=78×24=1872 m³. Therefore the requested answer is 1872.

Geometry with a cost constraint

Objective: Solve for a dimension using an area or volume constraint.

A multistep measurement model can combine material unit cost, total cost, and geometric formulas. Work backward from cost to area or volume, then isolate the requested dimension and verify units. Common error: dividing cost by length before accounting for the per-area price. Restrictions and interpretation: Keep unit dimensions consistent and apply any count, packing, or feasibility restrictions after modeling. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

cost=unit cost×area; width=area/length

Worked examples

Try itFlooring costs $5 per square meter. A rectangular room is 8 meters long and costs $1040 to cover, with no waste or fees. What is its width in meters?Show answer

Area=cost/rate=1040/5=208 m². Since area=8w, divide by length 8: width=26 m. Substitution gives 5×8×26=$1040. Therefore the requested answer is 26.

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