SAT — Volumes and 3D Shapes

Topic Overview & Frequency

The SAT Math section tests volumes and 3D shapes in 2-4 questions per test. Most questions plug directly into a formula provided on the SAT formula sheet — so this is among the most reliable scoring opportunities.

The digital SAT shows the formula sheet at the start of each module. Students who know how to use the formulas (rather than memorise them) tend to score highest.

Key Concepts You Must Know

Prioritized by SAT frequency:

Volume formulas for cube, rectangular box, cylinder, cone, sphere, pyramid.
Surface area of cube, rectangular box, cylinder.
Composite shapes — combining two or more 3D figures.
Cross-sections — what shape do you get cutting a cylinder/cone/sphere?
Inscribed/circumscribed — sphere in cube, cube in sphere, etc.
Density and volume — mass = density × volume problems.

Important Formulas (from SAT Formula Sheet)

Rectangular box: $V = lwh$ Cube: $V = s^3$ Cylinder: $V = \pi r^2 h$ Sphere: $V = (4/3)\pi r^3$ Cone: $V = (1/3)\pi r^2 h$ Pyramid: $V = (1/3)Bh$ ( $B$ = base area)

Cube: $SA = 6s^2$ Rectangular box: $SA = 2(lw + lh + wh)$ Cylinder: $SA = 2\pi r^2 + 2\pi r h$ Sphere: $SA = 4\pi r^2$

Solved Examples

Example 1 — Direct Volume Calculation

A cylindrical water tank has radius $3 \, \text{m}$ and height $4 \, \text{m}$ . What is its volume?

Solution: $V = \pi r^2 h = \pi(9)(4) = 36\pi \, \text{m}^3 \approx 113.1 \, \text{m}^3$ .

Example 2 — Composite Shape

A cylinder of radius $5 \, \text{cm}$ and height $10 \, \text{cm}$ has a hemisphere of radius $5 \, \text{cm}$ on top. Find the total volume.

Solution: Cylinder: $V_c = \pi(25)(10) = 250\pi$ . Hemisphere: $V_h = (1/2)(4/3)\pi(125) = (2/3)\pi(125) = 250\pi/3$ . Total: $250\pi + 250\pi/3 = 750\pi/3 + 250\pi/3 = 1000\pi/3 \approx 1047 \, \text{cm}^3$ .

Example 3 — Cross-Section Reasoning

If a cone with vertex pointing down is cut by a horizontal plane halfway up, what fraction of the cone’s volume is below the cut?

Solution: A cone of full height $h$ has volume $V = (1/3)\pi r^2 h$ . The “smaller cone” formed below (with vertex at the original vertex) is similar to the original. If we cut at half height, the smaller cone has half the height and half the radius (similar triangles).

Smaller volume: $(1/3)\pi (r/2)^2 (h/2) = V/8$ .

So 1/8 of the cone is below the cut. (The remaining 7/8 is the frustum on top.)

Example 4 — Density Problem

A solid sphere of radius $2 \, \text{cm}$ has density $7.8 \, \text{g/cm}^3$ . Find its mass.

Solution: Volume: $V = (4/3)\pi(8) = 32\pi/3 \, \text{cm}^3 \approx 33.5 \, \text{cm}^3$ . Mass: $m = \rho V = 7.8 \times 33.5 \approx 261 \, \text{g}$ .

Strategy Tips

Tip 1: Always check the formula sheet — but use it efficiently. Know which formulas you need before looking; saves seconds.

Tip 2: For composite shapes, break into known pieces. Add or subtract volumes carefully.

Tip 3: Similar 3D figures: if linear scale factor is $k$ , volume scales as $k^3$ , surface area as $k^2$ . This shortcut handles many “what fraction” problems instantly.

In the digital SAT, volumes/3D problems often pair with word problems (filling tanks, manufacturing parts). Read carefully — extract the geometric scenario and plug into formulas.

Common Traps

Trap 1: Forgetting the $1/3$ factor for cones and pyramids. Cylinders and prisms don’t have it; cones and pyramids do.

Trap 2: Confusing radius and diameter. The formula uses radius — if given diameter, halve it first.

Trap 3: For composite shapes, double-counting overlapping regions. Always check whether shapes share a base (which doesn’t add to total volume).

Trap 4: Mixing up linear and squared/cubed scaling. If a sphere’s radius doubles, volume increases by $2^3 = 8$ times, NOT by 2 times.

This topic is among the lowest-effort, highest-reward areas of SAT Math. Memorise the 6 main volume formulas and practice 20-30 problems — that locks in 2-4 marks every test.