Before we get into triangles, we need to take a very quick look at the ingredients of a triangle: line segments and angles. Please tell me you already know this stuff:
We good? Cool. Prove it:
- In the figure above, AE, BS, CG, DS, and FS intersect at point S. Which of the following pairs of angles must be congruent?
(A) ∠ASF and ∠BSF
(B) ∠ASG and ∠CSE
(C) ∠ASG and ∠FSG
(D) ∠ASB and ∠ASG
(E) ∠ASC and ∠BSE
What to do, what to do? It’s often possible to guesstimate on a question like this: they’re asking you which angles are congruent, and the diagram is drawn to scale, so look at the damn thing. Enough choices look plausible here that we can’t take that shortcut, although we can use it to eliminate choice (C) if we like.
Which angles are vertical angles? We know that all these segments meet at point S, but which ones actually go through it? AE and CG both go all the way through, so they’ll create a set of vertical angles: ∠ASG and ∠CSE. We know vertical angles are always congruent, so (B) is the answer. You’re really going to want to make sure you’re solid on this kind of question.
OK, that was fun, right? Now let’s talk triangles. There are only a few things you need to know about angles and triangles for the SAT, many of which are given to you at the beginning of each math section. This post is going to be long (I know what you’re thinking: it’s already long) because there are a lot of different kinds of questions you might be asked and I want you to see them, but don’t be daunted by its length; I’m willing to bet you already know pretty much everything you need to know.
Now, as always, you just need to study the scouting report: know what the SAT will throw at you, and you’ll have a better chance of knocking it out of the park. Note also that there is a separate post for the special case of Right Triangles, just to keep the length of this post somewhat manageable. Lastly, please forgive me if my drawings look janky.
The sum of the measures of the angles in a triangle is 180°:
The area of a triangle can be found with A=½bh:
In an isosceles triangle, the angles across from the equal sides are also equal:
In an equilateral triangle, all the angles are 60°, and all the sides are of equal length:
The bigger the angle, the bigger its opposite side:
No side of a triangle can be as long as or longer than the sum of the other two sides:
This is called the Triangle Inequality Theorem and there’s a cool demonstration of it here. The basic thrust is this: if one side were longer than the other two, then how would those two shorter ones connect to form the triangle? They couldn’t. And if one side was equal to the sum of the other two, would you have a triangle? No, you’d just have a straight line.
To drive this home: imagine your forearms (apologies to my armless friends) are two sides of a triangle, and the imaginary line that connects your elbows is the third side. If you touch your fingertips together and pull your elbows apart, eventually your fingertips have to disconnect…that’s when the length between your elbows is longer than the sum of the lengths of your forearms. Neat, huh?
There’s an awful lot to know about non-right triangles if you’re doing math, but on the SAT, the preceding basically covers it. You will never need trigonometry of any kind, nor will you need any knowledge about things like circumcenters or orthocenters of triangles (Google them if you’re curious). Occasionally it’ll be nice to know triangle congruence/similarity rules, but I’m not going to recreate them here because honestly, it’s incredibly rare that you’ll actually need one. There are some who think you need to know things like the external angle rule, or rules about alternate interior angles and opposite exterior angles in transversals. I say: if you know a straight line is 180° and you know a triangle’s angles add up to 180°, you already know them! Don’t overcomplicate your life with overlapping rules.
I’m in the business of preparing you for a very specific test. The above is, in my expert opinion, the salient stuff. Isn’t it awesome how little of it there is?
Let’s try one more example together:
Note: Figure not drawn to scale.
- If y = 180 – x and all lengths are integers, which of the following could be the perimeter of the triangle in the figure above?
Wait, what? We don’t know anything about any of the lengths, how are we supposed to figure out anything about the perimeter? Relax, friend. We’ll get through this.
We know y = 180 – x. How can we use that? Well, note that the red arc in the image to our right here is a straight line, which means it’s 180°. Therefore, the unlabeled angle in our original figure must be equal to 180 – y. Do a little algebra now:
[unmarked angle measure] = 180 – y
[unmarked angle measure] = 180 – (180 – x)
[unmarked angle measure] = x
WTF. It doesn’t look like it, but we’ve got an equilateral triangle on our hands: all the angles inside it are equal so they must all be 60°! If all our lengths are integers, then, the perimeter must be a multiple of 3. Only one of our choices is: (D) 24.