Struggling with 3-D geometry problems on the SAT, like the problems about cubes and spheres? In this lesson, you’ll learn how to crack them by breaking them down into more familiar 2-D shapes and solving from there.
Unlike 2-D geometry, three-dimensional geometry, or 3-D geometry, deals with objects that have three measurable dimensions: length, width, and height.
You can see this if you compare a 2-D shape, like a square, to its 3-D equivalent, a cube. The square has only length and width, but the cube has length, width, and height. The biggest trick to mastering 3-D geometry on the SAT is to realize that most of it really isn’t about 3-D shapes at all. Instead, it’s about collections of 2-D shapes that just so happen to be arranged into something we recognize as a cube or a sphere.You could see this shape below as a cube, but you could also see it as six squares stuck together, and that’s a lot easier to work with.
When you’re working with 3-D shapes, this ability to break them down into their component 2-D parts will give you a big boost on the test. In this lesson, we’ll cover one of the most important 2-D breakdown shapes: hidden triangles. Then, we’ll walk through a basic example and a second example with a cylinder to show how triangles can even help with rounded shapes if you know how to use them.
Hidden triangles are one of the biggest 3-D tricks on the SAT. These are most commonly right triangles but can sometimes be other types of triangles as well. Just to show you how this works, take a look at some of the triangles lurking inside just one cube. Here’s one that you could use to find the distance between points A and B, or the diagonal, of one side of the cube.
Here’s another that you could use to find the distance between points C and D, drawing a line right through the center of the cube.
First, think of the line segment AB as the longest side of a right triangle, like below. We already have one side of the triangle, which is 6. But we need the other side, the distance from the corner of the square to point B.
6^2 + 6^2 = x^2, so x^2 = 72. That makes x equal to 6(sqrt2). B is right in the middle of the bottom face, so we’ll cut this in half to find the distance just from the corner to B: it’s 3(sqrt2).
Now see what we have? A right triangle with two known sides.
We can solve for the third to get the distance from A to B: (3*sqrt2)^2 + 6^2 = x^2.
So we know that x^2 = 54. We can simplify the square root of 54 to 3(sqrt6). And now we have our answer.
Perfectly possible. It just takes some clever use of triangles to make it happen.
As well as hiding triangles in cubes and spheres, the SAT will also ask you questions about cylinders, but here, again, you can crack the problem by breaking the cylinder down into 2-D shapes. If you think about a cylinder, it’s really a rectangle rolled into a circle with caps at both ends. If this doesn’t make any sense, think about an ordinary tin can.
The top and bottom are circles, and if you peel away the label carefully, you’ll end up with a rectangular piece of paper. So a cylinder is really just two circles plus a rectangle. Not so scary after all.To illustrate this, let’s look at an example problem. A pencil is to be fit inside a tin can 12 inches tall with a volume of 300 pi inches cubed. What is the length of the longest pencil that will fit completely within the can?
So we’ll plug this in: 300 pi = (area of the base)(12). Divide both sides, and we get 25 pi = area of the base.
We know that the area of a circle equals pi times the radius squared, so we can solve to get a radius of 5. Can you guess what happens next?That’s right: it’s hidden triangles again! There’s a nice, friendly right triangle hiding in this cylinder once we figure out the diameter of the base. Just remember to use the diameter instead of the radius since the bottom side of our triangle stretches all the way across the base.
|This obtuse angle, where an obtuse angle is