Geometry

Angle sums of polygons

Every triangle’s angles add to the same number. Does every pentagon, every hexagon, every shape with straight sides have a number of its own, and if so, what is it?

You need

Triangle angle sumMath 8 QuadrilateralsGrades 3 to 5

Unlocks

Regular polygons and their anglesGeometry TessellationsGeometry Inscribed polygons and cyclic quadrilateralsGeometry

Where you are 3 in Polygons

Geometry11

Two-dimensional shapes15

Polygons3

Naming and classifying polygons

Angle sums of polygons

Tessellations

Why this exists

The angles of a triangle always add to 180^\circ, whatever its shape. That is the triangle angle sum, and it is what lets you find a third angle from the other two.

Most shapes with straight sides are not triangles. A floor tile, a field boundary, the outline of a nut on a bolt: each has four, five, six or more corners, and the same question comes up. If you know all the angles but one, is the last one forced? It is, and the reason is that every polygon can be cut into triangles. Count the triangles and you have the angle sum.

The same counting gives two more facts this lesson also covers: how many diagonals a polygon has, and what its exterior angles add to. The second has a surprise in it. The interior angles add to more and more as the sides increase, but the exterior angles always add to the same thing.

Definition

A polygon is a closed figure made of straight segments, its sides, joined end to end. Each side meets the next at a vertex, sides meet only at the vertices they share, and no two neighboring sides lie along one straight line. A polygon with n sides has n vertices and is called an n-gon; the everyday names, pentagon, hexagon and the rest, are in naming polygons.

The interior angle at a vertex is the angle between the two sides that meet there, measured on the inside of the polygon. The angle sum of a polygon is the total of its n interior angles.

A polygon is convex when every interior angle is less than 180^\circ. It is concave when at least one interior angle is more than 180^\circ, a reflex angle. A concave polygon has a notch, like the arrowhead in Figure 1, where the angle at C opens wider than a straight line.

Figure 1A concave quadrilateral. The angle at C is reflex, and the diagonal \overline{BD} lies outside the shape.

A diagonal is a segment joining two vertices that are not neighbors. In a convex polygon every diagonal lies inside. In a concave polygon a diagonal can run outside, as \overline{BD} does in Figure 1, while \overline{AC} (not drawn) lies inside.

For a convex polygon, an exterior angle at a vertex is the angle between one side and the extension of the other side past that vertex. In Figure 2 each side is extended past the corner where it ends, going round the same way, which gives one exterior angle at each vertex.

Figure 2One exterior angle at each vertex, taken going the same way round.

There are two ways to extend at each vertex, one for each side, and the two exterior angles they make are vertical angles, so they are equal. Either one forms a linear pair with the interior angle, so

\text{interior angle} + \text{exterior angle} = 180^\circ

at every vertex of a convex polygon.

Intuition

Stand at one vertex of a convex polygon and draw every diagonal you can from it. The polygon falls apart into triangles that all share that corner, like the folds of a fan.

Figure 3Three diagonals from one vertex cut a hexagon into four triangles.

In the hexagon of Figure 3 there are three diagonals from A and four triangles. Look at where the triangles’ corners sit. Every corner of every triangle is at a vertex of the hexagon, and at each vertex the triangles’ corners fit together, with no gaps and no overlaps, to make exactly the hexagon’s angle there. So the hexagon’s six angles add to the same as the four triangles’ twelve angles, which is 4 \times 180^\circ = 720^\circ.

One more side means one more triangle in the fan, so each extra side adds 180^\circ.

The exterior angles have a different picture. Walk around the polygon along its sides. At each corner you turn, and the amount you turn is the exterior angle there. When you are back where you started you face the way you began, having turned once all the way round, and a full turn is 360^\circ. That does not depend on how many corners there were.

Results

Angle sum of a quadrilateral. The interior angles of any quadrilateral add to 360^\circ.

Polygon angle sum theorem. The interior angles of an n-gon add to

(n - 2) \cdot 180^\circ

This holds for concave polygons as well as convex ones.

n	polygon	triangles	angle sum
3	triangle	1	180^\circ
4	quadrilateral	2	360^\circ
5	pentagon	3	540^\circ
6	hexagon	4	720^\circ
7	heptagon	5	900^\circ
8	octagon	6	1080^\circ
10	decagon	8	1440^\circ
12	dodecagon	10	1800^\circ

Diagonals. From one vertex of an n-gon there are n - 3 diagonals, and in a convex polygon they cut it into n - 2 triangles. Altogether an n-gon has

\frac{n(n - 3)}{2}

diagonals.

Exterior angle sum theorem. Take one exterior angle at each vertex of a convex polygon. Whatever the number of sides, they add to

360^\circ

Why it is true

The quadrilateral

Draw a diagonal that lies inside the quadrilateral. In a convex quadrilateral either diagonal will do; in a concave one, use the diagonal from the reflex vertex, like \overline{AC} in Figure 1. It cuts the quadrilateral into two triangles.

At each end of the diagonal, the quadrilateral’s angle is split into two parts, one in each triangle, and by angle addition the parts add back to the whole angle. At the other two vertices the quadrilateral’s angle is a triangle’s angle. So the four angles of the quadrilateral are made of exactly the six angles of the two triangles, and they add to 2 \times 180^\circ = 360^\circ.

Any convex polygon

From vertex A of a convex n-gon, a diagonal goes to every vertex except A itself and its two neighbors, which are joined to A by sides already. That is n - 3 diagonals. They lie inside, since the polygon is convex, and they cut it into n - 2 triangles: one between each pair of neighboring segments from A, and there are n - 1 segments from A (two sides and n - 3 diagonals) with n - 2 gaps between them.

As in the quadrilateral, the triangles’ angles fill the polygon’s angles exactly, so

\text{angle sum} = (n - 2) \cdot 180^\circ

Any polygon, convex or not

In a concave polygon the fan can fail, because a diagonal from the chosen vertex may run outside. The argument then uses a fact: every polygon with more than three sides has at least one diagonal lying completely inside it.

Here is why. Take the lowest vertex V (the leftmost of them, if several are equally low). The whole polygon lies on or above the level line through V, so the angle at V is less than 180^\circ. Call its neighbors U and W. If \overline{UW} lies inside the polygon, it is the diagonal we want. If not, some other vertex lies inside triangle UVW or on \overline{UW}; of those, take the one closest to V measured straight out from line UW, and call it Z. The line through Z parallel to UW cuts off a small triangle at V with no vertex inside it. A side that crossed \overline{VZ} would have to enter that small triangle across its third edge and, being straight and unable to cross \overline{VU} or \overline{VW}, end inside it, at a vertex. There is none, so \overline{VZ} is a diagonal lying inside.

Cut along such a diagonal. The n-gon splits into two smaller polygons, with p and q sides say. The diagonal becomes a side of both, so p + q = n + 2, and each has fewer sides than the original. If the formula is already known for polygons with fewer sides, the two pieces have angle sums (p - 2) \cdot 180^\circ and (q - 2) \cdot 180^\circ, and their angles fill the original polygon’s angles exactly as before:

(p - 2 + q - 2) \cdot 180^\circ = (n - 2) \cdot 180^\circ

Since the formula is true for triangles, it climbs from there to quadrilaterals, pentagons and every n. This way of reasoning is strong induction.

Another way to cut: a point inside

Pick a point P inside a convex polygon and join it to every vertex, as in Figure 4.

Figure 4A point inside makes n triangles, but the angles around P are not angles of the polygon.

Now there are n triangles, with angles adding to 180n degrees. But this time not every triangle’s angle is an angle of the polygon: the n angles at P go once around a point and add to 360^\circ. Take them away:

180n - 360 = 180(n - 2)

The two ways of cutting give the same formula, written differently.

Counting diagonals

Each of the n vertices is the end of n - 3 diagonals, which makes n(n - 3) ends. Every diagonal has two ends, so it was counted twice, once from each end. The number of diagonals is n(n - 3)/2.

The exterior angles

At each of the n vertices, the interior and exterior angles add to 180^\circ. Over all the vertices, the interior and exterior angles together add to 180n degrees. The interior angles account for 180(n - 2) of that, which leaves

180n - 180(n - 2) = 360

degrees for the exterior angles. The walk in the Intuition section is the same fact seen as turning.

Worked examples

Example 1

The angle sum of an octagon and of a 20-gon

Find the angle sum of an octagon, and of a polygon with 20 sides.

Solution

An octagon has n = 8 sides. A fan of diagonals from one vertex cuts it into 8 - 2 = 6 triangles, so

(8 - 2) \cdot 180^\circ = 6 \cdot 180^\circ = 1080^\circ

A 20-gon cuts into 18 triangles, and nobody needs to draw them to count them:

(20 - 2) \cdot 180^\circ = 18 \cdot 180^\circ = 3240^\circ

Example 2

The fourth angle of a quadrilateral

Three angles of a quadrilateral measure 85^\circ, 110^\circ and 72^\circ. Find the fourth, and decide whether the quadrilateral is convex.

Solution

The four angles of any quadrilateral add to 360^\circ. The three known ones add to 85 + 110 + 72 = 267, so the fourth is

360^\circ - 267^\circ = 93^\circ

All four angles, 85^\circ, 110^\circ, 72^\circ and 93^\circ, are less than 180^\circ, so the quadrilateral is convex.

Example 3

All five angles in terms of x

The five angles of the pentagon in Figure 5 are written in terms of x. Find x and each angle.

Figure 5All five angles of a pentagon, written in terms of x.

Solution

A pentagon has n = 5, so its angles add to (5 - 2) \cdot 180^\circ = 540^\circ. Add the five expressions: the x terms give 5x + 6x + 4x + 3x + 7x = 25x, and the numbers give -10 + 40 + 35 - 25 = 40. So

\begin{aligned} 25x + 40 &= 540 \\ 25x &= 500 \\ x &= 20 \end{aligned}

Substitute back: 5(20) = 100, 6(20) - 10 = 110, 4(20) + 40 = 120, 3(20) + 35 = 95 and 7(20) - 25 = 115. The angles are 100^\circ, 110^\circ, 120^\circ, 95^\circ and 115^\circ.

Check. 100 + 110 + 120 + 95 + 115 = 540.

Example 4

From the angle sum back to the sides

A polygon’s interior angles add to 1980^\circ. How many sides does it have? Could a polygon’s angles add to 1000^\circ?

Solution

Write the formula with n unknown and solve:

\begin{aligned} (n - 2) \cdot 180 &= 1980 \\ n - 2 &= 11 \\ n &= 13 \end{aligned}

The polygon has 13 sides.

For 1000^\circ the same step gives n - 2 = 1000 / 180 = 50/9, which is not a whole number. A polygon has a whole number of sides, so its angle sum is always a multiple of 180^\circ, and 1000^\circ is not one. It falls between a heptagon’s 900^\circ and an octagon’s 1080^\circ. No polygon has that angle sum.

Example 5

Counting diagonals

How many diagonals does a decagon have? Which polygon has exactly 54 diagonals?

Solution

A decagon has n = 10. From one vertex there are 10 - 3 = 7 diagonals. Counting from all ten vertices gives 10 \times 7 = 70 ends, and each diagonal has two, so a decagon has

\frac{10 \cdot 7}{2} = 35

diagonals.

For 54 diagonals, set the formula equal to 54 and clear the fraction:

\begin{aligned} \frac{n(n - 3)}{2} &= 54 \\ n^2 - 3n - 108 &= 0 \\ (n - 12)(n + 9) &= 0 \end{aligned}

A number of sides cannot be -9, so n = 12: a dodecagon. Check. 12 \cdot 9 / 2 = 54.

Example 6

The reflex angle of a dart

The quadrilateral in Figure 1 has angles of 50^\circ at A and 35^\circ at B and at D. Find x, the angle at C.

Solution

The angle sum of a quadrilateral is 360^\circ whether or not it is convex, because the diagonal \overline{AC} lies inside and cuts it into two triangles. So

\begin{aligned} 50 + 35 + 35 + x &= 360 \\ x &= 240 \end{aligned}

The angle at C is 240^\circ, a reflex angle, as the notch in the figure shows.

Notice that the angle on the outside of the notch is 360^\circ - 240^\circ = 120^\circ, which is exactly 50^\circ + 35^\circ + 35^\circ. That is true of every dart: since all four angles add to 360^\circ, the outside angle at the notch equals the other three angles together.

Example 7

Exterior angles in terms of x

The exterior angles of a convex hexagon, one at each vertex, measure x^\circ, (2x)^\circ, (2x)^\circ, (3x)^\circ, (3x)^\circ and x^\circ. Find x and the six interior angles.

Solution

One exterior angle at each vertex of a convex polygon makes 360^\circ in total, however many sides it has:

\begin{aligned} x + 2x + 2x + 3x + 3x + x &= 360 \\ 12x &= 360 \\ x &= 30 \end{aligned}

The exterior angles are 30^\circ, 60^\circ, 60^\circ, 90^\circ, 90^\circ and 30^\circ. Each interior angle forms a linear pair with its exterior angle, so it is 180^\circ minus it: 150^\circ, 120^\circ, 120^\circ, 90^\circ, 90^\circ and 150^\circ.

Check. A hexagon’s angles should add to 720^\circ. The first three angles above add to 390, the last three to 330, and 390 + 330 = 720.

Example 8

No more than three acute angles

Show that a convex polygon can have at most three acute interior angles.

Solution

Suppose a convex polygon had four acute interior angles. Each is less than 90^\circ, so the exterior angle at each of those four vertices, being 180^\circ minus it, is more than 90^\circ. Those four exterior angles alone add to more than 4 \times 90^\circ = 360^\circ.

The polygon is convex, so every interior angle is less than 180^\circ and every exterior angle is positive. The other exterior angles can only add to the total, so all the exterior angles together would come to more than 360^\circ. But they add to exactly 360^\circ. The supposition is impossible, so there are at most three acute angles.

Three is possible: a triangle with angles of 60^\circ, 60^\circ and 60^\circ, or a quadrilateral with angles of 80^\circ, 80^\circ, 80^\circ and 120^\circ, which add to 360^\circ as they must. The exterior angle sum here is not used to find a number, but to rule a whole family of shapes out.

Boundaries

The sides must not cross. A five-pointed star drawn in one stroke, with its sides crossing each other, is not a polygon in this sense. Its five point angles in the regular case are 36^\circ each, adding to 180^\circ, not 540^\circ. The formula counts on the sides meeting only at their ends.

Concave polygons keep the angle sum but not the fan. (n - 2) \cdot 180^\circ is true for every polygon, but in a concave one a diagonal from the chosen vertex can run outside, and then the fan does not cut it into triangles. Choose the diagonals with care, as in the proof above.

Exterior angles need a convex polygon. At a reflex vertex, 180^\circ minus the interior angle is negative, and there is no ordinary angle between the extended side and the next side that plays the exterior angle’s part. The 360^\circ total can be saved by counting a turn the other way as negative, but as stated here the exterior angle sum is about convex polygons.

One exterior angle per vertex. Each vertex has two exterior angles, equal to each other. Take one from each vertex; taking both doubles the total to 720^\circ.

A straight angle is not a corner. If a point on a side is called a vertex, with an angle of 180^\circ there, the count of sides goes up by one and so does the sum, by 180^\circ. That is why the definition rules out two neighboring sides on one line.

Flat surfaces only. The whole argument rests on the triangle angle sum, which holds in the plane. On a curved surface such as a sphere, a triangle’s angles add to more than 180^\circ, and none of these totals survive.

The diagonal count includes diagonals outside. In a concave polygon some diagonals lie partly or wholly outside, and n(n - 3)/2 still counts them, because a diagonal is defined by its endpoints, not by where it runs.

Where this goes next

When all the angles of a polygon are equal, the angle sum shares out evenly, and each angle of a regular n-gon is (n - 2) \cdot 180^\circ / n. Turned the other way, each exterior angle is 360^\circ / n. Both are worked out in regular polygons and their angles.

Those angles decide which shapes can cover a floor. Copies of a polygon fit around a point only if their angles there add to 360^\circ, which is why squares, equilateral triangles and regular hexagons tile and regular pentagons do not; that is the subject of Tessellations.

The quadrilateral sum of 360^\circ sits under a great deal of later work. In a parallelogram, neighboring angles add to 180^\circ (properties of a parallelogram), and in a quadrilateral drawn inside a circle, opposite angles add to 180^\circ, which is the cyclic quadrilateral theorem.

The diagonal count is also a counting fact in disguise. Choosing any two of n vertices gives n(n - 1)/2 segments, the combinations of n things taken two at a time, and taking away the n sides leaves n(n - 3)/2 diagonals.

Exercises

Angle sums

Find the angle sum of a hexagon.
Find the angle sum of a nonagon, a polygon with 9 sides.
Find the angle sum of a polygon with 100 sides.

Missing angles

Three angles of a quadrilateral measure 95^\circ, 78^\circ and 102^\circ. Find the fourth.
Four angles of a pentagon measure 120^\circ, 95^\circ, 130^\circ and 105^\circ. Find the fifth.
Five angles of a hexagon each measure 125^\circ. Find the sixth.
The angles of a quadrilateral measure x^\circ, (2x)^\circ, (3x)^\circ and (4x)^\circ. Find all four.
The angles of a pentagon measure (2x)^\circ, (x + 60)^\circ, (3x - 30)^\circ, (2x + 5)^\circ and (x + 55)^\circ. Find x and all five angles.
The angles of a hexagon are in the ratio 2 : 3 : 3 : 4 : 4 : 4. Find all six.

Working backwards

A polygon’s angles add to 2520^\circ. How many sides does it have?
Can a polygon’s angles add to 2000^\circ? Explain.
A convex polygon’s interior angles add to three times as much as its exterior angles, one at each vertex. How many sides does it have?

Diagonals

How many diagonals can be drawn from one vertex of an 11-gon, and into how many triangles do they cut it, if it is convex?
How many diagonals does an octagon have?
How many diagonals does a polygon with 15 sides have?
Which polygon has exactly 65 diagonals?
Can a polygon have exactly 30 diagonals? Explain.

Exterior angles and concave polygons

Four exterior angles of a convex pentagon, one at each of four vertices, measure 62^\circ, 80^\circ, 71^\circ and 75^\circ. Find the exterior angle at the fifth vertex.
The exterior angles of a convex quadrilateral, one at each vertex, measure (x + 10)^\circ, (2x)^\circ, (3x - 20)^\circ and (2x + 50)^\circ. Find x and the four interior angles.
A concave quadrilateral shaped like an arrowhead has three angles of 30^\circ, 35^\circ and 40^\circ. Find its fourth angle.

Reasoning

Pick a point P inside a convex n-gon and join it to every vertex. Explain why this shows the interior angles add to 180n - 360 degrees, and show that this is the same as (n - 2) \cdot 180.
Use the formula n(n - 3)/2 to count the diagonals of a triangle and of a quadrilateral. Explain from the definition why a triangle has none.
Explain why a quadrilateral cannot have four obtuse angles.
Can a quadrilateral have two reflex angles? Explain.
Explain, without the formula, why adding one more side to a polygon adds 180^\circ to its angle sum.

Answers

720^\circ. A hexagon has n = 6, and (6 - 2) \cdot 180 = 720.
1260^\circ. (9 - 2) \cdot 180 = 7 \cdot 180, which is 1260.
17640^\circ. (100 - 2) \cdot 180 = 98 \cdot 180, which is 17640.
85^\circ. The four angles add to 360^\circ; the three known ones add to 275, and 360 - 275 = 85.
90^\circ. A pentagon’s angles add to 540^\circ; the four known ones add to 450, and 540 - 450 = 90.
95^\circ. A hexagon’s angles add to 720^\circ; the five known ones add to 5 \cdot 125 = 625, and 720 - 625 = 95.
36^\circ, 72^\circ, 108^\circ and 144^\circ. From x + 2x + 3x + 4x = 360, 10x = 360 and x = 36.
x = 50, and the angles are 100^\circ, 110^\circ, 120^\circ, 105^\circ and 105^\circ. The x terms add to 9x and the numbers to 90, so 9x + 90 = 540 and 9x = 450.
72^\circ, 108^\circ, 108^\circ, 144^\circ, 144^\circ and 144^\circ. Write the angles as 2k, 3k, 3k, 4k, 4k, 4k; they add to 720^\circ, so 20k = 720 and k = 36.
16 sides. (n - 2) \cdot 180 = 2520 gives n - 2 = 14.
No. (n - 2) \cdot 180 = 2000 gives n - 2 = 100/9, not a whole number. Angle sums are multiples of 180^\circ, and 2000^\circ falls between 1980^\circ (13 sides) and 2160^\circ (14 sides).
8 sides. The exterior angles add to 360^\circ, so the interior angles add to 3 \cdot 360 = 1080 degrees, and (n - 2) \cdot 180 = 1080 gives n - 2 = 6.
8 diagonals and 9 triangles. From one vertex, 11 - 3 = 8 diagonals; they cut a convex 11-gon into 11 - 2 = 9 triangles.
20. \dfrac{8 \cdot 5}{2} = 20.
90. \dfrac{15 \cdot 12}{2} = 90.
A 13-gon. n(n - 3)/2 = 65 gives n^2 - 3n - 130 = 0, which factors as (n - 13)(n + 10) = 0, and a number of sides cannot be -10. Check: 13 \cdot 10 / 2 = 65.
No. A 9-gon has 9 \cdot 6 / 2 = 27 diagonals and a 10-gon has 10 \cdot 7 / 2 = 35, and the count only grows with n, so no polygon has exactly 30. In algebra, n^2 - 3n - 60 = 0 has no whole-number solution.
72^\circ. The five exterior angles add to 360^\circ; the four known ones add to 288, and 360 - 288 = 72.
x = 40. The exterior angles add to 360^\circ: 8x + 40 = 360, so 8x = 320. They are 50^\circ, 80^\circ, 100^\circ and 130^\circ, so the interior angles, each 180^\circ minus its exterior angle, are 130^\circ, 100^\circ, 80^\circ and 50^\circ. Check: 130 + 100 + 80 + 50 = 360.
255^\circ. The angles of any quadrilateral add to 360^\circ; the three known ones add to 105, and 360 - 105 = 255. It is the reflex angle at the notch.
The n triangles have angles adding to 180n degrees. At the vertices of the polygon their angles fill the polygon’s angles exactly, but the n angles at P are extra: they go once around a point and add to 360^\circ. Taking them away leaves 180n - 360, and 180n - 360 = 180(n - 2) by the distributive property.
A triangle: 3 \cdot 0 / 2 = 0. A quadrilateral: 4 \cdot 1 / 2 = 2, its two diagonals. In a triangle every vertex is a neighbor of both others, so there is no pair of vertices that are not neighbors, and a diagonal needs such a pair.
Obtuse angles are each more than 90^\circ, so four of them would add to more than 360^\circ. The angles of a quadrilateral add to exactly 360^\circ.
No. Two reflex angles are each more than 180^\circ, so together they are more than 360^\circ, already more than the whole angle sum, and the other two angles must be positive.
Take a polygon with n + 1 sides and cut off one vertex with the diagonal joining its two neighbors, choosing a vertex where that diagonal lies inside (in a convex polygon any vertex will do; that every polygon has such a vertex takes more proof). The piece cut off is a triangle, and what is left is a polygon with n sides. Their angles fill the angles of the original, so the larger polygon’s sum is the smaller one’s plus a triangle’s 180^\circ.