Find surface area and volume

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Difficulty: Intermediate
Estimated time: 34 minutes
Domains: Geometry and Trigonometry
Techniques: VolumeSurface-areaExposed-surfacesFormula-rearrangementComposed-solidsPyramid-slant-height

What you’ll learn

Tell surface area from volume by the question’s wording and units.
Pick the formula for a prism, cylinder, pyramid, cone, or sphere.
Match each given length to its job in the formula before you put numbers in.
Let Desmos do formula arithmetic that isn’t quick to do in your head.
Work backward through a formula to a missing radius or height.
Build a pyramid’s face slant height from its perpendicular height and half of the matching base side.
Add or subtract solids for volume, and count only the outside surfaces for surface area.
Keep an answer exact in terms of $\pi$ when the question wants it that way.

Prerequisites

Find perimeter and area of plane figures0%
Choose plane-figure formulas and substitute the dimensions that play each role.

Why this matters on the SAT

Decide whether the problem asks for covering or capacity

Any solid gives you two different things to measure. Surface area is the covering on the outside, measured in square units. Volume is the space inside, measured in cubic units. The same box with the same numbers has both, so the question’s wording decides which one you find. Ask yourself: wrap it or fill it?

SAT example

A right rectangular prism has length $9$ centimeters, width $4$ centimeters, and height $5$ centimeters. Which choice gives the volume of the prism, in cubic centimeters?

A
$18$
B
$90$
C
$180$
D
$360$

Solution to the example

The question asks for volume, so use $V=\ell wh$ . The three numbers are already labeled length, width, and height, so type the whole product (9)(4)(5) into Desmos.

V=\ell wh=(9)(4)(5)=180.

The answer is C. The question asks for cubic centimeters, which confirms you were right to find volume.

Calculator loads as you approach

You pick the formula. Desmos multiplies it out in one line.

Recognize the measurement before the solid

Start with what the question wants. Words like capacity, contains, filled, displaced, inside, and cubic units point to volume. Words like cover, paint, wrap, material, outside, exposed, and square units point to surface area.

Then name the solid and label its lengths. A radius runs from the center of a circular base to its edge. A diameter goes all the way across through the center, so it’s twice the radius, and $r=\frac d2$ . The height of a prism or cylinder is the perpendicular distance between its two matching, parallel bases. A cone or pyramid also uses a perpendicular height for volume, measured straight from the tip to the base, not the slant height along its side.

Check your understanding:

A closed cylindrical can is $12$ centimeters tall with radius $3$ centimeters. One question asks how much liquid the can holds, and another asks how much metal covers the can. Which measurement and units belong to each?

Common mistake:

Mixing up the units: square units on a volume, or cubic units on a surface area. Attach units as you set up. A face’s area is in square units, and base area times height gives cubic units.

Build volume from a base

The SAT asks about five familiar solids, and the reference sheet gives a volume formula for each: rectangular prism, cylinder, sphere, rectangular pyramid, and cone. Your job is to name the solid, then match each length to its job in the formula.

The five solids and their volumes. Writing the base area as

B

lets the prism and pyramid formulas work for any base shape.

The reference sheet gives $V=\ell wh$ for a rectangular prism. Here $\ell w$ is the area of the rectangular base, so you can write the same formula as

V=Bh,

where $B$ is the area of one base and $h$ is the perpendicular distance between the bases. That version isn’t on the sheet, but it works for any right prism, whatever shape its base is. For a cube, all three lengths are $s$ , so $V=s^3$ .

A cylinder works the same way. Its base is a circle with area $\pi r^2$ , so

V=\pi r^2h.

Pyramids and cones narrow to a point, so they hold less. The sheet’s rectangular-pyramid formula uses the rectangle’s length and width; replace that base area with $B$ and it works for any pyramid. A pyramid or cone holds one-third as much as a prism or cylinder with the same base and the same perpendicular height:

V_{\text{pyramid}}=\frac13Bh \qquad\text{and}\qquad V_{\text{cone}}=\frac13\pi r^2h.

A sphere has no base or height, only a radius. Its volume is

V=\frac43\pi r^3.

Once the formula is set up, decide how to do the arithmetic:

If what’s left is quick, like one-third of $150\pi$ , do it in your head.
If the formula leaves a chain of products, powers, or sums, type the whole number part into Desmos as one expression.
If the answer should stay in terms of $\pi$ , find the number in front of $\pi$ and then put $\pi$ back. Don’t swap in a decimal.

Check your understanding:

A cylinder and a cone have the same circular base and the same perpendicular height. If the cylinder’s volume is $150\pi$ cubic units, what is the cone’s volume?

Common mistake:

Leaving out the $\frac13$ for a cone or pyramid, which makes the answer three times too big. If the solid comes to a point, check that your formula has the $\frac13$ .

Count exposed surfaces

Surface area isn’t a new kind of space. Picture painting the solid: its surface area is the total area of every surface the paint can reach.

A closed rectangular prism has three pairs of matching faces:

SA=2\ell w+2\ell h+2wh.

A cube has six identical square faces:

SA=6s^2.

A closed cylinder has two circular bases and one curved side. Unroll the curved side, and it lies flat as a rectangle.

The curved side unrolls into a rectangle

2\pi r

long and

h

tall, with the same area.

The rectangle’s bottom edge used to wrap once around the circular base, so its length is the base’s circumference, $2\pi r$ . Its height is still $h$ . So the curved side’s area, called the lateral area, and the whole surface are

\begin{aligned} \text{lateral area}&=2\pi rh,\\[1.4em] \text{total surface area}&=2\pi r^2+2\pi rh. \end{aligned}

For a sphere,

SA=4\pi r^2.

For a pyramid or cone, add the base only if it’s exposed. Then add a pyramid’s triangular faces, or a cone’s curved side, $\pi r\ell$ . Here $\ell$ is the slant height, measured along the surface from the tip down to the edge of the base. For a closed cone,

SA=\pi r^2+\pi r\ell.

None of these surface-area formulas are on the reference sheet. If you forget one, rebuild it: ask which flat or curved pieces cover the outside, and add their areas.

Common mistake:

Putting a cone’s slant height $\ell$ into its volume formula. Volume uses the perpendicular height $h$ , and slant height belongs only to the curved surface, $\pi r\ell$ . Label both before you put in numbers.

Construct a pyramid’s face slant height

A right pyramid’s perpendicular height runs from the tip, or apex, straight down to the center of its base. A triangular face needs a different height: the face slant height, which runs from the apex down the middle of that face to the midpoint of its bottom edge.

Those two heights form a right triangle with a third segment, which lies flat in the base. The SAT often gives you the perpendicular height and the base and leaves this triangle for you to find. The flat segment runs from the center of the base to the midpoint of the edge you picked, and its length is half of the other side of the rectangle. So an edge of length $L$ uses $\frac W2$ , and an edge of length $W$ uses $\frac L2$ .

The flat leg goes from the center of the base to the midpoint of the edge you picked.

Start with the easier case, a right square pyramid with base side $s$ and perpendicular height $H$ . All four faces have the same slant height:

\ell=\sqrt{H^2+\left(\frac s2\right)^2}.

Each face is a triangle with area $\frac12s\ell$ , so the four faces together have

\text{lateral area}=4\left(\frac12s\ell\right)=2s\ell.

For example, if $s=10$ and $H=12$ , then

\ell=\sqrt{12^2+5^2}=13.

One face has area $\frac12(10)(13)=65$ , and all four faces have a lateral area of $260$ .

A right rectangular pyramid takes more care, because its faces aren’t all the same. Say its base has length $L$ and width $W$ , and its perpendicular height is $H$ .

A face whose bottom edge is $L$ uses the flat distance $\frac W2$ , so

\ell_L=\sqrt{H^2+\left(\frac W2\right)^2}.

A face whose bottom edge is $W$ uses the flat distance $\frac L2$ , so

\ell_W=\sqrt{H^2+\left(\frac L2\right)^2}.

There are two faces of each kind, so the lateral and total surface areas are

\begin{aligned} \text{lateral area} &=2\left(\frac12L\ell_L\right) +2\left(\frac12W\ell_W\right)\\[1.4em] &=L\ell_L+W\ell_W,\\[1.4em] \text{total surface area} &=LW+L\ell_L+W\ell_W. \end{aligned}

Check your understanding:

A right rectangular pyramid has perpendicular height $8$ , base length $30$ , and base width $12$ . What are the two face slant heights?

Common mistake:

Using one slant height for every face of a rectangular pyramid. Pair each face’s bottom edge with half of the other side of the base, and you’ll get two slant heights. All four match only when the base is a square.

Example: recover a cylinder’s radius

Worked example

Volume is given, so the unknown length is the radius

r

A right circular cylinder has a volume of $450\pi$ cubic inches and a height of $18$ inches. What is the radius, in inches, of the cylinder?

Step 1

Pick the formula from what’s given

The solid is a cylinder, and you’re given its volume. So start with

V=\pi r^2h.

The radius is squared in this formula, so we’ll get $r^2$ alone first and then take a square root.

Step 2

Put in what you know

Put in $V=450\pi$ and $h=18$ :

450\pi=\pi r^2(18).

Both sides have a factor of $\pi$ , so divide it out:

450=18r^2.

Step 3

Solve for the radius

\begin{aligned} 450&=18r^2\\[1.4em] 25&=r^2\\[1.4em] 5&=r. \end{aligned}

Algebra alone would allow $r=5$ or $r=-5$ , but a radius is a length, so it has to be positive. The cylinder’s radius is $\boxed{5}$ inches.

Step 4

Check with the original volume

Put $r=5$ and $h=18$ back into the formula:

\pi(5)^2(18)=\pi(25)(18)=450\pi.

That matches the given volume. And since the answer is a radius, it’s in plain inches, not square or cubic inches.

Check your understanding:

If the same cylinder had height $50$ inches but kept volume $450\pi$ cubic inches, what would its radius be?

Combine solids without double-counting

Some solids are built from familiar pieces. The key question is whether each piece adds material or takes it away.

For volume, add pieces that are joined on, and subtract a hole or a piece that’s cut out.
For surface area, count only the surfaces you could touch on the finished solid. A face where two pieces are joined is hidden inside, so it doesn’t count.

A tank: a cylinder with a hemisphere on top. The dashed circle where they meet is inside the tank.

This tank is two pieces joined together, so add their volumes. The cylinder holds

\pi(3)^2(8)=72\pi.

The top is a hemisphere, which is half a sphere:

\frac12\left(\frac43\pi(3)^3\right)=18\pi.

So the tank’s volume is

72\pi+18\pi=90\pi

cubic units. To do it faster, type both number parts into Desmos as one expression, (3^2)(8)+(1/2)(4/3)(3^3), and put $\pi$ back on the result.

What about the dashed circle where the pieces meet? It doesn’t change the volume, because the two pieces don’t overlap. But it’s sealed inside the tank, so it isn’t part of the surface area.

Check your understanding:

For the same tank, what is the exposed surface area, in square units? Exclude the shared circle where the cylinder and hemisphere meet.

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Cylinder plus hemisphere, without the pi. The output is 90, so the volume is 90 pi.

Choose the geometry, then calculate

Most of these questions go the same way. You do the geometry, and Desmos can do the arithmetic:

Decide whether the question wants a volume, a surface area, or a missing length.
Name the solid, or the familiar pieces it’s built from.
Label each length by its job: radius or diameter, perpendicular height or slant height, and base area.
Write the formula before you put in any numbers.
If the arithmetic isn’t quick, type the whole expression into Desmos.
Check that you counted only exposed surfaces, used the right units, kept every length positive, and kept $\pi$ if the answer needs it.

Desmos can’t make the geometry decisions for you. It doesn’t know whether a shared face is exposed, whether a $6$ is a radius or a diameter, or whether the question wants square or cubic units. Make those calls first. After that, one complete expression beats working out several in-between numbers by hand.

Short steps are still faster in your head, like dividing both sides by $\pi$ in the radius example.

Try it yourself:

Before each practice problem, write down what it asks for and the formula you’ll use, with no numbers yet. If the choices have $\pi$ in them, keep $\pi$ exact all the way through.

Practice problems

Set up the geometry first, then let Desmos handle any arithmetic that isn’t quick. The pyramid problem is the trickiest, because it needs two different slant heights.

Use the cone formula

Practice problem

A right circular cone has radius $6$ centimeters and perpendicular height $7$ centimeters. Which choice gives its volume, in cubic centimeters?

Calculator loads as you approach

Find the number in front of pi, then put pi back.

Move from volume to surface area

Practice problem

A closed cylinder has two circular bases and one curved surface.

A closed right circular cylinder has height $8$ units and volume $128\pi$ cubic units. Which choice gives the total surface area of the cylinder, in square units?

Calculator loads as you approach

Find the radius first, then add both bases and the curved side in one expression.

Construct two pyramid slant heights

Practice problem

A right rectangular pyramid has perpendicular height $12$ inches. Its rectangular base measures $18$ inches by $10$ inches. The apex is directly above the center of the base. Which choice gives the total surface area of the pyramid, in square inches?

Calculator loads as you approach

Find both slant heights first, then add the base and all four faces in one expression.

Subtract a cylindrical hole

Practice problem

The remaining volume is the block minus the cylindrical hole.

A rectangular block measures $12$ centimeters by $10$ centimeters by $8$ centimeters. A straight cylindrical hole with diameter $4$ centimeters is drilled completely through the block from top to bottom. The height of the cylindrical hole is therefore $8$ centimeters.

Which choice gives the volume, in cubic centimeters, of the remaining solid?

Calculator loads as you approach

The block minus the hole. Keep pi exact.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Surface area is the outside covering, in square units. Volume is the space inside, in cubic units.
A right prism has $V=Bh$ , a rectangular prism has $V=\ell wh$ , and a cylinder has $V=\pi r^2h$ .
A pyramid has $V=\frac13Bh$ , a cone has $V=\frac13\pi r^2h$ , and a sphere has $V=\frac43\pi r^3$ .
Surface area adds up every exposed surface.
A right pyramid’s face slant height comes from its perpendicular height and half of the other base side. A rectangular base can give two different slant heights.
A closed cylinder has $SA=2\pi r^2+2\pi rh$ , and a sphere has $SA=4\pi r^2$ .
Cone volume uses the perpendicular height $h$ ; the cone’s curved side uses the slant height $\ell$ .
For combined solids, add joined volumes, subtract removed ones, and leave hidden inside faces out of the surface area.
You decide the solid, the lengths, and the exposed surfaces. Desmos can’t.
Type arithmetic that isn’t quick into Desmos as one complete expression, and do quick facts and short steps in your head.
Keep $\pi$ exact when the answer needs it, and finish with the right units.

Next lesson

Use scale factors in two and three dimensions

Connect a change in length to the related changes in area, surface area, and volume.

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312 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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