3D Figures, Surface Area, and Volume | SAT Geometry | SAT Math

Introduction

Three-dimensional figures involve a number of formulas and make you think about the figures’ characteristics in several ways. Whereas questions about two-dimensional figures like rectangles and circles ask about area and perimeter, 3-D figures involve questions about volume and surface area. At least one of these two quantities is present in nearly every SAT 3-D solids question.

The normal geometry reminder to draw the figures applies here, perhaps even more so because 3-D figures can be harder to visualize than 2-D ones. But ultimately, SAT problems involving volume usually require using and manipulating the volume formula for the shape in question. Thankfully, several volume formulas are provided for you in the SAT reference box! We’ve provided it immediately below for your reference.

Approach Question

What is the volume of the largest possible cylinder that can be inscribed in a cube with a volume of $64$ $c m^{3}$ ? The cylinder’s flat top and bottom touch the top and bottom faces of the cube.

A. $16 π$
B. $12 π$
C. $8 π$
D. $4 π$

Explanation

The largest possible cylinder in this case would touch all four side faces of the cube (in other words, all faces that are neither the top nor the bottom of the cube). This means that the diameter of the cylinder’s circular bases would be equal to the length of one of the cube’s edges. (If you haven’t yet attempted to draw this figure, do so now in order to visualize what this paragraph is describing.) Meanwhile, the height of the cylinder would also be limited by the length of the cube’s edge and equal to that length.

What is the length of one of the cube’s edges? We are speaking of a consistent edge length because we know that all the dimensions of a cube–length, width, and height–are the same. So we need a number that, when used for each of the three dimensions, multiplies to $64$ . Students often mistakenly think we need to divide $64$ by $3$ in this situation, but it’s actually the cube root of $64$ we need. (Consider that the volume formula for a cube is $e d g e^{3}$ and imagine reversing that process.) The cube root of $64$ is $4$ . So both the diameter of the cylinder’s bases and its height are equal to $4$ cm.

We are almost ready to plug these numbers into the volume formula for the cylinder, which is $π r^{2} h$ . (Think about how this makes sense: $π r^{2}$ should sound familiar as the circle’s area, and we need to multiply the area of that circular base times its height.) But we first need the radius, which must be half of the diameter, or $2$ inches in this case. Now we can plug in: $π \times 2^{2} \times 4 = 16 π$ (just as with circle, the SAT will typically leave pi in the answer choices rather than making you calculate it). The answer is $16 π$ .

Definitions

Edge: A line segment connecting two vertices of a 3-D figure. Edges are used overwhelmingly in discussing cubes because the formula for volume of a cube is simply $e d g e^{3}$ . (It’s length x width x height just like any rectangular prism, but a cube has all equal edges, so the formula can be put more simply.)
Face: The two-dimensional “side” of any 3-D figure. A cube is made up of six faces, all squares. A triangular prism has two faces that are triangles and three that are rectangles. A rectangular prism has six faces that are rectangles.
Prism: A 3-D shape with two congruent opposite faces. A triangular prism has two congruent triangles on its end (considered its base); a rectangular prism has rectangles for its two bases; a square prism has two square bases.

Topics for Cross-Reference

Variations

The hardest problems with 3-D figures may require you to combine two figures (such as stacking two rectangular prisms on top of each other). Being able to draw these figures well can be important for visualizing such problems correctly. As you work through problems like this, don’t hesitate to draw the figures you’re working on!

Strategy Insights

If asked to estimate the value of $π$ , always assume that $π = 3$ . The SAT will not likely ask you to get more precise about $π$ than this.
We discussed the principle of dimensions as it relates to quadrilaterals and circles. The principle extends to 3-D figures as well, but instead of squaring any multiplier to a line segment’s length, we’ll cube it in the case of three dimensions. For example, if you triple the radius of a sphere, it doesn’t become $3$ times as big or even $9$ times as big, but rather $27$ times as big, because the tripling is “cubed” (that is, it happens three times, one for each dimension). Watch out for this exponential effect in 3-D and volume questions.

Flashcard Fodder

Circle formulas provided by the SAT on test day

Volume of a sphere = $\frac{4}{3} π r^{3}$

Volume of a prism = $(l e n g t h) (w i d t h) (h e i g h t)$

Volume of a cylinder = area of base x height = $π r^{2} h$

Volume of a pyramid = $\frac{1}{3} lw h$

Volume of a cone = $\frac{1}{3} π r^{2} h$

Circle formulas NOT provided by the SAT on test day

Volume of a cube = $e d g e^{3}$ (this is a corollary of the prism volume formula, given that length, width, and height are all identical in a cube)

Surface area of a cube = $6$ x $e d g e^{2}$ (if you think about how a cube is made up of six squares, you may not need to memorize this)

Density = mass/volume (see practice problem below)

Sample Questions

Difficulty 1

If a right rectangular prism has a length of $6$ $c m$ , a width of $4$ $c m$ , and a volume of $360$ $c m^{3}$ , what is the height of the prism? (Note: this is a free-response question.)

(spoiler)

The answer is 15. The SAT formula reference tells us that the volume of a rectangular prism is equal to $(l e n g t h) (w i d t h) (h e i g h t)$ . A more straightforward problem would provide all three of these values and ask you to find the volume; this question adds the slight wrinkle of providing the volume and asking you to calculate the height. This is best done by making the height the unknown variable in an equation:

$(6) (4) h = 360$

Multiplying $6 \times 4$ and dividing both sides by $24$ gives us the answer.

Difficulty 2

A spherical balloon has a volume of $288 π$ inches. What is the radius of the balloon?

A. 4 inches
B. 6 inches
C. 8 inches
D. 12 inches

(spoiler)

The answer is 6 inches. As with the previous problem, we are given the volume; since this is a sphere and not a prism, though, this time we are asked for the radius. Students often get stuck here; it’s always a good idea to write out the formula to start. In this case, it’s $V = \frac{4}{3} π r^{3}$ . Now, let’s plug $288 π$ in for $V$ . Knowing that $π$ is a number, we have only one unknown variable left: $r$ . Further, we can cancel the $π$ on both sides by dividing by $π$ . That leaves us with $288 = \frac{4}{3} r^{3}$ .

We need to divide both sides by 4/3, recalling in the process that dividing by a fraction is the same thing as multiplying by its reciprocal. The left side of the equation is now $(288) (3/4)$ . That value can be found with the calculator. An alternate approach is to plug the original equation $288 π = \frac{4}{3} π r^{3}$ into Desmos; the graphing calculator can solve one-variable equations for you. You simply need to be aware that you will read the answer at the x-intercept; Desmos will graph for you the entire line $x = 6$ , not just one point.

Difficulty 3

A right circular cone has a height of $18$ $c m$ and a base with a radius of $7$ $c m$ . The cone has a volume of $aπ$ $c m$ . What is the value of $a$ ? (Note: this is a free-response question.)

(spoiler)

The answer is 294. This problem is complicated slightly by the inclusion of the unknown $a$ , but this is just another way of asking what number should accompany $π$ in our final calculation. The problem is straightforwardly about applying the cone formula given the SAT formula reference; that formula is $\frac{1}{3} π r^{2} h$ . Plugging in $7$ for $r$ and $18$ for $h$ and (as usual) leaving $π$ as it is, we discover that the volume of this cone is $294 π$ . That means $a = 294$ .

Difficulty 4

A right circular cylinder has a volume of $175$ cubic inches. If the length of the diameter of the cylinder’s base is $10$ inches, what is the exact height of the cylinder, in inches?

A. $\frac{7}{2}$
B. $\frac{7}{2} π$
C. $7$
D. $\frac{7}{π}$

(spoiler)

The answer is $\frac{7}{π}$ . To answer this question, we need to use the cylinder formula, being careful to plug in $175$ for $V$ and leaving h as the unknown. But what should we do for $r$ ? The radius is half the length of the diameter, so $r = 5$ . We can now render the equation as follows and solve:

$1751757 \frac{7}{π} = (5^{2}) hπ = 25 hπ = hπ = h$

It may seem odd to end up with pi in the denominator, but remember that the original volume was given without a $π$ , so with $π$ on the other side of the formula from $V$ , at some point both sides have to be divided by $π$ .

Difficulty 5

A sample of iron in the shape of a cube has a density of $8$ grams per cubic centimeter. Each edge of the cube has a length of $0.8$ centimeters. What is the mass, in grams, of this sample of iron?

A. 4.096
B. 6.4
C. 10.0
D. 19.5

(spoiler)

The answer is 4.096. Now and then, the SAT will assume you remember some algebraic formula necessary to solve a higher-difficulty problem. In the case of three-dimensional figures, the test will sometimes include density in the question, since density is a property of three-dimensional matter. Do you remember the formula for density? It’s $d = m / v$ , where $m$ is mass and $v$ is volume. It makes sense if you think about it: the more stuff (mass) there is, or the less space (volume) there is, the more dense something will be.

Now we can use the density formula to solve for something unknown; using the UnCLES method and keeping the question itself in mind, we note that mass is the unknown quantity. That means we must know the value of the density and the volume. The density is given to us, but not the volume; we have to calculate the volume from the edge length of $0.8$ . Since we know that the volume of a cube equals $e d g e^{3}$ , we calculate a volume of $0.4512$ . We are now ready to set up the equation:

$d 8 (8) (0.4512) m = m / v = \frac{m}{0.4512} = m = 4.096$

Having progressed this far in the Achievable SAT course, you will now that unit conversion is sometimes a crucial factor in SAT math. But in this case, the units (grams and centimeters) stay consistent, we don’t have to do any conversions. We do, however, have to avoid the mistake of simply multiplying $8$ times $0.8$ , which yields one of the trap answers. If we manipulate the equation wrongly and end up dividing $8$ by a number instead of multiplying $8$ by a number, we have trap answers present for that as well.

For Reflection

How will you approach three-dimensional solid questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
The SAT reference box is generous with the formulas it provides. But that doesn’t mean you can take those formulas for granted; make sure you know how to use them, especially the process of employing them to set up, and then solve, equations.

Introduction

Approach Question

What is the volume of the largest possible cylinder that can be inscribed in a cube with a volume of $64$ $c m^{3}$ ? The cylinder’s flat top and bottom touch the top and bottom faces of the cube.

A. $16 π$
B. $12 π$
C. $8 π$
D. $4 π$

Explanation

Definitions

Edge: A line segment connecting two vertices of a 3-D figure. Edges are used overwhelmingly in discussing cubes because the formula for volume of a cube is simply $e d g e^{3}$ . (It’s length x width x height just like any rectangular prism, but a cube has all equal edges, so the formula can be put more simply.)
Face: The two-dimensional “side” of any 3-D figure. A cube is made up of six faces, all squares. A triangular prism has two faces that are triangles and three that are rectangles. A rectangular prism has six faces that are rectangles.
Prism: A 3-D shape with two congruent opposite faces. A triangular prism has two congruent triangles on its end (considered its base); a rectangular prism has rectangles for its two bases; a square prism has two square bases.

Topics for Cross-Reference

Variations

Strategy Insights

If asked to estimate the value of $π$ , always assume that $π = 3$ . The SAT will not likely ask you to get more precise about $π$ than this.
We discussed the principle of dimensions as it relates to quadrilaterals and circles. The principle extends to 3-D figures as well, but instead of squaring any multiplier to a line segment’s length, we’ll cube it in the case of three dimensions. For example, if you triple the radius of a sphere, it doesn’t become $3$ times as big or even $9$ times as big, but rather $27$ times as big, because the tripling is “cubed” (that is, it happens three times, one for each dimension). Watch out for this exponential effect in 3-D and volume questions.

Flashcard Fodder

Circle formulas provided by the SAT on test day

Volume of a sphere = $\frac{4}{3} π r^{3}$

Volume of a prism = $(l e n g t h) (w i d t h) (h e i g h t)$

Volume of a cylinder = area of base x height = $π r^{2} h$

Volume of a pyramid = $\frac{1}{3} lw h$

Volume of a cone = $\frac{1}{3} π r^{2} h$

Circle formulas NOT provided by the SAT on test day

Volume of a cube = $e d g e^{3}$ (this is a corollary of the prism volume formula, given that length, width, and height are all identical in a cube)

Surface area of a cube = $6$ x $e d g e^{2}$ (if you think about how a cube is made up of six squares, you may not need to memorize this)

Density = mass/volume (see practice problem below)

Sample Questions

Difficulty 1

If a right rectangular prism has a length of $6$ $c m$ , a width of $4$ $c m$ , and a volume of $360$ $c m^{3}$ , what is the height of the prism? (Note: this is a free-response question.)

(spoiler)

$(6) (4) h = 360$

Multiplying $6 \times 4$ and dividing both sides by $24$ gives us the answer.

Difficulty 2

A spherical balloon has a volume of $288 π$ inches. What is the radius of the balloon?

A. 4 inches
B. 6 inches
C. 8 inches
D. 12 inches

(spoiler)

Difficulty 3

A right circular cone has a height of $18$ $c m$ and a base with a radius of $7$ $c m$ . The cone has a volume of $aπ$ $c m$ . What is the value of $a$ ? (Note: this is a free-response question.)

(spoiler)

Difficulty 4

A right circular cylinder has a volume of $175$ cubic inches. If the length of the diameter of the cylinder’s base is $10$ inches, what is the exact height of the cylinder, in inches?

A. $\frac{7}{2}$
B. $\frac{7}{2} π$
C. $7$
D. $\frac{7}{π}$

(spoiler)

$1751757 \frac{7}{π} = (5^{2}) hπ = 25 hπ = hπ = h$

Difficulty 5

A sample of iron in the shape of a cube has a density of $8$ grams per cubic centimeter. Each edge of the cube has a length of $0.8$ centimeters. What is the mass, in grams, of this sample of iron?

A. 4.096
B. 6.4
C. 10.0
D. 19.5

(spoiler)

$d 8 (8) (0.4512) m = m / v = \frac{m}{0.4512} = m = 4.096$

For Reflection

How will you approach three-dimensional solid questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
The SAT reference box is generous with the formulas it provides. But that doesn’t mean you can take those formulas for granted; make sure you know how to use them, especially the process of employing them to set up, and then solve, equations.