SAT Math – Geometry and Trigonometry
Area and Volume
Mastering formulas for 2D shapes and 3D solids
Area and volume questions test your ability to calculate measurements of two-dimensional shapes and three-dimensional solids. On the SAT, you'll apply formulas for rectangles, triangles, circles, trapezoids, cylinders, spheres, cones, and composite figures—often requiring decomposition of complex shapes into simpler components.
Success requires memorizing essential formulas, recognizing when to use Pythagorean theorem, understanding how scaling affects area and volume, and breaking down composite figures systematically. These calculations aren't just academic—they model real construction projects, container capacity, land measurement, material costs, and any situation requiring space quantification.
Essential Area and Volume Formulas
Two-Dimensional Areas
Area measures space inside flat shapes (square units).
Triangle: \(A = \frac{1}{2}bh\) (½ base × height)
Circle: \(A = \pi r^2\) (π × radius²)
Trapezoid: \(A = \frac{1}{2}(b_1 + b_2)h\) (average of bases × height)
Three-Dimensional Volumes
Volume measures space inside solids (cubic units).
Cylinder: \(V = \pi r^2 h\) (π × radius² × height)
Sphere: \(V = \frac{4}{3}\pi r^3\) (4/3 × π × radius³)
Cone: \(V = \frac{1}{3}\pi r^2 h\) (1/3 × π × radius² × height)
Surface Area
Surface area measures total outside covering (square units).
Cylinder: \(SA = 2\pi r^2 + 2\pi rh\) (two circles + lateral)
Sphere: \(SA = 4\pi r^2\)
Key: Add all face areas
Scaling Effects
Understand how dimensions affect area and volume.
Triple dimension: Area increases 9×, Volume increases 27×
General rule: Scale by k → Area by k², Volume by k³
Critical insight: Different rates for different measures
Formula Reference Guide
Circles
Circumference: \(C = 2\pi r\) or \(C = \pi d\)
Area: \(A = \pi r^2\)
Sector area: \(A = \frac{\theta}{360°} \times \pi r^2\) (θ in degrees)
Remember: Radius is half diameter
Triangles
Standard: \(A = \frac{1}{2}bh\)
Right triangle: \(A = \frac{1}{2}ab\) (legs as base and height)
Perimeter: Sum of all three sides
Key: Height must be perpendicular to base
Prisms and Cylinders
Volume pattern: Base area × height
Rectangular prism: \(V = lwh\)
Cylinder: \(V = \pi r^2 h\)
Cube: \(V = s^3\) (side³)
Pyramids and Cones
Volume pattern: 1/3 × Base area × height
Cone: \(V = \frac{1}{3}\pi r^2 h\)
Pyramid: \(V = \frac{1}{3}Bh\) (B = base area)
Remember: One-third of corresponding prism
Common Pitfalls & Expert Tips
❌ Using diameter instead of radius
Circle formulas use RADIUS, not diameter. If given d = 10, then r = 5. Don't forget to divide by 2!
❌ Forgetting the 1/2 in triangle area
\(A = \frac{1}{2}bh\), not just bh. Must multiply by one-half!
❌ Mixing up area and volume units
Area uses square units (ft²), volume uses cubic units (ft³). Check what the question asks for!
❌ Not recognizing composite figures
Complex shapes = sum of simpler shapes. Break them down into rectangles, triangles, circles you can calculate!
✓ Expert Tip: Draw and label diagrams
Visualize the problem! Draw the shape, mark dimensions, and identify which formula applies. Prevents errors!
✓ Expert Tip: Check unit consistency
All dimensions must use same units. If mixing feet and inches, convert first before calculating!
✓ Expert Tip: Remember the 1/3 rule
Cone/pyramid volume = 1/3 of cylinder/prism with same base and height. Easy way to remember formulas!
Fully Worked SAT-Style Examples
A circle has a diameter of 14 cm. What is its area?
Solution:
Find radius:
Diameter = 14 cm
Radius = 14 ÷ 2 = 7 cm
Apply area formula:
\(A = \pi r^2 = \pi(7)^2 = 49\pi\) cm²
Answer: \(49\pi\) cm² (or approximately 153.94 cm²)
A triangle has a base of 12 inches and a height of 8 inches. What is its area?
Solution:
Apply triangle area formula:
\(A = \frac{1}{2}bh = \frac{1}{2}(12)(8)\)
\(= \frac{96}{2} = 48\) square inches
Answer: 48 square inches
A box has dimensions 5 ft by 3 ft by 2 ft. What is its volume?
Solution:
Apply volume formula:
\(V = lwh = 5 \times 3 \times 2 = 30\) ft³
Answer: 30 cubic feet
A cylindrical tank has a radius of 4 meters and height of 10 meters. What is its volume?
Solution:
Apply cylinder volume formula:
\(V = \pi r^2 h = \pi(4)^2(10)\)
\(= \pi(16)(10) = 160\pi\) m³
Answer: \(160\pi\) m³ (or approximately 502.65 m³)
A sphere has a radius of 3 cm. What is its volume?
Solution:
Apply sphere volume formula:
\(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3)^3\)
\(= \frac{4}{3}\pi(27) = 36\pi\) cm³
Answer: \(36\pi\) cm³ (or approximately 113.10 cm³)
A cone has a base radius of 6 inches and height of 9 inches. What is its volume?
Solution:
Apply cone volume formula:
\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6)^2(9)\)
\(= \frac{1}{3}\pi(36)(9) = \frac{324\pi}{3} = 108\pi\) in³
Answer: \(108\pi\) in³ (or approximately 339.29 in³)
A shape consists of a rectangle (10 cm × 6 cm) with a semicircle on top (diameter 6 cm). What is the total area?
Solution:
Calculate rectangle area:
\(A_{rect} = 10 \times 6 = 60\) cm²
Calculate semicircle area:
Radius = 6 ÷ 2 = 3 cm
\(A_{semi} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(3)^2 = \frac{9\pi}{2}\) cm²
Add areas:
\(A_{total} = 60 + \frac{9\pi}{2} = 60 + 4.5\pi\) cm²
Answer: \(60 + 4.5\pi\) cm² (or approximately 74.14 cm²)
A cube has volume 27 cm³. If each dimension is doubled, what is the new volume?
Solution:
Find original side length:
\(s^3 = 27\) → \(s = 3\) cm
New side length:
New s = 2 × 3 = 6 cm
Calculate new volume:
\(V_{new} = 6^3 = 216\) cm³
Scaling principle:
Doubling dimensions multiplies volume by 2³ = 8
Check: 27 × 8 = 216 ✓
Answer: 216 cm³
Formula Quick Reference
2D Shapes
Rectangle: \(A = lw\)
Triangle: \(A = \frac{1}{2}bh\)
Circle: \(A = \pi r^2\)
Trapezoid: \(A = \frac{1}{2}(b_1+b_2)h\)
3D Solids
Prism: \(V = Bh\)
Cylinder: \(V = \pi r^2h\)
Sphere: \(V = \frac{4}{3}\pi r^3\)
Cone: \(V = \frac{1}{3}\pi r^2h\)
Area and Volume: Quantifying Space in Two and Three Dimensions
Area and volume calculations represent fundamental spatial reasoning skills essential across disciplines—from architecture to engineering, from land surveying to container design. The SAT tests these formulas because they're foundational to quantitative literacy: understanding that doubling dimensions quadruples area but octuples volume explains why larger vehicles are disproportionately inefficient, why small animals have higher surface-to-volume ratios affecting metabolism, and why packaging costs don't scale linearly with product size. Memorizing formulas is just the beginning—true mastery requires recognizing which formula applies, decomposing complex shapes into manageable components, maintaining unit consistency, and understanding relationships between linear dimensions and area or volume measures. The distinction between perimeter and area, between surface area and volume, reflects different measurement purposes: fencing requires perimeter, painting requires surface area, and filling requires volume. Composite figures demand strategic decomposition—recognizing that L-shapes combine rectangles, that rounded shapes involve circles or sectors, and that irregular solids may require subtraction of volumes. The scaling principle—that linear scaling by factor k multiplies area by k² and volume by k³—explains countless phenomena from why elephants can't jump like fleas to how heat loss scales with size. These calculations transcend classroom exercises, empowering you to estimate construction materials, compare container capacities, understand why larger organisms need specialized circulatory systems, calculate medication dosages that scale with body surface area, and recognize when spatial measurements matter in daily decisions from furniture shopping to gardening to meal planning.
