Volume of a Prism with Fractional Edge Lengths

Whole cubes stop fitting

Volume counts unit cubes. A box $4$ by $3$ by $2$ holds $24$ of them, and the formula counts the same cubes a layer at a time.

$$V = l \times w \times h$$

Now take a box $\tfrac{3}{2}$ by $\tfrac{1}{2}$ by $1$.

No whole unit cube fits. The width is half a unit, so any cube you place sticks out of the box. Counting whole cubes has run out.

The fix is to count in smaller cubes.

Cubes with unit-fraction edges

A cube whose edge is $\tfrac{1}{2}$ has volume:

$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

That matches the picture: eight of these half-cubes stack into one unit cube, two along each edge, so each is one eighth of it.

Cube edge	Its volume	Cubes per unit cube
$\tfrac{1}{2}$	$\tfrac{1}{8}$	$8$
$\tfrac{1}{3}$	$\tfrac{1}{27}$	$27$
$\tfrac{1}{4}$	$\tfrac{1}{64}$	$64$

The count grows fast because cutting each edge in half makes eight pieces, not two.

Packing the box

Choose a cube whose edge divides every side exactly. For $\tfrac{3}{2}$, $\tfrac{1}{2}$ and $1$, the edge $\tfrac{1}{2}$ works for all three.

Count how many fit along each side:

Side	Length	Half-cubes along it
length	$\tfrac{3}{2}$	$3$
width	$\tfrac{1}{2}$	$1$
height	$1$	$2$

$$3 \times 1 \times 2 = 6 \text{ half-cubes}$$

Each is worth $\tfrac{1}{8}$, so the volume is:

$$6 \times \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$$

Why the formula gives the same answer

Multiply the edge lengths directly:

$$\frac{3}{2} \times \frac{1}{2} \times 1 = \frac{3}{4}$$

The same $\tfrac{3}{4}$. That agreement is not luck, and it is worth seeing why.

Counting cubes multiplies three counts: $3 \times 1 \times 2$. Each count is a side length measured in half-units, so each is the side divided by $\tfrac{1}{2}$. Multiplying by $\tfrac{1}{8}$ at the end undoes all three divisions at once.

$$\left(\frac{l}{1/2} \times \frac{w}{1/2} \times \frac{h}{1/2}\right) \times \frac{1}{8} = l \times w \times h$$

Packing and multiplying are the same arithmetic in a different order. So $V = lwh$ needs no separate rule for fractions — it was already counting cubes, whatever size they are.

The base-times-height form

Multiplying the first two edges gives the area of the base:

$$V = B \times h$$

For the box above, the base is $\tfrac{3}{2} \times \tfrac{1}{2} = \tfrac{3}{4}$, and the height is $1$:

$$V = \frac{3}{4} \times 1 = \frac{3}{4}$$

This form is quicker when the base area is already known, and it is the form that generalizes to prisms whose base is not a rectangle.

Working with mixed numbers

Convert mixed numbers to improper fractions before multiplying. Multiplying the whole parts and the fraction parts separately misses two of the four products.

A box is $2\tfrac{1}{2}$ by $1\tfrac{1}{2}$ by $3$ inches.

$$2\tfrac{1}{2} = \frac{5}{2} \qquad 1\tfrac{1}{2} = \frac{3}{2}$$ $$V = \frac{5}{2} \times \frac{3}{2} \times 3 = \frac{45}{4} = 11\tfrac{1}{4} \text{ in}^3$$

Check the size. The box is a bit more than $2$ by $1$ by $3$, so a volume a bit over $6$ was expected — and $11\tfrac{1}{4}$ is nearly double that. Both of the first two sides grew by half, and $\tfrac{5}{2} \times \tfrac{3}{2}$ is $\tfrac{15}{4}$, almost twice $2 \times 1$. The answer stands.

Worked examples

Common mistakes

Practice problems