12.3 Volume of prisms, cylinders, pyramids, and cones

Volume Formulas and Calculations

Volume tells you how much three-dimensional space a solid occupies. For this section, you need to know the volume formulas for four solids: prisms, cylinders, pyramids, and cones. The big idea connecting all of them is that pyramids and cones are exactly one-third the volume of their corresponding prism or cylinder with the same base and height.

more resources to help you study

practice questions

Volume Formulas for Prisms and Cylinders

Both prisms and cylinders share the same core idea: they have a uniform cross-section from bottom to top. That means every horizontal "slice" through the solid looks identical. Because of this, the volume formula is the same concept for both.

Prism:

$V = Bh$

where $B$ is the area of the base and $h$ is the height (the perpendicular distance between the two bases). The base could be a triangle, rectangle, hexagon, or any polygon. Calculate that base area first, then multiply by height.

Cylinder:

$V = \pi r^2 h$

This is really the same formula as the prism. The base of a cylinder is a circle, so $B = \pi r^2$. Plug that in for $B$ and you get $V = \pi r^2 h$.

• $r$ = radius of the circular base
• $h$ = height of the cylinder (perpendicular distance between the two circular bases)

Quick example: A triangular prism has a base that's a right triangle with legs 5 cm and 12 cm, and the prism's height is 20 cm.

1. Find the base area: $B = \frac{1}{2}(5)(12) = 30$ cm²
2. Multiply by height: $V = 30 \times 20 = 600$ cm³

Volume Calculations for Pyramids and Cones

Pyramids and cones taper to a point, so they hold less material than a prism or cylinder with the same base and height. How much less? Exactly one-third.

Pyramid:

$V = \frac{1}{3}Bh$

• $B$ = area of the base (any polygon)
• $h$ = height measured perpendicular from the base to the apex

Cone:

$V = \frac{1}{3}\pi r^2 h$

• $r$ = radius of the circular base
• $h$ = perpendicular height from the base to the tip

Steps for calculating volume of a pyramid or cone:

1. Identify whether the base is a polygon (pyramid) or a circle (cone).
2. Calculate the base area $B$. For a cone, that's $\pi r^2$. For a pyramid, use the appropriate polygon area formula.
3. Make sure you're using the perpendicular height, not the slant height. If you're given slant height instead, you'll need the Pythagorean theorem to find $h$.
4. Plug into $V = \frac{1}{3}Bh$ and simplify.

Common mistake: Confusing height with slant height. The height $h$ goes straight down from the apex to the base at a right angle. The slant height runs along the lateral face. If a problem gives you slant height and the radius (or half the base edge), use $h = \sqrt{\ell^2 - r^2}$ to find the actual height.

Real-World Applications of Volume Formulas

When you encounter a word problem, follow this process:

1. Identify the shape. What solid does the object resemble? A silo is a cylinder, a tent might be a cone or pyramid, a box is a rectangular prism.
2. Extract the dimensions. Pull out the base measurements, height, and radius. Watch for diameter vs. radius (a very common trap).
3. Check your units. If the problem mixes units (inches and feet, cm and m), convert everything to the same unit before plugging into the formula.
4. Apply the formula and calculate.
5. Interpret and convert if the problem asks for a different unit (like liters instead of cubic meters).

Example: A cylindrical water tank has a diameter of 6 m and a height of 10 m. How many liters of water can it hold? (Use $1 \text{ m}^3 = 1000 \text{ L}$)

1. Shape: cylinder
2. Dimensions: diameter = 6 m, so radius = 3 m; height = 10 m
3. Apply the formula: $V = \pi (3)^2 (10) = 90\pi \approx 282.74 \text{ m}^3$
4. Convert: $282.74 \times 1000 \approx 282{,}740 \text{ L}$

That tank holds roughly 282,740 liters of water.

Volume Relationships Between Solid Shapes

This is one of the most important ideas in the unit. The $\frac{1}{3}$ factor isn't a coincidence; it reflects a precise geometric relationship.

• A pyramid has exactly $\frac{1}{3}$ the volume of a prism with the same base and height: $V_{\text{pyramid}} = \frac{1}{3} V_{\text{prism}}$
• A cone has exactly $\frac{1}{3}$ the volume of a cylinder with the same base and height: $V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}}$

You can verify this experimentally: if you fill a cone with water and pour it into a cylinder of the same base and height, it takes exactly three cones to fill the cylinder.

Concrete example: Suppose a cylinder and a cone both have a radius of 4 cm and a height of 9 cm.

• Cylinder: $V = \pi(4)^2(9) = 144\pi \approx 452.39 \text{ cm}^3$
• Cone: $V = \frac{1}{3}\pi(4)^2(9) = 48\pi \approx 150.80 \text{ cm}^3$

Notice that $150.80 \times 3 = 452.39$. The relationship holds exactly.