*Three-dimensional figures* involve a number of formulas and make you think about their characteristics in several ways. Whereas two-dimensional figures like rectangles and circles are concerned with concepts like 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 CLT 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.

A cylinder is inscribed in a cube with a volume of $64cm_{3}$; the cylinder’s circular bases touch the top and bottom faces of the cube. What is the volume of the largest possible cylinder in this case?

A. $16π$

B. $12π$

C. $8π$

D. $4π$

The largest possible cylinder in this case would touch all four side faces of the cube. This means that 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 edge^3 and imagine reversing that process.) The cube root of 64 is 4 (know your cube roots of 8, 27, 64, and 125 at least for the CLT!). 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 $πr_{2}h$. (Think about how this makes sense: $πr_{2}$ 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: $π×2_{2}×4=16π$ (just as with circle, the CLT will typically leave $π$ in the answer choices rather than making you calculate it). The answer is **A**.

As illustrated in the practice questions, the CLT will sometimes refer to a *cross-section* to introduce a two-dimensional aspect into a 3D figure. Cross-sections on the CLT will be parallel to the base and create simple shapes like circles and rectangles.

**Circle formulas provided by the CLT on test day**

Volume of a sphere = $(4/3)$ $pi$ $r_{3}$

Surface area of a sphere = $4$ $pi$ $r_{2}$

**Circle formulas NOT provided by the CLT on test day**

Volume of a prism = $length$ x $width$ x $height$

Volume of a cube = $edge_{3}$

Volume of a cylinder = $areaofbase$ x $height$ = $pi$ $r_{2}$ x $height$

Surface area of a cylinder = $2$ $pi$ $r_{2}$ + $2$ $pi$ $r$ $h$

Volume of a prism = $areaofbase$ x $height$

Volume of a pyramid = $(⅓)$ $(areaofsquarebase)$ $(height)$

Volume of a cone = $(⅓)$ $pi$ $r_{2}$ $h$

The planet Mercury has a radius of approximately 1,500 miles. Which of the following is closest to the surface area of Mercury? (Assume, for purposes of this question, that Mercury is a perfect sphere).

A. $4.500π$ square miles

B. $9,000π$ square miles

C. $2,250,000π$ square miles

D. $9,000,000π$ square miles

(spoiler)

The answer is **D**. Good news about spheres: both the volume and surface formulas are provided to you! It’s never a bad idea to memorize them, but you can consult the reference list and find that the formula for a sphere’s surface area is $4πr_{2}$. It can be tricky to calculate this without a calculator, but let’s use the “look” step of the UnCLES method and make a couple of observations. First off, as is typical, all the answers include $π$, so we don’t need to worry about the pi part of the calculation. Second, answers C and D are much larger than the first two choices. Which magnitude is correct? We know we need to square the radius. Imagine squaring $1,000$ instead of $1,500$. A thousand thousands is a million, so it looks like we need to be in the millions. Choices A and B are out.

To continue, focus on the nonzero part of the radius: essentially the number $15$. If you know your perfect squares up to 15 (as you should for the CLT!), you know that $15_{2}=225$. From that alone, it appears that the answer is $C$, but watch out! The formula for surface area includes a $4$, so the right answer is $4$ times as much as choice $C$. $C$ is the trap, but $D$ is the answer.

Notice how we got to the answers without precisely calculating the number of zeroes. We didn’t need to! This kind of numerical reasoning can help you move faster on the CLT. Also, if you notice that choice $D$ is $4$ times as much as choice $C$, you might suspect that choice $C$ is a trap for students who forget to multiply by $4$.

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

A. 4 inches

B. 6 inches

C. 8 inches

D. 12 inches

(spoiler)

The answer is **B**. The challenge to this problem compared to the Difficulty 1 problem is that now we are given the volume and have to find 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 $V=(4/3)πr_{3}$. Now, let’s plug $288π$ 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 $288=(4/3)r_{3}$.

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 $(288)(¾)$. We can simply by dividing $288$ by $4$ first. $4$ should go into $280$ $70$ times, so it must go into $288$ $72$ times. Now we multiply $72$ by $3$, getting $216$. So our radius cubed is equal to $216$ $in_{3}$. Do we need to figure out the cube root of $216$? It’s great if we know it, but if not, why not look at the answers? We should know that $4_{3}$ is $64$; that’s too small. $12_{2}$ is already $144$, so $12_{3}$ will certainly be too big. What about $8$? $8x8=64$; multiply that by another $8$ and we’ll be well over $216$. The answer must be $6$.

Notice that you were asked for the cube root of $216$ in that question. Learning your cubes up to $10_{3}$ would be a smart move: $2_{3}=8$, $3_{3}=27$, etc.

A company that manufactures mini-globes detaches the base of the globes from their spherical parts before shipping them to retailers. If the manufacturer packs as many spherical globes with radius $3$ cm into boxes with dimensions $15cm$ x $16cm$ x $30cm$, how much empty space (space not covered by the spheres, including their insides) is there in each box?

A. $7200−720πcm_{3}$

B. $7200−1440πcm_{3}$

C. $720−36πcm_{3}$

D. $720−720πcm_{3}$

(spoiler)

The answer is **A**. To find the empty space in the box, we have to find the volume of both the box and of each individual globe; we also need to know how many globes we can fit in the box. Let’s start with the last question first. It’s important to recognize that we can’t just take the large volume and divide it by the small volume. Not only will there be extra space between the spheres, but there might also be space *left over* in each dimension. Importantly, we need to use the *diameter* (not the radius) to conceive of how we fit the globes into the box. Each globe has a diameter of 6 cm, so we have to divide that value into each of the box’s dimensions. Taking the last dimension first, it’s helpful that 6 goes evenly into $30$, yielding a quotient of $5$. We can fit $5$ boxes into that dimension of the box. But the other two dimensions don’t divide evenly by $5$, and we can’t pack *part* of a globe, so we’ll have to leave any remainder unused. The quotient of $15×6$ and $16×6$ tells us that we can only fit two globes in each of these dimensions. (Don’t ask us why the company chose such ill-fitting boxes in which to ship their globes!). $2$ times $2$ times $5$ means that we can put a total of $20$ globes in each box.

Now let’s find the volumes. The box is simpler: $length$ x $width$ x $height$. That gives us a volume of $7200$ (be careful not to lose a zero in the process, which would send you to wrong answers $C$ and $D$). For the globes, we’ll use $(4/3)πr_{3}$. Yet again, $π$ is already accounted for in the answers, so we can ignore that. We’ll first cube our radius to get $27$, then multiply by $4/3$. Let’s work smarter here and divide $27$ by $3$ first, getting $9$. Multiplying by $4$ gives us $36$. So disregarding the $π$, we need to account for $20$ globes times $36cm_{3}$ per globe. $20×36=720$, so the answer must be $A$.

For a cylinder with a radius of 5 cm and a height no larger than its diameter, which of the following must be false?

A. A cross-section of the cylinder has an area of 25pi square cm.

B. The volume of the cylinder is larger than the volume of a cube with an edge length of 9 cm.

C. The cylinder’s volume is 750 cubic centimeters.

D. The cylinder’s surface area is 750 cubic centimeters.

(spoiler)

The answer is **D**. Using the UnCLES reminder to look at the answers, we can see that volume comes up twice. So let’s begin by calculating the cylinder’s maximum volume. Its diameter is $10$, so its height is a maximum of $10$. Using the formula $πr_{2}h$ (*note*: this is not included in the CLT reference and must be memorized!), we get $π×25×10$ or $250π$. Unlike most CLT questions involving $π$, this problem will require us to estimate the value of $π$; to do so, we should always assume that $π=3$. This means the volume of the cylinder is about $750$ cubic centimeters (actually a little more since $π$ is about $3.14$). That means we can eliminate answer choice $C$; if the volume can be more than $750$, then it can certainly be exactly $750$.

Is the volume of the cylinder bigger than that of the cube in choice $B$? The volume of a cube is equal to $edge_{3}$. We should know that $9x9=81$; what about $81×9$? Let’s take $8×90$ first; since $8×9=72$, $8∣times90$ should equal $720$. Then $1×9=9$, so the total volume is $729$. It is possible that the cube’s volume is bigger than $729$, so B is out.

Let’s address the cross-section next. Although this answer choice doesn’t specify that the cross-section is cut out parallel to the circular base, such a cross-section is certainly possible and is the simplest to deal with it, so let’s assume a parallel cross-section is present. This means we are cutting out a circle identical to the cylinder’s circular base. That base has an area of $5×5×π=25π$, so choice $A$ is true and therefore *not* our answer.

We know by process of elimination that the answer is $D$, but how can we prove it? The surface area of a cylinder is somewhat complicated; the formula is $2πr_{2}+2πrh$. The $2πr_{2}$ corresponds to the combined area of the two circular bases. The rest of the formula comes from the fact that if we “unwrap” the side of the cylinder, we get a rectangle whose dimensions are equal to the cylinder’s height times the circumference of its base. We’ll leave the calculations to you using the formula, but the short answer is that the surface area of this cylinder can be no larger than $150π$. Estimating a value for $π$, we can see that this overall surface area is less than $500$, so it certainly can’t be as large as $750$.

The surface area of a cylinder comes up rarely on the CLT, so you can deprioritize it among your formulas, but we’ll include it in the formulas for this chapter. We encourage you to think conceptually about how the surface area of a cylinder is composed; that way, even without the formula, you may well be able to visualize and calculate (or at least estimate) such a surface area.

A student is studying triangular prisms and comes up with the following conjecture:

The volume of a triangular prism, in cubic units, is always less than its surface area, in square units.

Which of the following is a counterexample that disproves the above statement?

A. A triangular prism with a right angle base with legs of 6 and 8 and a height of 10.

B. A triangular prism with an equilateral triangular base with an area of $3 $ and a height of 4.

C. A triangular prism with a right angle base with legs of 5 and 12 and a height of 8.

D. A triangular prism with an equilateral triangular base of $253 $ and a height of 20.

(spoiler)

The answer is **D**. For this problem, let’s begin by defining a prism as a *three-dimensional solid with two parallel bases*. A triangular prism has triangles for bases, a rectangular prism has rectangles, etc. For the volume of a prism, the master principle is that, just as with a cylinder, we are thinking about $base$ x $height$. But in this case, we mean the *area* of the base, so in this case we’ll need to account for the area of each of our triangle’s bases. For surface area, meanwhile, we’ll begin by doubling the area of the triangular base because there are two such bases. What makes up the rest of the prism? Drawing the figure out as necessary to visualize it, you can see that the rest of the prism is made up of three rectangles; each rectangle’s dimension are the height of the prism and one of the sides of the triangle. So we’ll need to calculate the area of each rectangle separately and add them together.

Let’s begin, taking the choices one by one. (At least we don’t have to deal with pi in this scenario!) For choice area, we know we’ll eventually need to know the hypotenuse of the base triangle; knowing our Pythagorean triples saves us time; the hypotenuse is $10$! We also know from our triangles lesson that a right triangle’s legs are the same as its base and height, so we know the area of the triangular base is $(½)(6)(8)=24$. For the rectangles, we use the height of $10$ and multiply $10×6$, $10×8$, and $10×10$, adding them together to get $240$. Our surface area is $240+24+24=288$. The volume is the area of the base ($24$) times the height of $10$, so $240$. This is *not* a counterexample; the volume is less than the surface area.

Let’s skip to choice $C$ because it also has a right triangle base. (Always seek to be efficient with your evaluation of the choices.). This is also a Pythagorean triple; the hypotenuse is $13$. So the area of the base is $30$; the areas of the rectangles are $20$, $48$, and $52$. The surface area, then, is $30+30+20+48+52=180$. The volume is $30$ times the height of $4$; that’s only $120$. The volume is once again less than the surface area and choice $C$ is eliminated.

To assess choices $B$ and $D$, we need to remember that the formula for the area of an equilateral triangle is $s_{2}3/4 $, where $s$ is the length of the triangle’s side. Let’s take choice $B$ and set the formula equal to the given area of $3 $.

$s_{2}3/4 =3 $ (now divide both sides by $3 $)

$s_{2}=1$

$s=1$

This helps us know that the areas of the three rectangles in the prism will all be $1x8=8$. That means the surface area is equal to $24+23 $. Remembering that $3 $ is approximately $1.7$, this gives us about $27.4$. The volume, meanwhile, is the height of $8$ times the base area, which is only $83 $ or about $13.6$. Choice $B$ is out as well. It must be $D$, but let’s check! We can use the same equilateral triangle area formula to determine that the sides of the triangular base are $10$ each. This means that each of the rectangles has an area of $200$. The surface area, then is $600+503 $, or about $685$. The volume is the height of $20$ times the base of $253 $, which is $5003 $, or approximately $850$. It turns out that if we made the dimensions of the prism large enough, eventually its volume would outstrip its surface area. Fascinating!

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