ACT Math · Geometry · free lesson

Surface Area and Volume

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The topic in one minute

Press Play: the idea and a worked example, explained aloud (about 1:09).

Surface Area and Volume

One layer holds $12$ cubes. The box holds $24$ .

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Read what is said

Volume is the space a solid fills, counted in unit cubes. It shows up on the ACT in boxes, cylinders and tanks. This box is four by three by two: one layer holds twelve cubes, and two layers hold twenty-four.
Any solid with the same cross-section all the way up has volume equal to the area of the base times the height. For a cylinder the base is a circle, so the base area is pi r squared.
Here is a water tank. Its diameter is ten feet and its height is seven feet. The formula needs the radius, so the first step is to halve the diameter.
Now square the radius and multiply by the height. Keep pi out front, because answer choices on the ACT often keep it.
The question also asks for gallons. One cubic foot holds about seven point four eight gallons, so multiply the volume by that. Use the decimal only at the end.
Here is the trap. If you use the diameter as the radius, you get seven hundred pi. That is four times too large, and it is always among the choices.
Remember: the formulas use the radius, so when a question gives the diameter, halve it first. Now practice it with new numbers.

The idea

Volume counts cubes

The volume of a solid is the amount of space it fills, counted in unit cubes. It is measured in cubic units such as cubic feet. The box below is 4 units long, 3 units deep and 2 units high. Its bottom layer holds 4 rows of 3 cubes, and there are 2 layers.

\begin{aligned} B &= lw \\ &= 4 \times 3 \\ &= 12 \\ V &= hB \\ &= 2 \times 12 \\ &= 24 \end{aligned}

One layer holds $12$ cubes and the box holds $24$ . The same idea gives the volume of every solid that has the same cross-section all the way up: the area of the base, times the height.

Rule.

A prism or a cylinder has volume $V = Bh$ , where $B$ is the area of the base. For a box that is $V = \ell w h$ . A cylinder of radius $r$ has a circle for a base, so $V = \pi r^2 h$ .

In each solid the base is shaded red, and the height is measured straight up from it. A cylinder of radius 3 and height 5: the engine works out the number in front of $\pi$ , then a decimal.

\begin{aligned} k &= r^{2}h \\ &= 3^{2} \times 5 \\ &= 45 \\ D &= \operatorname{decimal}(k\pi, 1) \\ &= \operatorname{decimal}(45\pi, 1) \\ &= 141.4 \end{aligned}

The volume is $45$ $\pi$ cubic units, about 141.4. Answer choices on the ACT often keep $\pi$ .

Solids that come to a point, and the sphere

Solid	Volume	Surface area
Rectangular prism (box)	$\ell w h$	$2(\ell w + \ell h + wh)$
Cylinder	$\pi r^2 h$	$2\pi r^2 + 2\pi r h$
Cone	$\frac{1}{3}\pi r^2 h$	(given when needed)
Pyramid	$\frac{1}{3}Bh$	the base plus the triangles
Sphere	$\frac{4}{3}\pi r^3$	$4\pi r^2$

The ACT has no reference sheet. Know the box and the cylinder; a question that needs the cone, the pyramid or the sphere usually prints its formula. A cone fills one third of the cylinder it stands in, and a pyramid one third of the prism with the same base and height.

With $r = 3$ and $h = 5$ again, and a pyramid on a 6 by 4 rectangle with height 5:

\begin{aligned} C &= \frac{1}{3}r^{2}h \\ &= \frac{1}{3} \times 3^{2} \times 5 \\ &= 15 \\ S &= \frac{4}{3}r^{3} \\ &= \frac{4}{3} \times 3^{3} \\ &= 36 \\ Q &= 6 \times 4 \\ &= 24 \\ P &= \frac{1}{3}Qh \\ &= \frac{1}{3} \times 24 \times 5 \\ &= 40 \end{aligned}

The cone holds $15$ $\pi$ cubic units, a third of the cylinder's $45$ $\pi$ . The sphere holds $36$ $\pi$ , and the pyramid $40$ cubic units.

Radius counts twice

Drag the sliders and watch the volume of the cylinder. Doubling the height doubles the volume. Doubling the radius makes the volume 4 times as large, because the radius is squared.

Warning.

The formulas use the radius. When a question gives the diameter, halve it first. Using a diameter of 6 as the radius makes a cylinder's volume 4 times too large.

Surface area: unfold the solid

The surface area is the total area of the outside, in square units: the amount of paint, wrapping paper or sheet metal it takes. Cut a box along some edges and lay it flat, and you get its net: six rectangles in three equal pairs. Watch a 4 by 3 by 2 box unfold into its net.

\begin{aligned} A &= 2(lw + lh + wh) \\ &= 2 \times (4 \times 3 + 4 \times 2 + 3 \times 2) \\ &= 52 \end{aligned}

The two blue faces are 4 by 3, the two red faces 4 by 2 and the two green faces 3 by 2, so the box has $52$ square units of surface. Count only the faces that are really outside: a box without a lid has five faces, and two solids glued together lose the two faces that touch.

A cylinder unrolls into two circles and one rectangle. The long side of the rectangle wraps once around the circle, so it is the circumference, $2\pi r$ .

For the cylinder of radius 3 and height 5, the number in front of $\pi$ in its surface area is

\begin{aligned} T &= 2r^{2} + 2rh \\ &= 2 \times 3^{2} + 2 \times 3 \times 5 \\ &= 48 \end{aligned}

so the surface area is $48$ $\pi$ square units. The side alone, the lateral area, is $30$ $\pi$ : a can's label, or a pipe open at both ends.

Working backward, filling and scaling

When the volume is known and one length is missing, put the numbers into the formula and solve. A tank holds water at a rate: the time to fill it is the volume divided by the rate. An object sunk in water raises the level, and the volume of the object is the volume of water it pushes up: base area of the tank times the rise.

\begin{aligned} Bh &= V \\ 16h &= 80 \\ h &= 5 \end{aligned}

A cylinder whose base has area 16 $\pi$ and whose volume is 80 $\pi$ has height $5$ .

Rule.

If every length of a solid is multiplied by $k$ , its surface area is multiplied by $k^{2}$ and its volume by $k^{3}$ . Doubling every edge of a box makes it hold 8 times as much.

Note.

On the ACT the traps are four: the diameter used as the radius, the lateral area given where the whole surface is asked (or a lid counted that is not there), the area factor $k^{2}$ used for a volume, and units not changed (cubic feet into gallons, minutes into hours). The wrong choices are often exactly those numbers.

Worked examples

Each example is one of the kinds of question the ACT asks about surface area and volume, from one formula to the many-step questions about tanks, pools and changing units. The numbers in the working are found by the engine.

Example.

1. A tank given by its diameter. A cylindrical water tank has a diameter of 10 feet and a height of 7 feet. What is its volume, in cubic feet? About how many gallons does it hold, if 1 cubic foot holds about 7.48 gallons?

The formula $\pi r^2 h$ needs the radius, half the diameter. The engine finds the number in front of $\pi$ , then the volume in gallons.

\begin{aligned} r &= \frac{d}{2} \\ &= \frac{10}{2} \\ &= 5 \\ k &= r^{2}h \\ &= 5^{2} \times 7 \\ &= 175 \\ G &= \operatorname{round}(7.48k\pi) \\ &= \operatorname{round}(7.48 \times 175\pi) \\ &= 4112 \end{aligned}

The volume is $175$ $\pi$ cubic feet, about $549.8$ , and the tank holds about $4112$ gallons. Using the diameter as the radius gives $700$ $\pi$ , four times too large: it is always among the choices.

Result.

$175$ $\pi$ cubic feet, about $4112$ gallons.

Example.

2. An aquarium with no lid. A glass aquarium is a rectangular box 30 inches long, 12 inches wide and 16 inches high, open at the top. How many square inches of glass does it take?

A closed box has three pairs of equal faces. With no top, the bottom is counted once: the bottom, two long sides and two ends.

\begin{aligned} b &= lw \\ &= 30 \times 12 \\ &= 360 \\ F &= 2lh \\ &= 2 \times 30 \times 16 \\ &= 960 \\ E &= 2wh \\ &= 2 \times 12 \times 16 \\ &= 384 \\ A &= b + F + E \\ &= 360 + 960 + 384 \\ &= 1704 \\ S &= 2b + F + E \\ &= 2 \times 360 + 960 + 384 \\ &= 2064 \end{aligned}

The aquarium takes $1704$ square inches of glass. The closed box would be $2064$ , the number to expect among the wrong choices.

Result.

$1704$ square inches of glass.

Example.

3. A cone with a scoop on top. An ice cream cone is a cone of radius 3 centimeters and height 10 centimeters, filled with ice cream and topped with a half sphere of the same radius. What is the total volume of ice cream? (A cone has volume $\frac{1}{3}\pi r^2 h$ and a sphere $\frac{4}{3}\pi r^3$ .)

A composite solid is added piece by piece. The half sphere is half of $\frac{4}{3}\pi r^3$ . Work with the numbers in front of $\pi$ .

\begin{aligned} C &= \frac{1}{3}r^{2}h \\ &= \frac{1}{3} \times 3^{2} \times 10 \\ &= 30 \\ S &= \frac{4}{3}r^{3} \\ &= \frac{4}{3} \times 3^{3} \\ &= 36 \\ H &= \frac{S}{2} \\ &= \frac{36}{2} \\ &= 18 \\ k &= C + H \\ &= 30 + 18 \\ &= 48 \end{aligned}

The cone holds $30$ $\pi$ , the half sphere $18$ $\pi$ , and the whole $48$ $\pi$ cubic centimeters, about $150.8$ . Counting a whole sphere on top would give $66$ $\pi$ .

Result.

$48$ $\pi$ cubic centimeters.

Example.

4. A stone in a glass of water. A cylindrical glass has an inside radius of 4 centimeters and holds some water. A stone is dropped in and sinks completely, and the water level rises by 1.5 centimeters. What is the volume of the stone?

The stone pushes up exactly its own volume of water. That water is a short cylinder: the base of the glass, as tall as the rise. The water that was there before does not matter.

\begin{aligned} k &= r^{2}e \\ &= 4^{2} \times \frac{3}{2} \\ &= 24 \end{aligned}

The stone's volume is $24$ $\pi$ cubic centimeters, about $75.4$ . A common slip is to find the volume of all the water, which the question never gives.

Result.

$24$ $\pi$ cubic centimeters.

Example.

5. Filling a pool. A rectangular swimming pool is 30 feet long and 15 feet wide, and it is filled to a depth of 5 feet. Water flows in at 150 gallons per minute, and 1 cubic foot is about 7.5 gallons. How many hours does it take to fill the pool from empty?

Four steps, each with its own unit: the volume in cubic feet, the volume in gallons, the time in minutes, and the time in hours.

\begin{aligned} V &= lwh \\ &= 30 \times 15 \times 5 \\ &= 2250 \\ G &= 7.5V \\ &= 7.5 \times 2250 \\ &= 16875 \\ m &= \frac{G}{q} \\ &= \frac{16875}{150} \\ &= 112.5 \\ t &= \frac{m}{60} \\ &= \frac{112.5}{60} \\ &= 1.875 \end{aligned}

The pool holds $2250$ cubic feet, or $16875$ gallons, which take $112.5$ minutes: $1.875$ hours. The choices that stop early, at $112.5$ (minutes, not hours), or divide the cubic feet straight by the rate, at $15$ , are the traps.

Result.

$1.875$ hours.

Example.

6. Two solids of the same shape. Two cylinders have the same shape. The surface area of the larger is $\frac{9}{4}$ times the surface area of the smaller, and the smaller holds 40 cubic inches. How much does the larger hold?

Areas grow by the square of the length factor $k$ and volumes by its cube. The ratio of the areas is $k^{2}$ : take the square root to get $k$ , then cube it.

\begin{aligned} k &= \sqrt{q} \\ &= \sqrt{\frac{9}{4}} \\ &= \frac{3}{2} \\ m &= k^{3} \\ &= (\frac{3}{2})^{3} \\ &= \frac{27}{8} \\ V &= mv \\ &= \frac{27}{8} \times 40 \\ &= 135 \end{aligned}

The lengths grow by $\frac{3}{2}$ , the volume by $\frac{27}{8}$ , and the larger cylinder holds $135$ cubic inches. Multiplying 40 by the area factor $\frac{9}{4}$ gives $90$ , the answer the question is built to catch.

Result.

$135$ cubic inches.

Practice

Answer ACT-style questions on surface area and volume. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise surface area and volume with new numbers