Posts tagged with: 3D

Test 4 Section 4 Number 18

This is a volume question. The first thing we need to do is figure out the volume of liquid each of these glasses will hold. You need not have any formulas memorized—they’re in the beginning of the section. The volume of a right circular cylinder can be found using this formula: V=\pi r^2h.

In this case, we’re told that the height is 6 inches and the internal diameter is 3 inches. If the diameter is 3 inches, then the radius is 1.5 inches. So the volume of the glasses:

    \begin{align*}V_\text{glass}&=\pi (1.5\text{ inches})^2(6\text{ inches})\\V_\text{glass}&\approx 42.4\text{ cubic inches}\end{align*}

He has 231 cubic inches—how many glasses can he fill all the way up?

    \begin{align*}\frac{231}{42.4}\approx 5.4\end{align*}

If he can fill 5.4 glasses, then he can only fill 5 all the way up.

Hey Mike, I have a question concerning an example problem in the PWN SAT Guide, 4th edition. On p. 280 in the Working in 3-D chapter (Example 3), it says that a right triangle was formed by the dotted lines in the cube. I’m kind of confused because the dotted lines don’t seem to contain a right angle.

Thanks for asking this!

The dotted lines and the edge of the cube don’t appear to form a right triangle in the figure because of the orientation of the cube, but that is indeed a right triangle. If you follow the edge of the cube up from point A to the top corner, and then you head towards point B, you must make a 90° turn.

If you could reorient the cube so that you were looking straight at the triangle, it would be more clear. The triangle in 2-D looks like this:

Test 1 #35 (calculator section)

You can find the formula for the volume of a right circular cylinder at the beginning of the section; it’s V=\pi r^2h. Since the question tells you the volume (72π cubic yards) and the height (8 yards), you can solve for the radius:

    \begin{align*}72\pi&=\pi r^2(8)\\9&=r^2\\3&=r\end{align*}

Now, be careful! The question asks for the diameter, not the radius. Even though they underline the word diameter, I bet a lot of people still miss this question by putting the radius.

If the radius is 3, then the diameter is 6.


The cube above is made from 27 small cubes, each with an edge of length 1. If the shaded cube is removed what will be the surface area of the remaining solid?

I’m assuming the shaded cube is a corner cube, right? So when you take it away, you’re removing 3 faces from the solid? Like this?

rubiks cube


Look! When you remove that cube, you’re taking away 3 faces, but you’re also exposing 3 more previously unexposed faces! And those newly exposed faces have the exact same areas as the faces you took away. So the surface area, amazingly enough, remains the same.

You’ve got a 3\times 3\times 3 cube there. The original area of each face was 9, and of course all cubes have 6 faces. So the original cube had a surface area of 54.

When you remove the corner piece, you’ve got 54 – 3 + 3 = 54.

Does that help?

Bonus question: what would happen if the removed cube came from the middle of a face, instead of a corner?

Each of the 8 edges of a pyramid with a square base is 4 inches long and each edge of a cube is 4 inches long. The base of the pyramid is set on one face of the cube so that their vertices coincide. The new solid that is formed has how many faces?

The cube has 6 faces. The pyramid has 5 faces. When you put them together, then you lose one face from each (the bottom of the pyramid and the top of the cube become part of the inside of the new solid), and end up with 5 + 4 = 9 faces.

A cube with edge of length 4 is divided into 8 identical cubes. How much greater is the combined surface area of the 8 smaller cubes than the surface area of the original cube?

A) 48
B) 56
C) 96
D) 288
E) 384

The surface area of the original cube is 6\left(4^2\right)=96.

sube cut into 8 cubes

When the cube is cut into 8 smaller cubes (accomplished by cutting it in half along each axis, see above) then each cube has an edge length of 2, so each has a surface area of 6\left(2^2\right)=24. Because there are 8 of them, the total surface area is 8(24)=192.

The question asked “how much greater,” so we subtract: 192-96=96. This makes sense in an abstract way, if you think about it. By cutting a cube in half along each axis, you’re doubling the surface area.

A large cube with edge of length 3 units is built from 6 small blue unit cubes and 21 small white unit cubes. What is the greatest possible fraction of the surface area of the large cube that could be blue?
A) 1/9
B) 2/9
C) 1/3
D) 4/9
E) 2/3

The surface area will be 9 for each of the 6 faces for a total surface area of 54. If you put all the blue cubes in corners (you can do this–there are 8 corners in a cube and only 6 blue cubes) then you’re maximizing the amount of blue exposed: each corner cube contributes 3 to the surface area.* So you’ll have 6 blue cubes time 3 faces exposed equals 18 square units of blue out of a total surface area of 54.

18/54 = 1/3

* I just said that bit like it’s no big deal, but there’s a fair amount of insight required to see that without drawing it. Look:

  • Red: a corner cube will have 3 faces exposed
  • Green: a cube on the edge but not on the corner will have 2 faces exposed
  • Yellow: a cube that’s not on an edge will have only one face exposed.

I’m having a lot of fun playing around with some geometry drawing software this week (nyeeerd!), so I figured I’d use it again to make another “fun” 3-D problem for the weekend challenge. This is a bit tougher than you’d find on the SAT, but the underlying concepts, as always, are important for the SAT.

The prize this week inspired by my current situation: If there should ever come a time in your life where you’re trying to move from Brooklyn to the Bronx, you’ll be smart enough not to bother trying to move yourself in your Toyota Yaris and instead just hire movers. And when you do, they won’t break your stuff.

Point B is in the center of the top face of the cube in the figure above, and point A is one of the cube’s vertices. If the distance between points A and B is d, then what is the cube’s volume in terms of d?

Put your answers in the comments, and I’ll post the solution Monday. Good luck!

UPDATE: Nice work, JD. You were wrong at first, but you corrected yourself before I did. Solution below the cut.

The most difficult 3-D problems you’ll come across on the SAT will require you to work with right triangles, usually on the same plane as one of the solid’s sides, and then on a plane that cuts through the solid. In this question, you’re going to want to make a right triangle whose hypotenuse is AB (A.K.A. d).

To do so, we’re going to have to first figure out the distance from B to the cube’s corner. Let’s assume the cube has sides of length c:

Since we’re so good with right triangles, we know that the diagonal of the square on the right is c2, and since B is the midpoint of that segment, the distance from the corner to B must be c2/2. Now we know the lengths of both legs of the right triangle we actually care about, the one with hypotenuse d. We can use it to solve for d in terms of c. (“But wait,” you say, “aren’t we looking for the volume in terms of d?” Yes, we are. We’ll get there. Promise.)

Now that we’ve got d in terms of c, we can put c in terms of d, and cube it to get the volume. Wahoo!

Some people like to stop here, and others like to rationalize the denominator (as JD did in his response). The SAT has been known to swing both ways.


Note: you can check your work here by plugging in. Say c = 2, so the volume of the cube is 8. That would make d = 2√3/√2. Plug that in for d in our solution. Does it give you 8? Yup.

It’s not uncommon for a question or two involving three-dimensional shapes to appear on the SAT. Luckily, most of the time these questions either deal directly with the simple properties of three-dimensional shapes (like surface area and volume), or are just 2-D questions in disguise. It’s pretty rare to come across a truly difficult 3-D question — but you know I’m gonna give you some in this post because I care about you so.


Generally speaking, the SAT will give you every volume formula that you need, either in the beginning of the section (rectangular solid — V = lwh; right circular cylinder — V = πr2h) or in the question itself in the (exceedingly) rare case where you’ll have to deal with the volume of a different kind of solid. It’s worth mentioning, though, that the volume of any right prism* can be calculated by finding the area of its base, and multiplying that by its height.

For example, if you needed to calculate the volume of a prism with an equilateral triangle base, you’d find the area of an equilateral triangle:

And multiply that by the height of the prism:

You almost definitely won’t need this particular formula on the SAT, but it’s nice to know how to find the volume of a right prism in general: just find the area of the base, and multiply it by the height.

Most volume questions you’ll see on the SAT will require you to deftly maneuver between the volume of a solid and its dimensions. Let’s see an example (and showcase my fresh new drawing software):

  1. If the volume of the cube in the figure above is 27, what is the length of AF?
    (A) 3
    (B) 3√2
    (C) 3√3
    (D) 3√5
    (E) 6

Remember that a cube is the special case of rectangular solid where all the sides are equal, so the volume of a cube is the length of one edge CUBED:

V = 27 = s3
s = 3

So far, so good, right? Now it’s time to do the thing that you’re going to find yourself doing for almost every single 3-D question you come across: work with one piece of the 3-D figure in 2-D.

The segment we’re interested in is the diagonal of the square base of the cube. If we look at it in 2 dimensions, it looks like this:

The diagonal of a square is the hypotenuse of an isosceles right triangle, so we can actually skip the Pythagorean Theorem here since we’re so attuned to special right triangles. AF = 3√2. That’s choice (B).

Surface Area

The surface area of a solid is simply the sum of the areas of each of its faces. Easy surface area problems are really easy. Trickier surface area problems will often also involve volume, like this example:

  1. If the volume of a cube is 8s3, which of the following is NOT a value of s for which the value of the surface area of the cube is greater than the value of the volume of the cube?
    (A) 0.5
    (B) 1
    (C) 2
    (D) 2.25
    (E) 3

Yuuuuck. What to do? Well, to find the surface area of a solid, you need to know the areas of its faces. To find those areas, you need to know the lengths of the sides of the solid. Luckily for us, it’s pretty easy to find the lengths of the sides of this cube, since we know that the volume is 8s3. Take the cube root of the volume to find the length of one side of the cube:

If a side of the cube is 2s, then the area of one face of the cube is (2s)2, or 4s2. There are 6 sides on a cube, so the surface area of the cube is found thusly:

6 × 4s2 = 24s2

From here, it’s trivial to either backsolve, or solve the inequality spelled out in the question:

24s2 > 8s3
3s2 > s3
3 > s

The answer must be (E), the one choice for which the inequality is NOT true.

* Right circular cylinders and rectangular solids are both special cases of right prisms — a right prism is any prism whose top lines up directly above its bottom.

Break it down.

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