8.1 Area, Volume, and Valuation Calculations
Key Takeaways
- Area of a rectangle is length times width; convert all measurements to the same unit before multiplying.
- One acre equals 43,560 square feet, the single most-tested conversion on the national exam.
- Volume equals length times width times height, used for cubic-yard concrete and storage problems.
- The cost approach values property as land value plus replacement cost of improvements minus accrued depreciation.
- Straight-line depreciation spreads cost evenly over the economic life: annual depreciation equals cost divided by life in years.
Area and the acre
Area problems are the foundation of real-estate math. For a rectangle, Area = Length × Width, and the answer is always in square units (square feet, square yards). The single most common error is mixing units, so convert everything to feet first.
The conversion you must memorize: 1 acre = 43,560 square feet. Examiners hide this number inside lot-size problems. If a rectangular lot measures 200 ft by 300 ft, the area is 60,000 sq ft, which is 60,000 ÷ 43,560 ≈ 1.38 acres.
Triangles and irregular lots
For a triangle, Area = ½ × Base × Height. Irregular lots are split into a rectangle plus a triangle, and you add the two areas. A lot that is a 100 ft × 80 ft rectangle with a triangular front (base 100 ft, height 20 ft) totals 8,000 + (½ × 100 × 20) = 8,000 + 1,000 = 9,000 sq ft.
Watch the trap: the height of a triangle is the perpendicular distance, not the slanted side length. Exam writers supply an extra slant figure to lure you.
Common conversions table
| Measurement | Equals |
|---|---|
| 1 acre | 43,560 sq ft |
| 1 square yard | 9 sq ft |
| 1 cubic yard | 27 cu ft |
| 1 mile | 5,280 ft |
| 1 township | 36 sq miles (36 sections) |
| 1 section | 640 acres |
The section = 640 acres fact links to government-survey questions: a quarter-quarter (¼ of ¼) of a section is 640 ÷ 16 = 40 acres.
Volume
Volume answers "how much space" and uses three dimensions: Volume = Length × Width × Height, expressed in cubic units. A warehouse 50 ft long, 30 ft wide, and 12 ft high holds 50 × 30 × 12 = 18,000 cubic feet.
Concrete and fill problems ask for cubic yards, so divide cubic feet by 27. A driveway slab 30 ft × 12 ft × 0.5 ft = 180 cu ft, then 180 ÷ 27 ≈ 6.67 cubic yards of concrete.
Square footage and price-per-foot questions
Exam items often combine area with a unit price. To find a home's living area when given outside dimensions, multiply length by width per floor and add floors: a two-story home that is 40 ft x 30 ft per floor has 40 x 30 x 2 = 2,400 sq ft. At a list price of $360,000 that is $360,000 ÷ 2,400 = $150 per square foot — a figure used to compare listings and support sales-comparison adjustments.
Watch the trap where the problem gives lot dimensions but asks for building area, or mixes interior and exterior measurements. Always pair the number you compute with the unit the question actually requests (lot vs. living area, square feet vs. acres).
Reproduction vs. replacement and a unit-cost build-up
The cost approach can be built from a unit-in-place or square-foot cost. If a builder's cost is $140 per square foot for a 2,500-sq-ft house, the improvement's reproduction cost is 2,500 x $140 = $350,000. Subtract accrued depreciation, then add land value to reach the cost-approach indicator.
| Concept | Meaning |
|---|---|
| Reproduction cost | Exact replica, same materials and design |
| Replacement cost | Equivalent utility, modern materials |
| Effective age | Apparent age based on condition, not actual age |
Examiners exploit effective age: a well-maintained 30-year-old home may have an effective age of 15, lowering its accrued depreciation and raising value above what its chronological age suggests.
Multi-step area-to-cost problem
Real exam items chain area into a dollar answer. Suppose a developer buys a rectangular tract measuring 660 ft by 660 ft and must know both its acreage and a per-acre price. Area = 660 x 660 = 435,600 sq ft, and 435,600 ÷ 43,560 = exactly 10 acres. If the tract sold for $850,000, the per-acre price is $850,000 ÷ 10 = $85,000 per acre.
Now layer a build cost: the developer plans 4 homes averaging 2,000 sq ft at $135 per square foot. Total construction = 4 x 2,000 x $135 = $1,080,000. Adding the $850,000 land gives a project cost of $1,930,000 before soft costs. Item writers test whether you keep land and improvement costs separate — land is added at full value and never depreciated, while only the $1,080,000 of improvements is subject to later depreciation.
A rectangular parcel measures 435.6 feet by 500 feet. How many acres does it contain?
Valuation: the cost approach
The cost approach estimates value as: Land Value + Replacement (or Reproduction) Cost of Improvements − Accrued Depreciation. It is most reliable for new or special-purpose buildings (schools, churches) that rarely sell.
Replacement cost reproduces equivalent utility with modern materials; reproduction cost duplicates the exact structure. Land is valued separately because land does not depreciate — only improvements do.
Straight-line depreciation
The exam uses straight-line ("age-life") depreciation: spread the improvement cost evenly across its economic life.
- Annual depreciation = Cost ÷ Economic life (years)
- Accrued depreciation = Annual depreciation × Effective age
Example: a building cost $300,000 with a 40-year life. Annual depreciation = $300,000 ÷ 40 = $7,500/year. After 10 years, accrued depreciation = $7,500 × 10 = $75,000, so the depreciated improvement value is $225,000. Add $80,000 land and the cost-approach value is $305,000.
Trap: depreciation applies only to the improvement cost, never to land. If a problem gives a combined figure, subtract land first.
A building cost $400,000 to construct and has an economic life of 50 years. Using straight-line depreciation, what is the accrued depreciation after 8 years?