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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