Polygon Angles and Tessellations

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Learning goals

Derive the interior angle sum of a convex polygon by triangulation
Explain why one exterior turn per vertex totals a full turn
Calculate regular polygon angles and check whether a proposed side count is possible
Compare local angle fits with repeating tilings that cover the plane

Triangles inside a polygon

A polygon’s interior angle sum can be found by counting triangles. For a convex hexagon, draw diagonals from one vertex to the three vertices that are not its neighbors. Four triangles fill the hexagon, so their angle sums give $720^\circ$ .

The dashed diagonals stay inside the convex hexagon and divide it into four triangles.

In a convex polygon, every interior angle is less than a straight angle, and a segment joining two points in or on it stays in or on it. A regular polygon has equal sides and equal interior angles; return to Classifying Polygons for those definitions and their examples.

Each triangle has angle sum $180^\circ$ , as proved in Angles in a Triangle. At each hexagon vertex, the triangle angle or angle pieces fill exactly the polygon’s interior angle, so adding the four triangle sums counts every interior angle once.

Interior angle sum of a convex polygon#

For a convex pentagon, diagonals from one vertex make three triangles. Their angles fill the five interior angles without adding an angle at a new interior point.

S=3\times180^\circ

S=540^\circ

Now take any convex polygon with $n$ sides, where $n$ is a whole number at least $3$ . Choose a vertex $A$ . Join $A$ to each nonadjacent vertex. Convexity keeps each diagonal inside the polygon. The diagonals meet only at $A$ , so they do not cross each other and they divide the polygon without gaps or overlaps.

Every side except the two sides ending at $A$ is the opposite side of one triangle. There are $n-2$ such sides, so there are $n-2$ triangles. Their angles fill the polygon’s interior angles exactly, as in the pentagon. Nothing in this count depended on having five sides.

S=(n-2)180^\circ

This proof uses convexity. In a concave polygon, a diagonal from an arbitrarily chosen vertex can pass outside the polygon, so that fan construction cannot be assumed to work. All angle-sum calculations below concern convex polygons.

Worked example 1 Count triangles before angles

Problem. A convex octagon has eight sides. How many triangles are formed by drawing every diagonal from one vertex? Find its interior angle sum.

Solution. The two sides touching the chosen vertex are excluded from the opposite-side count. The remaining six sides give six triangles.

S=6\times180^\circ

S=1080^\circ

Six nonoverlapping triangles account for the octagon’s interior angles.

The total does not require the octagon to be regular.

Check your understanding

A convex polygon is divided into seven triangles by all the diagonals from one vertex. How many sides does it have?

5

7

9

10

Exterior angles are turns

Walking around a square requires four quarter-turns, totaling $360^\circ$ . At each corner, continue the incoming side straight ahead before turning toward the next side. The angle between that forward extension and the outgoing side is the exterior turn we use.

Follow the boundary in one direction and count exactly one turn at each vertex.

The exterior turn and the interior angle form a linear pair. If their measures are $e$ and $i$ , then

i+e=180^\circ.

For a convex polygon, all these turns go in the same direction. Choose one traversal direction and one exterior turn at every vertex, including the starting vertex when you return. Do not add both exterior angles that could be drawn beside an interior angle.

Exterior turns of a convex polygon#

A convex hexagon has six linear pairs. Together they total $1080^\circ$ . Its interior angles use $720^\circ$ of that amount, leaving

E=1080^\circ-720^\circ

E=360^\circ.

For a convex $n$ -gon, all $n$ linear pairs total $180n$ degrees. Subtract the interior sum just proved:

\begin{aligned} E&=180n^\circ-180(n-2)^\circ\\ E&=180n^\circ-180n^\circ+360^\circ\\ E&=360^\circ. \end{aligned}

This is one complete turn, independent of the number of sides. The square’s quarter-turns were one instance; the subtraction works for every convex polygon.

Worked example 2 A turn and its neighboring angle

Problem. One interior angle of a convex pentagon measures $126^\circ$ . Find the exterior turn there, then the sum of the exterior turns at the other four vertices.

The dashed extension and the outgoing side bound the exterior turn.

Solution. Subtract the interior angle from a straight angle.

e=180^\circ-126^\circ

e=54^\circ

All five turns total $360^\circ$ , so the remaining four total

360^\circ-54^\circ=306^\circ.

The other turns need not be equal.

Check your understanding

Five exterior turns of a convex hexagon measure $40^\circ$ , $55^\circ$ , $60^\circ$ , $65^\circ$ , and $70^\circ$ . What is the sixth turn?

70^\circ

110^\circ

180^\circ

430^\circ

Regular polygons and possible side counts

In a convex regular decagon, ten equal turns share a full turn. Each exterior turn is $36^\circ$ , and each interior angle is $144^\circ$ . Equal interior angles have equal supplements, so the exterior turns are equal too.

For a convex regular $n$ -gon, let $e$ be one exterior turn and $i$ one interior angle. Sharing the full turn and then taking a supplement gives

e=\frac{360^\circ}{n}.

i=180^\circ-\frac{360^\circ}{n}.

Dividing the total interior sum equally gives the same interior angle:

i=\frac{(n-2)180^\circ}{n}.

Worked example 3 Two angles of a regular polygon

Problem. Find one exterior turn and one interior angle of a convex regular $15$ -gon.

Equal interior angles give equal exterior turns.

Solution. Divide the full turn among the fifteen vertices.

e=\frac{360^\circ}{15}

e=24^\circ

Take the supplement to find the interior angle.

i=180^\circ-24^\circ

i=156^\circ

A proposed regular polygon angle must give a whole number of sides at least three. For instance, an exterior turn of $40^\circ$ gives nine sides, but $50^\circ$ gives $7.2$ sides, which cannot describe a polygon. Rounding a side count would change the angle.

Worked example 4 Which specifications are possible?

Problem. Decide whether each proposed convex regular polygon exists. If it does, find its side count.

Case A: each interior angle measures $165^\circ$ .

Case B: each interior angle measures $158^\circ$ .

Solution. Case A has exterior turn

e=180^\circ-165^\circ

e=15^\circ.

Its side count is

n=\frac{360}{15}

n=24.

This is a whole number at least three, so a convex regular $24$ -gon has the required angle.

Case B has exterior turn

e=180^\circ-158^\circ

e=22^\circ.

The required count would be

n=\frac{180}{11}.

That is not a whole number, so Case B is impossible.

Check your understanding

A convex regular polygon has exterior turn $16^\circ$ . Which conclusion is justified?

A It has

16

sides B It has

22

sides C It has

23

sides D No such polygon exists

From corners to a whole tiling

A tessellation, or tiling, covers the whole plane with no gaps and no overlapping interiors. Here a regular tessellation uses congruent copies of one convex regular polygon, meeting whole side to whole side. A small patch illustrates the repeating rule; the rule must allow continuation beyond the patch.

Four squares fit at a vertex because four right angles total $360^\circ$ . Rows and columns of squares also demonstrate a way to continue that fit throughout the plane.

A square grid repeats in both directions, pairing every boundary edge.

If $k$ congruent regular polygon corners meet at a tiling vertex, their angles must fill a full turn:

ki=360^\circ.

Here $k$ must be a whole number at least three. A total below $360^\circ$ leaves a gap, while a total above it forces overlap. This angle condition is necessary at a vertex; a vertex fit alone does not demonstrate a whole tiling.

Worked example 5 Test the corners

Problem. Each interior angle of a regular pentagon is $108^\circ$ . Compare three and four congruent regular pentagons meeting at one vertex. Find the gap or excess angle in each case. Can regular pentagons form a regular tessellation as defined here?

Solution. Three corners leave a gap.

3\times108^\circ=324^\circ

360^\circ-324^\circ=36^\circ

Three whole pentagon corners leave an unfilled sector.

Four corners require overlap.

4\times108^\circ=432^\circ

432^\circ-360^\circ=72^\circ

The dashed fourth pentagon overlaps the first near the shared vertex.

No integer number of pentagon corners totals $360^\circ$ : three are too few and four are already too many. A regular tessellation by these pentagons is impossible.

The angle test leaves only three candidates. An equilateral triangle has a $60^\circ$ interior angle, a square has $90^\circ$ , and a regular hexagon has $120^\circ$ . They require six, four, and three corners respectively.

To see that there are no other regular candidates, the interior-angle formula gives at least $60^\circ$ per corner. Thus the integer $k$ lies from three through six. Substitute each possibility into the angle condition and solve for the side count:

Corners at a vertex	Required interior angle	Required sides
$3$	$120^\circ$	$6$
$4$	$90^\circ$	$4$
$5$	$72^\circ$	$10/3$ , impossible
$6$	$60^\circ$	$3$

These remaining candidates really do tile. Squares use the grid above. Equilateral triangles fit in alternating orientations along a straight row; a matching row joins along all its exposed sides, and the same arrangement repeats.

Matching triangle rows extend the six-corner fit throughout the plane.

Regular hexagons fit in staggered columns. Along each column, horizontal sides match; neighboring columns match along the slanted sides. Repeating the same column placement fills each adjacent strip, with three $120^\circ$ corners at every vertex.

Each new hexagon column matches the previous one along complete sides.

The argument has two jobs: angle sums rule out impossible candidates, and these repeating constructions establish the three possible regular tessellations. Mixed tiles need their own angle and continuation checks.

Check your understanding

Two regular octagon corners and one square corner have angles $135^\circ$ , $135^\circ$ , and $90^\circ$ . What does their sum establish?

A The corners leave a

90^\circ

gap B The corners fit at one vertex; a whole tiling still needs a construction C Regular octagons tile by themselves D Every order of mixed regular tiles extends to a whole tiling

Polygon Angles and Tessellations

Learning goals

Triangles inside a polygon

Exterior angles are turns

Regular polygons and possible side counts

From corners to a whole tiling

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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