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# Number of Diagonals in a Polygon - Problem 2

###### Brian McCall

###### Brian McCall

**Univ. of Wisconsin**

J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

By using the formula for the number of diagonals of a polygon with n sides, you can determine how many sides a polygon has if you know the number of diagonals it has. For example, if a polygon has 54 diagonals, find how many sides it has.

Remember, the formula is:

Plugging in the known information, we know that diag = 54. So,

Then, solve for n using algebra. First multiply both sides by 2.

^{2}- 3n

Notice there is a squared term. This is a quadratic equation, so we can solve it either by factoring or using the quadratic formula. Here, we'll use factoring.

^{2}- 3n - 108

So, n = 12 or -9. But, since a polygon cannot have a negative number of sides, we know that this polygon must have 12 sides.

Let’s look at a problem where you know the number of diagonals but you don’t know how many sides it has. In this problem it says if a polygon has 54 diagonals, how many sides does it have?

Now the key thing here is we’ve been talking about vertices then in the polygon the number of vertices would be the same as the number of sides. So I guess you could even erase this in your mind and say how many vertices does it have? And what’s our formula? Well we said that the number of diagonals is equal to the number of sides times the quantity of n minus 3 or the number of sides minus 3 all divided by 2.

Well what are you given here? You’re not given the number of sides that’s what we’re solving for. You’re given the number of diagonals so I’m going to substitute in 54 for the number of diagonals with n times n minus 3 all divided by 2. So at this point in your mind you might be thinking, "Hey that’s an easy quadratic," or you might be thinking. "Oh no, I have to remember algebra and geometry!" I’m going to show you both ways just incase you want to see both ways.

To solve this algebraically first we’re going to multiply by 2 on both sides, 2 times 54 is 108 and I’m also going to do something that Math teachers don’t like to do very often, but since I’m running out of space I’m going to two steps in one here. I’m going to distribute that n. So n times n is n², n times -3 is -3n. Now I see that I have a quadratic, if I want to solve I have to have 0 on one side of my equation. So I’m going to subtract 108 on both sides of my equation and I’m going get 0 equals n² minus 3n minus 108.

A couple of ways you can solve this, you could use a quadratic formula but since you’ve got a really large number this 108 that’s negative, it’s going to make it a little messy. The easiest way is to solve by factoring. So if I can factor this into two binomials, I can use this zero product property to find my answers.

So I’m looking for two numbers that multiply to -108 and add to -3. To start off with I know that n times n will give us n². If it multiplies to a negative and adds to a negative we have opposite signs and our negative number is larger. So these two numbers are -12 and +9, so the zero product property says that if n is -9 then that binomial will be 0. So that’s our first answer, n equals -9. Zero product property says that n is 12 here, 12 minus 12 is 0, so n could also equal 12.

We have a problem here we have two answers. In geometry most of the time we’re going to be talking about a distance, number of sides, number of vertices, so we’re always going to be taking the positive number. I’m ignoring the -9 because it doesn’t make sense for a polygon to have negative nine sides. So what we did we do, we solved using the zero product property and we found our two answers but we took the positive one.

Now let’s rewind a little bit, let’s say you got to this point right here, so I’m going to draw in a little box here to separate this. Let’s say you got to this point, you said okay Mr. McCall, I can substitute in that 54 into my equation. So we got n times n minus 3 all divided by 2, and I have no idea how to solve a quadratic anymore, that was last year. Here’s what you’re going to do.

You’re going to take your calculator and you’re going to substitute in numbers for n. Notice that this is n for both of these so that number has to be the same. A good place to start for this problem is at 10. So you’d say that 54 is equal to 10 times 10 minus three all divided by 2. So if you wanted to you could take out your calculator or you could say that 10 times 7 is 70, 70 divided by 2 is 35. So you get 54 equals 35. This equation isn’t true so you know that 10 can’t be correct.

So then you just kind of do a guess and check which is a valid way of solving a problem, it’s not very efficient but here you’d have 11 times 8, so you 88, 88 divided 2 would give you 44. And you’d say well moving in the right direction not quite there, so let’s try 12. So we have 12, 12 minus 3 is 9 divided 2 and that is 54 equals 54, we got the right answer and we can say n equals 12.

They’re both valid, you might get extra credit if you do it this way but I would caution you, I would say last resort use guess and check, ideally you’re going to use algebra.

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###### Brian McCall

B.S. in Chemical Engineering, University of Wisconsin

J.D. University of Wisconsin Law School (magna cum laude)

He doesn't beat around the bush. His straightforward teaching style is effective and his subtle midwestern accent is engaging. There's never a dull moment with him.

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## Mike · 1 year ago

If you have to rely on guess and check, be careful as some teachers won't allow calculators on exams. Also, if you plan on taking algebra 2 or attending college, don't forget any algebra you learn! Algebra is used in trig, geometry, and especially calculus.