Brian was a geometry teacher through the Teach because that America program and started the geometry regimen at his school




You are watching: How to find how many diagonals are in a polygon

A diagonal is a segment the connects 2 non-consecutive vertices in a polygon. The number that diagonals in a polygon that deserve to be attracted from any vertex in a polygon is three much less than the number of sides. To find the total number of diagonals in a polygon, multiply the variety of diagonals every vertex (n - 3) by the number of vertices, n, and also divide by 2 (otherwise each diagonal is counted twice).


How numerous diagonals are a polygon through 300 sides? well wouldn't make a entirety lot of feeling to attract a polygon v 300 sides and draw on every those diagonals. There needs to be a shortcut or a formula.Well, an initial let's back up. What is a diagonal? A diagonal line is any type of line segment the connects 2 non-consecutive vertices. Therefore if us look at a triangle. If i look at every single vertex, again the peak is where two ends meet two political parties meet. There's no means for me to draw on a diagonal here due to the fact that for this peak both of these sides are consecutive. So there's no method for us to have any diagonals.If ns look at a square however, I deserve to see that there is one non-consecutive crest if ns look at this vertex. I look at one more vertex there's just one non-consecutive vertex. Therefore let's view if us can number out the pattern. To perform that, we're walk to usage this table over here where I have actually three columns; one for the variety of vertices, one because that the number of diagonals every vertex and the total variety of diagonals the we view in a polygon.So we've already started v two various polygons. We've talked about a triangle. So, variety of vertices in a triangle well, that's simply three. The variety of diagonals we stated was zero since there is no way for united state to attract in a diagonal. Which means our total diagonals is tho zero. Okay?Let's go earlier and look at the square. The square us said, there space 1, 2, 3, 4 vertices. This vertex appropriate here has actually only one diagonal, this vertex right here has actually only one diagonal line so we're 4 vertices, each vertex has actually one diagonal yet we only see two of them. So we watch that there's going to it is in some type of division that's walk to have to go top top here.Last, let's look at a pentagon. If i look at this vertex, i can attract in one, two diagonals. And also I'm walking to watch that because that every vertex, I'm walk to be able to draw in, two different diagonals. So number of vertices right here is five, number of diagonals per vertex is two and also the full of diagonals below we have a little star so we have 5 diagonals. For this reason I want to know first for n vertices since I'm walk to attract dot dot dot for n vertices, what will certainly be the complete number?Well I view that if i multiply 3 times 0, so us make a dot here. 3 times 0 is 0 for this reason we're okay there. Here we have actually 4 time 1, yet that does not equal 2. Therefore what we're going to need to do is I'm walking to have to take 4 time 1 and divide the in half. 5 times 2 divided in fifty percent is same to 5. So ns look at the number of vertices we have three therefore we're going to contact that n. Below we have number of diagonals per vertex, right here we have actually 0, 1 and 2 and I view that to get from 3 to 0 I'm going to subtract 3 to get from 5 to 1, i subtract 3 from 5 come 2 ns subtract 3. So we have n times the amount of n minus 3 all divided by 2.So two vital things around this formula right below which tells you the number of diagonals and also I'm walking to abbreviation DIAG.So the variety of diagonals, there's two crucial things I desire to allude out. The first is this n-3. Wherein is n-3 come from? well if we have 5 vertices here. We're no going to counting the peak that's itself since you can't draw a peak to itself plus there space two an ext consecutive vertices for complete of 3 vertices in this polygon the we're no really counting.Second key part right here is this division by 2. Why carry out we have to divide this by 2? If ns go earlier to this to the square, if i look at this vertex, I've attracted in one diagonal. Indigenous this vertice's perspective, I've only attracted one. Native this vertex's perspective I've drawn another diagonal.

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But it's the exact same diagonal. So every vertex and every diagonal that we draw from a vertex will certainly be counted double which is why we need to divide ours formula by two.