Students love to skip over the basics, asking questions like: How many vertices does an octagon have? How many diagonals does an octagon have? What’s the difference between a regular octagon and, well, an octagon? And GRE geometry does delve into some complex polygon math. But before we get to that, I will begin with two challenging GRE math problems.

## GRE Geometry: Polygon Problems

1) Regular pentagon P has all five diagonals drawn. What is the angle between two of these diagonals where they meet at a vertex of the pentagon?

(A) 12°

(B) 36°

(C) 54°

(D) 60°

(E) 72°

2) How many diagonals does a regular 20-sided polygon have?

(A) 60

(B) 120

(C) 170

(D) 240

(E) 400

Explanations to these practice problems will appear at the end of this blog article. Jump ahead by clicking here.

## Polygons

First, some basic terminology to begin this discussion. A **polygon** is any geometric shape all of whose sides are straight line-segments. Any **triangle** is a polygon. Any **quadrilateral** (including trapezoids, parallelograms, rhombuses, rectangles and squares) are polygons. A **pentagon** is a 5-sided polygon. A **hexagon** is a 6-sided polygon. An **octagon** is an eight-sided polygon. A circle or parabola or anything with a curved side is __not__ a polygon.

A point where two of the sides of a polygon meet is called a **vertex**. The number of vertices a polygon has is always equal to the number of sides it has.

Another important polygon fact concerns the sum of angles. You may know that the sum of the three angles in any triangle is 180°. You may even know that the sum of the four angles in any quadrilateral is 360°. This pattern generalizes. The sum of all n angle in any n-sided polygon is:

**sum of angles = (n – 2)*180°**

Thus, any pentagon (n = 5) would have angles that add up to 3*180 = 540°. Any hexagon (n = 6) would have angles that add up to 4*180 = 720°. Any octagon (n = 8) would have angles that add up to 6*180 = 1080°. (See the blog on GRE Geometric Formulas)

Finally, there is this paradoxical word “**regular**.” In everyday language, “regular” means “ordinary, unexceptional, commonplace.” In geometry, it connotes the exact opposite! A shape is regular if and only if it is both equilateral and equiangular—that is, if and only if all the sides have the same length and all the angles are equal. The “regular” version of any polygon is the most elite, most symmetrical version possible of that polygon. The “regular triangle” would be what we known as an equilateral triangle. The “regular quadrilateral” is the square. For higher polygons, you are most likely to see the regular version on the GRE, because the test (like all mathematicians) loves symmetry.

## Diagonals of a Polygon

Now, we can talk about diagonals. A **diagonal** is any line through the interior of a polygon that connects two non-adjacent vertices. What does this mean? Well, first of all, starting from any vertex, an adjacent vertex is either vertex connected to the starting vertex by one side of the polygon.

Consider an irregular quadrilateral:

Let’s start at vertex A. Starting from vertex A, we are connected by sides of this quadrilateral to both B and D; vertices B & D are the ones that are adjacent to vertex A. The only vertex not connected to A by a side of the quadrilateral is C. C is A’s only non-adjacent vertex, and A is C’s. Thus, one diagonal goes from A to C. It’s not hard to see that the other goes from B to D. Any quadrilateral has just two diagonals.

Notice that **triangles NEVER have diagonals**: if we start from any vertex in a triangle, the other two vertices are adjacent. There simply are no non-adjacent vertices in a triangle, so diagonals are not possible. Among quadrilaterals, there are special rules for the diagonals of a parallelogram and the categories within parallelograms:

The diagonals of a parallelogram bisect each other: that is to say, the intersection point of the two diagonals is the midpoint of each one.

A rhombus is a parallelogram with four equal sides. The diagonals of a rhombus bisect each other and are perpendicular.

A rectangle is a parallelogram with four 90° angles. The rectangle of a rhombus bisect each other and have equal length. This is related to an old trick among carpenters. When a carpenter cuts two pairs of equal lengths to make the sides of a door or window frame, he knows he has a parallelogram because of the equal lengths, but how does he know whether he has a rectangle? Without precise equipment, it’s very hard to measure the difference between, say, an 89° or 90° angle. Well, all the carpenter has to do is measure the two diagonals: if these two easy-to-measure lengths are equal, then it’s guaranteed that he has four right angles!

A square is a parallelogram, a rectangle, and a rhombus. It is a regular quadrilateral with four equal sides and four 90° angles. The diagonals of a square bisect each other, have equal length, and are perpendicular.

## Diagonals of a Regular Pentagon

A **pentagon** is any five-sided polygon, and the sum of its angles is 540°, as we saw above. The only pentagon you are likely to meet on the GRE is the most symmetrical, the regular pentagon. Since the angles are equal, we can divide the sum of the angles by five.

540°/5 = 108°

That’s the angle of each of the five angles in the pentagon.

Here’s a regular pentagon with the five diagonals drawn.

How many diagonals does a pentagon have? Any pentagon has exactly five diagonals. These diagonals trace out the shape of a classic five-pointed star, such as that those on the Flag of the United States of America. The lengths and divisions of this star are intimately related to that magical and mystical number, the Golden Ratio; Sacred Geometry is purported to give insight into the meaning of life, but you don’t need to know any of that for the GRE!

How would we find any angles in this shape? Well, we know each big angle of the pentagon is 108°. Look, for example, triangle ABC. This triangle is an isosceles triangle, because AB = BC, and we know that angle ABC = 108°. The other two angles must be equal: call them x.

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

This means that angle BAC = angle BCA = 36°, and so do many other symmetrically related angles around the shape. We could subtract (angle BAC) from (angle BAE) to get (angle CAE)

angle CAE = (angle BAE) – (angle BAC) = 108° – 36° = 72°

From that, we could find many other angles inside the shape. We could use analogous means to find angles involving the diagonals of any higher polygon.

## Diagonals of a Regular Hexagon

A **hexagon** is any six-sided polygon, and the sum of its angles is 720°, as we saw above. In a regular hexagon,

each angle = 720°/6 = 120°

How many diagonals does a hexagon have? Starting from one vertex, two other vertices are adjacent, so 3 vertices are non-adjacent, making possible three diagonals from one vertex. From A, we can draw diagonals to C, D, and E.

From each vertex, there are three diagonals. Since there are six vertices, you might think there would be a total of 3*6 = 18 diagonals, but that counting method double-counts everything. You see, the diagonal from A to C would get counted once as a diagonal from A and again as a diagonal from C to A. Thus, the number of diagonals in a hexagon is 18/2 = 9. These can be groups into two kinds. The six shorter diagonals together make a six-sided star, the Magen David. The three longer diagonals form just three symmetrically criss-crossing segments, what is called in mathematics a “degenerate six-pointed star.”

These two diagrams show the nine diagonals of a regular hexagon. Of course, the six-pointed star simply consists of two overlapping equilateral triangles, pointing in opposite directions. (That geometric fact led to extensive mystical speculation in about the Star of David in the Kabbalah, but again, you don’t need to understand any mysticism for the GRE!)

## Diagonals of a Regular Heptagon

A **heptagon** is any seven-sided polygon (n = 7). Sometimes it is called a “septagon,” but “heptagon” is the preferred mathematical name. The sum of its angles would be

(n – 2)*180° = 5*180° = 900°

This means that each of the seven angles in a regular heptagon would have a measure of

each angle = 900°/7 = 128.5714286…°

The angle measures are not integers! This is why the GRE is exceptionally unlikely to ask you anything about the regular heptagon, and it’s also why you probably never talked much about regular heptagons in high school geometry. It’s why most people aren’t even clear on the proper name for this beast! Their non-integer angle measure makes them the first “black sheep” in the regular polygon family! I won’t say anything else about them, because they almost never make an appearance on the GRE, but I will show you the two possible seven pointed stars from their diagonals: these stars are hauntingly beautiful, because of their idiosyncratic symmetry.

## Diagonals of a Regular Octagon

An **octagon** is any eight-sided polygon, and the sum of its angles is 1080°, as we saw above. In a regular octagon,

each angle = 1080°/8 = 135°

That angle is the supplement of a 45° angle. The regular octagon is the typical stop sign shape in many parts of the world.

### How many diagonals does an octagon have? How many vertices does an octagon have?

Starting from one vertex, two other vertices are adjacent, so five vertices are non-adjacent, making possible five diagonals from one vertex. From A, B & H are the symmetrical vertices, so we can draw diagonals to C, D, E, F, and G.

Similar logic as with the hexagon: five at each vertex, eight vertices, but that counts each diagonal twice, so the total number is 5*8/2 = 20. AC and AG are what we might call “3 vertex diagonals”: there are eight of these, which form a star. AF and AD, each parallel to two sides, are what we might call “4 vertex diagonals”: eight of these form another star. Finally, AE is like a diameter of the whole octagon, cutting across its center: there are four such lines, and they form a degenerate eight-pointed star.

Just as the six-pointed star consisted of two overlapping equilateral triangles, the first eight-pointed star on the left consists of two separate overlapping squares: square ACEG and square BDFG. (The names of these squares are reminiscent of the lines at W. 4th Street!) Between these three stars (counting the degenerate thing on the right as a “star”) we have all 20 of the regular octagon’s diagonals.

## Summary

Onward and upward! The methods discussed in this blog can be extended to apply to a nonagon (n = 9), a decagon (n = 10), or any higher polygon. Armed with this information, you should be able to answer anything the GRE asks you about a diagonal of a polygon! If you had any “aha” moments while reading this blog, you might want to give the practice problems on the top another peek before reading the explanations below.

## Practice Problem Explanations

1) Let’s take a look at the pentagon with its five diagonals.

An example of the angle between two diagonals at a vertex would be angle EBD, where diagonals BD and BE meet at vertex B.

We will follow the logic outlined above.

Triangle BCD is isosceles with BC = CD, and angle BCD = 108°. The other two angles are equal: call them each x.

108° + x + x = 180*

2x = 180° – 108° = 72°

x = 36°

So, angle CBD = 36°. Well, triangle ABE is in every way equal to triangle BCD, so angle ABE must also equal 36°. Thus, we can subtract from the big angle at vertex B.

(angle EBD) = (angle ABC) – (angle CBD) – (angle ABE)

(angle EBD) = 108° – 36° – 36° = 36°

Answer = **(B)**

2) If we start at one vertex of the 20-sided polygon, then there’s an adjacent vertex on each side. Not counting these three vertices, there would be 17 non-adjacent vertices, so 17 possible diagonals could be drawn from any vertex. Twenty vertices, 17 diagonals from each vertex, but this method double-counts the diagonals, as pointed out above.

# of diagonals = (17*20)/2 = 17*10 = 170

Answer = **(C)**

*Editor’s Note: This post was originally published in January, 2014, and has been updated for freshness, accuracy, and comprehensiveness.*

I want to know the length of the largest diagonal of a octagon.how can I?

Hi there,

In order to answer this question you would need more information–for example, the length of other diagonals that form a triangle, or the height of a square that encloses an octagon. This isn’t a question that the GRE would ask you to calculate without more context. If you have a specific question in mind, let us know! 🙂

Actually I have the side length of the regular octagon and each angle measure of the regular octagon.so now can I get the longest diagonal of the regylar octagon?if I can get the answer.

Then how?!can u please help me….

Hi Shubhranshu,

Thanks for your response 🙂 Here’s the formula to find d, the length of the longest diagonal, for a regular octagon:

d = a * √(4 + 2√2)

where a is the side length of the octagon.

We could derive this length in a couple of ways. One way is to split up the octagon into 8 isosceles triangles with base a and sides d/2, with all of the triangles sharing a vertex in the middle of the octagon. From there, can find the length of d/2 in terms of a and then solve for d 🙂

I hope this helps! 🙂

hi please help me to find how many different lengths of diagonals does a regular octagon have?

Hi Esme,

In order to ask about the lengths of diagonals, you would have to be given a lot more information!

However, we can use logic to tell us that in a regular octagon we will have three different lengths of diagonals. This is because each vertex of a regular octagon will have five diagonals connected to it, but some of the diagonals will have the same length. For example, in the figure given above, if we take the diagonals from A, we see that the six diagonals are AC, AD, AE, AF, AG. However, since it is regular we know that AC=AG and AD=AF, so there are only three distinct lengths. And all of the points will have the same situation (for example, AC=BD=CE…etc) Does that makes sense? So even though there are 20 diagonals in a regular octagon, they will all be one of three lengths.

Hi,

In the solution for practice problem #1, consider triangle EBD. Because EB is a diagonal, the angle between between EB and BD must be half of 108 degrees which is equal to 54 degrees. Hence, the angle between BD and DE must also equal 54 degrees. Now, in triangle EBD, we have the 2 angles ,each 54 degrees, summing upto 108 degrees. Since, the sum of all the angles in any triangle must be 180 degrees, angle B must then be equal to 72 degrees (180-108=72). But, the answer is 36. What am I missing?

Hi Megha,

I think the problem with your explanation is that (1) you claim that Angle DBE is half of Angle ABC, and (2) you equal Angle DBE to a non-similar angle, Angle BDE (the similar angle of Angle DBE is Angle BDA). So, both of the above claims are not true, which is leading you to an incorrect solution.

Take another look at how the problem was broken down in the “Diagonals of a Regular Pentagon” section of this post. This is a better approach to solve this problem. I hope this helps!

for the regular polygons last question in the video, Is it wrong to assume that AB and CH are parallel. I assumed this and since angle HAB was 67.5, i used parallel lines to deduce that angle x was also 67.5 which was really quick but I do not know if it was a coincidence that this is true and if I should not make this assumption

Hi Daniel,

Good question! Yes, you can use the fact that AB and CH are parallel to solve the practice problem but only if it is stated in the question that the octagon is regular. Otherwise, you cannot make that assumption.

How many line segments are in an Octagon ?

Hi Vanshika 🙂

Are you referring to the total number of line segments including those that connect two adjacent vertices? If that the case, we have 8 such line segments that constitute the 8 sides of the octagon. Additionally, every polygon has n(n – 3)/2 diagonals, where n is the number of sides o the polygon. In the case of the octagon, then, there are

8(8-3)/2 = 8*5/2 = 20 diagonals.

So, the total number of line segments (the sides of the octagon + the diagonals) of an octagon is 8 + 20 = 28.

Hope this helps!

Hello could you solve this for me if considering 8 vertices of octagon and it’s centre and T denotes number of triangles drawn and S denotes number of straight lines that can be formed with these 9 points then value of T-S

Hi Jatin 🙂

You can see the number of diagonals that we can draw from a single vertex in the figure mentioned in the body of the explanation above. With that in mind, to determine the total number of diagonals in a polygon using the following formula.

total number of diagonals = (n/2) * (n-3)

where n is the number of sides of the polygon.

So, for an octagon (n=8), we have

total number of diagonals = (n/2) * (n-3) = (8/2)*(8-3) = 4*5 = 20

I hope this helps! 🙂

Ya! Thanks for it!

Hey, Mike, I have a question on regular polygons and diagonals in general:

I noticed that you can divide a square into four equal triangles, but this is inconsistent with the (n-2)*180 formula. As well as that you can even divide a triangle into two equal right triangles with one diagonal, but this also violates the rule.

What exactly forms the backbone of this rule? Is it that with every added side (from triangle upwards) we increase the amount of triangles by one and only one? I.e., triangle has one (itself), quadrilaterals have two triangles, pentagons have three, and so on. Or is it that the diagonals within each figure must NOT intersect with each other in order for this rule to be valid?

Thanks,

Alex

Hi Alex,

The formula you refer to is to determine the sum of the interior angles of a polygon:

sum of interior angles = (n-2)*180

where n is the number of sides of the polygon.

For a square, n = 4 and we have

sum of interior angles = (4-2)*180 = 2*180 = 360˚

This is consistent with the definition of a square, which has four 90˚ angles.

With this in mind, let’s look into your question about where this formula comes from. As you’ve alluded to (n-2) indicates the number of triangles that can be formed from the given polygon. However, we have to be more specific. In reality, (n-2) is the number of triangles that can be formed by drawing line segments from 1 vertex to the other vertices of the polygon. In the case of the square, for example, only 1 such line segment can be drawn from a given vertex. This line segment divides the square into two congruent triangles. Thus (n-2) = 4-2 = 2 is consistent with this principle.

Following this idea, we cannot draw any line segments from a vertex of a triangle to the other vertices that divide the triangle into smaller triangles. This is consistent with the idea that for a triangle, (n-2) = 3-2 = 1.

Again, the principle is based on drawing line segments from 1 vertex to the other vertices of the given polygon.

I hope this clears up your doubts! 🙂

So we can make the assumptions about the diagonals being parallel to certain segments even though there is no parallel marker?

Dear Sam,

I’m happy to respond. 🙂 I want to make a distinction here: it is not an

assumption: rather, it is adeduction. If you examine carefully all the angles & the symmetries in this diagram, you will see that a certain diagonals, such as AD in the diagram above, are parallel to certain sides, here BC & FG. In much the same way, we are notassumingthat BC is parallel to FG: instead, we arededucingthat from the very nature of the octagon itself.One of the fundamental features of the regular octagon is its symmetry. For example, if we constructed the midpoint of BC, the midpoint of FG, and then connected these with a line segment, that would create one mirror line of the shape, and everything is symmetrical across that mirror line. That means that points A and D are equal distances down on respective sides, which means that AD must be perpendicular to the mirror line, just as BC & FG must be. Lines perpendicular to the same thing must be parallel, so this is why AD is parallel to both BC & FG. That’s one way to understand it.

To save you all the work of figuring it out yourself, I shared with you that the diagonal AD is parallel to the sides, as I share many geometry facts. I know these as a deductions. Therefore, you should understand them as a deductions, not an assumptions, even though you didn’t deduce them yourself.

In general, we only need parallel markers when the parallelism is an added feature, an assumption, not something we can deduce from the underlying geometry. For example, we would never draw parallel markers in a known square.

Does all this make sense?

Mike 🙂

Hi Mike,

Does this mean that AE is the only diagonal that is parallel to at least one side of the octagon, in this case HG and CD since it cuts through the centre?

Dear Michael,

First of all, I think you meant AF, not AE. AF is parallel to both CD & GH. I guess the way I would say it: the octagon has 3-vertex diagonals (such as AG & AC), 4-vertex diagonals (such as AD & AF), and 5-vertex diagonals (such as AE). Every 4-vertex diagonal is parallel to two sides, but the 3- and 5-vertex diagonals are not parallel to sides. For example, AD is parallel to BC & FG, and similarly for every other possible 4-vertex diagonal. Another way to say this is: the perpendicular bisector of any 4-vertex diagonal is also the perpendicular bisector of two sides, whereas the perpendicular bisector of a 3- or 5-vertex diagonal passes through two vertices of the octagon.

Does all this make sense?

Mike 🙂

An interesting challenge would be to figure out how many different triangles can be constructed by drawing lines that join the vertices of an octagon. The answer to this question never changes, though, so it would be a pretty poor GRE question. Nonetheless, it would be interesting…I wonder if there’s any general formula to calculate the number of triangles that can be made with n-sided polygons.

Jeff,

Yes, that’s true: that’s a good question. It depends on exactly what we mean by “different” — in particular, if congruent triangles of different orientations count as different (i.e. ABC and DEF are “different”), then any three points would determine a triangle, and the number of triangles simply would be 8C3 = 56, or nC3 for an n-sided polygon. That case is easy.

If we are counting all reflections & rotations of congruent triangles as the “same” triangle, then it gets a bit trickier. For case, I would fix one point (say A), and pick two other points — 7C2 = 21. Now, most of those are reflections: for example AFG is a reflection of ADC. Subtract the three symmetrical triangles — ADF, ACG, and ABH — that leaves us with 18. Divide that by two, so we are not counting the reflections — that’s 9 asymmetrical triangles and 3 symmetrical ones, for 12 total. That’s a procedure one could us for higher polygons. Of course, this is beyond what the GRE is going to ask.

Does all this make sense?

Mike 🙂

could you explain more why you divided the 18 by 2, please?

Dear Salib,

I’m happy to respond. 🙂 The first thing I will say, my friend, is that I think you need to draw this out yourself. Perhaps copy the diagram from this page into another document and print it. I will give you a verbal response here, but do not try to get from where you are to deeper understanding relying exclusively on words. You need to SEE geometry. On the hard copy, trace the triangles with a pencil or with your finger, so that you engage your kinesthetic intelligence as well. The human brain has many different modes for apprehending spatial relationships, and you want to bring them all to bear on the understanding of geometry.

OK, so the question I posed was about finding different triangles, that is, non-congruent triangles. If two triangles have the same shape but are in different orientations, we will not count them as different. For example Triangle ABG and Triangle DBE have the same shape, so they would not count as truly different.

I proposed fixing A as a starting point, and looking at pairs of points from there. For almost every pair of points we pick, say C & D to make triangle ACD, there’s another pair that would be a reflection over the segment AE — in this case, those other two point would be F & G, and triangle AFG is a mirror image of triangle ACD. They have the exact same shape, and are simply in different positions. If we counted those as two separate triangles, we would be over-counting, because they only should count as one shape. Of the 21 possible pairs, 3 are symmetrical, that is, each is its own reflection over the segment AE. For the other 18, there’s a one-to-one matching of pairs, each of which is the reflection of the other over AE, exactly as triangles AFG and ACD are. Each pair should count as only one shape. To eliminate all this redundancy, I divided by two, so each pair would be counted only once.

Does all this make sense?

Mike 🙂