3-D Volume

Master the formulas for calculating the volume of prisms, cylinders, pyramids, cones, and spheres: the essential skill for SSAT quantitative success.

SSAT Upper Level: Quantitative

What Is Volume?

Volume measures the amount of three-dimensional space enclosed by a solid figure. It answers the question: "How much stuff fits inside?"

Two Families of Solids

Every 3-D solid on the SSAT falls into one of two families, each with its own volume logic.

Prism-Type

Same cross-section stacked along a height: rectangular prisms, cubes, cylinders, triangular prisms. Formula pattern: \(V = B \cdot h\).

Pointed-Type

Taper to a point or curve inward: pyramids, cones, spheres. These use a fraction: \(\frac{1}{3}\) or \(\frac{4}{3}\).

Why Volume Matters on the SSAT

Volume questions appear in both quantitative sections of the SSAT Upper Level. They test whether you can:

1

Recall the correct formula for each solid shape.

2

Substitute accurately and handle \(\pi\), fractions, and exponents.

3

Solve multi-step problems such as finding dimensions first, then computing volume.

4

Compare volumes of different solids or determine how changing a dimension affects the total.

Prisms and Cylinders

Pyramids and Cones

The Sphere

Prism-Type vs. Pointed-Type

	Prism-Type	Pointed-Type
Shape	Uniform cross-section	Tapers to a point
Formula Pattern	\(V = Bh\)	\(V = \frac{1}{3}Bh\)
Examples	Box, Cylinder, Triangular Prism	Pyramid, Cone
Relationship	Pointed = \(\frac{1}{3}\) of Prism with same base and height

Step-by-Step Approach

1

Identify the shape. Is it a prism, cylinder, pyramid, cone, or sphere?

2

Write the formula. Recall the correct volume equation for that shape.

3

Find every measurement. Extract \(l\), \(w\), \(h\), or \(r\) from the problem. Watch for diameter vs. radius.

4

Substitute and compute. Plug values in, square or cube where needed, then multiply.

5

Check units. Your answer must be in cubic units.

Example 1: Rectangular Prism

A box has length \(8\) cm, width \(5\) cm, and height \(3\) cm. Find its volume.

Example 2: Cylinder

A cylinder has diameter \(10\) in and height \(7\) in. Find the volume in terms of \(\pi\).

Example 3: Cone

A cone has radius \(6\) cm and height \(10\) cm. Find the volume.

Example 4: Sphere

A basketball has a diameter of \(24\) cm. Find its volume to the nearest whole number.

Common Mistakes

✕ Diameter vs. Radius

Using the diameter where the formula requires the radius. Always divide diameter by 2 first.

✕ Forgetting \(\frac{1}{3}\)

Computing \(Bh\) for a cone or pyramid without the one-third factor. This triples your answer.

✕ Squaring vs. Cubing

Using \(r^2\) in the sphere formula instead of \(r^3\). Spheres cube the radius.

Volume Formula Reference

Shape	Formula	Variables
Rectangular Prism	\(V = lwh\)	\(l, w, h\)
Cube	\(V = s^3\)	\(s =\) side
Cylinder	\(V = \pi r^2 h\)	\(r, h\)
Pyramid	\(V = \frac{1}{3}Bh\)	\(B =\) base area, \(h\)
Cone	\(V = \frac{1}{3}\pi r^2 h\)	\(r, h\)
Sphere	\(V = \frac{4}{3}\pi r^3\)	\(r\)

Question 1 of 5

A rectangular box has dimensions \(4 \times 6 \times 9\) inches. What is its volume?

\(V = lwh = 4 \times 6 \times 9 = 216 \text{ in}^3\). Multiply all three dimensions together.

Question 2 of 5

A cylinder has a diameter of \(14\) cm and a height of \(10\) cm. What is its volume in terms of \(\pi\)?

The radius is \(14 \div 2 = 7\). Then \(V = \pi r^2 h = \pi(7)^2(10) = 490\pi \text{ cm}^3\). Don't forget to halve the diameter!

Question 3 of 5

A cone has a radius of \(3\) m and a height of \(12\) m. What is its volume?

\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(3)^2(12) = \frac{1}{3}\pi(108) = 36\pi \text{ m}^3\). Remember the \(\frac{1}{3}\) factor for cones.

Question 4 of 5

A sphere has a radius of \(6\) cm. What is its volume?

\(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi \text{ cm}^3\). Cube the radius, not square it!

Question 5 of 5

A square pyramid has a base edge of \(10\) ft and a height of \(15\) ft. What is its volume?

Base area \(B = 10^2 = 100 \text{ ft}^2\). Then \(V = \frac{1}{3}Bh = \frac{1}{3}(100)(15) = 500 \text{ ft}^3\). Choice (a) is the common mistake: forgetting the \(\frac{1}{3}\).