Geometry is heavily tested on the GRE Math section, and a thorough review of geometrical concepts is essential to a high score. Consider the following problem:
“If the length of an edge of a cube X is twice the length of an edge of cube Y, what is the ratio of the volume of cube Y to the volume of cube X?”
The easiest way to solve this is to pick a number for the initial edge length and plug it into the problem. For instance, let’s say cube X is a 4x4x4 cube. Cube X would have a volume of 64. Cube Y would have to be a 2x2x2 cube, since 2 is half of 4, and it would have a volume of 8. The ratio of the volume of cube Y to the volume of cube X would thus be 8 to 64, or 1/8.
However, you really should have known that to begin with. Imagine that cube X had edges that were three times as long as those of Cube Y. Then Cube X would now be a 6x6x6 cube if Cube Y remains a 2x2x2 cube, and the volume ratio would be 8 to 216, or 1/27. Notice something? 8 is 2 ^3, and 27 is 3^3. If the ratio of the sides is 1:4, the ratio of the volumes will be 1:64. If the ratio of the sides is 1:5, the ratio of the volumes will be 1:125. Since these are cubes, you just cube the ratios. 1^3 is 1, and 4^3 is 64; 5^3 is 125. If you know this simple property of the relationship between length and volume, it will take a problem that would take 30 seconds to solve and turn it into a problem that takes 5 seconds to solve. On a timed exam, that could be the difference between getting another, harder question right or wrong. Memorizing these kinds of mathematical facts is something that the GRE test writers expect top scorers to do, and they write the questions so that they can be solved quickly if you know them. It also pays to memorize the squares and cubes of the numbers 1 through 12.
So with cubes, you cube the ratio of the sides. What about squares? If you guessed that you square the ratio of the side lengths in order to get the ratio of the areas, you’d be right, as you can see from a quick demonstration. If the original square has side lengths of 1 and the new square has side lengths of 2, the side ratio is 1:2 and the area ratio is 1:4. If the new square has side lengths of 3, then the side ratio is 1:3 and the area ratio is 1:9. If the new square has side lengths of 4, then the side ratio is 1:4 and the area ratio is 1:16, and so on. Sure enough, you just square the original ratio.
So now you know about cubes and squares, but what about tesseracts? “Tessawhats?” you say? A tesseract is to a cube as a cube is to a square, just as a cube is to a square what a square is to a line. Still confused? Let me explain it this way: say you draw a line a foot long running from east to west. This line only exists in one dimension: east-west. Then, you decide to square it by adding three more lines: two perpendicular to it running north to south and one parallel to it running east to west. This square exists in two dimensions: east-west and north-south. Now you decide to turn the square into a cube by adding lines in the up-down dimension, so that each edge of the original square is now the edge of another square emanating from it. This cube exists in three spatial dimensions: east-west, north-south, and up-down. Now you take this cube you’ve made and decide to square it…in a fourth spacial dimension.
What is this fourth dimension? Who knows. We live in a world in which we experience only three spacial dimensions, so it is impossible for us to imagine what a four dimensional object would look like. That hasn’t stopped mathematicians from naming four-dimensional objects, and this hypercube I’ve just described to you is called a tesseract. As you know, even though a cube is a three dimensional object, it is possible to draw a cube on a piece of paper in only two dimensions by using perspective and all those other artistic illusions. Likewise, some have attempted to render tesseracts in three dimensions in order to give some approximation of what they might look like. Having never seen an actual tesseract, though, you might still find these representations confusing.
In terms of doing calculations, though, tesseracts are simple as can be. For a square with side lengths of 1 and another square with side lengths of 2, the ratio of side lengths is 1:2^1 (since sides are 1 dimensional), or 1:2, and the ratio of areas will be 1:2^2 (since squares are 2 dimensional) or 1:4. For a cube with side lengths of 1 and another cube with side lengths of 2, the ratio of volumes is 1:2^3 (since cubes are 3 dimensional), or 1:8. So, for a tesseract with side lengths of 1 and another tesseract with side lengths of 2, the ratio of hypervolumes(?) is 1:2^4 (since tesseracts are 4 dimensional), or 1:16. It just follows the pattern. Try not to think about it too much.
If you’re having trouble with tesseracts, don’t worry. They’re not on the test. I just wrote about them to mess with your head.
Remember, if you ever want extra help getting ready for the GRE, you can always study with experts like me through Test Masters. Until then, happy studying!