3D shapes

The three-dimensional (“3D” or “3-d”) shapes that you will need to understand for the GRE are cubes, rectangular solids, and right circular cylinders. Every 3D shape you encounter on the GRE will be some combination of these core building blocks. Even if the shape seems unfamiliar at a glance, you’ll be able to decompose it into some combination of these ones.

It’s good to memorize all these equations so you can recall and apply them quickly. In addition, it’s helpful to understand how to derive them yourself based on the fundamental characteristics of the shapes, as this will give you a better understanding of how to calculate the volume or surface area of a composite shape.

Cube

A cube is a squared square. Its base is a square, and then that square is extended upward to the height of that same square. Since it’s composed entirely of squares, all sides $s$ are equal in length.

The volume of a cube is $s^3$. This is fairly easy to derive once we step through it. The size of the footprint square on the base is $s\ast s=s^2$, and then it’s extended upwards for a height of $s$. The volume is the base size times the height, and $s^2\ast s=s^3$.

The surface area of a cube is $6s^2$, and is also fairly easy to derive. A cube has $6$ square surfaces, each with an area of $s^2$, so the total surface area is $6\ast s^2=6s^2$.

Let’s try a simple cube question:

What is the surface area of a cube that has a volume of 125?

Give it a try using the formulas above, then check your work!

We’re given the volume of the cube, and we can use that to solve for the length of a side.

$$\begin{array}{rl}V& =s^3\\ 125& =s^3\\ 5& =s\end{array}$$

Now that we have the length of a side, we can calculate the surface area.

$$\begin{array}{rl}\text{SA}& =6s^2\\ & =6(5^2)\\ & =6(25)\\ & =150\end{array}$$

This is a very straightforward question, but it illustrates the same general process that you’ll use for any question like this!

Rectangular solid

A rectangular solid has a rectangular base that has been extended up by some distance. Technically, a cube is also a rectangular solid that has all sides equal: $l=w=h$, but typically a rectangular solid will have varying dimensions.

The volume of a rectangular solid is $l\ast w\ast h$. We can arrive at this formula by taking the size of the base $l\ast w$, and since we’re extending it upward, multiplying by the height $h$ to get $lwh$.

A rectangular solid has six rectangular sides. They come in matching pairs:

• The top and the bottom ($2\ast lw$)
• The front and the back ($2\ast lh$)
• The close side and the further side ($2\ast wh$)

Adding these all up, the total surface area is $2lw+2lh+2wh$. The order of the variables doesn’t matter as long as we’re adding up all the sides.

Now let’s try a fundamental rectangular solid question:

What is the surface area of a rectangular solid that has a length of 10, a width of 12, and a volume of 720?

Solve it yourself, and then check your work!

We’re given the volume, length, and width, and we can use this to solve for the missing dimension: the height.

$$\begin{array}{rl}V& =lwh\\ 720& =10\ast 12\ast h\\ 720& =120h\\ 6& =h\end{array}$$

Now that we have all the dimensions, we just need to plug them in to calculate the surface area.

$$\begin{array}{rl}\text{SA}& =2lw+2wh+2lh\\ & =2\ast 10\ast 12+2\ast 12\ast 6+2\ast 10\ast 6\\ & =240+144+120\\ & =504\end{array}$$

Right circular cylinder

Cylinders might seem a little trickier to start, but you can break them down in the same way as the other shapes you’re more familiar with. A right circular cylinder is essentially just a shape that has a circle as the base, and that circle is extended upward for some height.

To arrive at the formula for the volume of a right circular cylinder, we can start with the size of the circular base. It’s just a normal circle, so its area is $\pi r^2$. This circle is extended straight up the entire height $h$ of the shape, so we just multiply by $h$ to get a volume of $\pi r^{2}h$.

The surface area is also seemingly trickier to start, but not so much once you break it down. A right circular cylinder has three sides: the top circle, the bottom circle, and a rectangle that wraps around to connect those two circles. Think about a tube of wrapping paper: the shape of the tube is a right circular cylinder, but the sheet of paper that comes off it is just an ordinary rectangle. Wrapping it around the tube doesn’t change the dimensions of that rectangular sheet. One dimension of the sheet is the height $h$, and the width of the sheet of paper is actually the circumference (i.e. the perimeter) of the base circle: $2\pi r$. So to get the area of this sheet of paper, we just multiply those together to get $2\pi rh$. Adding on the area of the top and bottom circles ($2\ast \pi r^2$), we can see the total surface area of the cylinder is $2\pi r^2+2\pi rh$.

Let’s try the same type of question as with the other shapes, but with a right circular cylinder.

Cylinders A and B have the same volume. Cylinder A has a height of 3 and Cylinder B has a radius of 3, and the radius of Cylinder A is equivalent to the height of Cylinder B. What is the surface area of Cylinder B?

Take your time with this one, but give it a try before you check the answer and explanation below.

This time we’re given the volume of another cylinder instead of a number, but the concept is still the same. The volume of Cylinder A is the same as Cylinder B, so we can just set them equal to each other and solve. We’ll use subscripts and parentheses in our equation to make the math a bit easier to read since we’re working with multiple similar dimensions.

$$\begin{array}{rl}{V}_A& =V_B\\ \pi (r_A^2)(h_A)& =\pi (r_B^2)(h_B)\\ (r_A^2)(h_A)& =(r_B^2)(h_B)\\ (r_A^2)(3)& =(3^2)(h_B)\\ (h_B{)}^2(3)& =(3^2)(h_B)\\ (h_B)(3)& =3^2\\ h_B& =3\end{array}$$

Now that we have all the dimensions of Cylinder B, we just need to plug them into the surface area formula using $h_B=3$ and $r_B=3$.

$$\begin{array}{rl}{\text{SA}}_B& =2\pi (r_B^2)+2\pi (r_B)(h_B)\\ & =2\pi (3^2)+2\pi (3)(3)\\ & =2\pi (9)+2\pi (9)\\ & =18\pi +18\pi \\ & =36\pi \end{array}$$

Just move through it slowly, ensuring you don’t mix up similar variables, and you’ll be able to solve any of these questions!