Does the exterior angle sum really stay 360° even for a polygon with 100 sides?

Yes. The interior sum grows with n, but the exterior sum never changes. Algebraically, the n interior/exterior linear pairs total 180n, and subtracting the interior sum (n-2)180 leaves exactly 360 every time. Intuitively, walking once around any convex polygon turns you through one complete rotation no matter how many corners you round.

What's the difference between the interior angle sum and one interior angle?

The sum (n-2)· 180° counts every angle in the polygon added together. One interior angle of a **regular** polygon is that sum divided by n. For a regular octagon the sum is 1080° but each angle is 135°. Read the question carefully — 'the sum of the interior angles' and 'the measure of an interior angle' are different requests.

Can I use these formulas on a concave polygon?

The interior angle sum (n-2)· 180° still works for any simple polygon, including concave ones, as long as you count the reflex angle (greater than 180°) at each inward-pointing vertex. The clean 360° exterior sum, however, is stated for convex polygons in this course, because concave vertices require signed turns that partly cancel. Stick to convex polygons unless your teacher says otherwise.

Which formula should I start with when a problem gives me an angle and asks for the number of sides?

Start with the exterior angle. Compute e = 180° - i if you were given an interior angle, then use n = 360°/e. The numbers are smaller and 360 divides evenly by many values, so mistakes are less likely than solving (n-2)180/n = i algebraically. Both methods give the same answer, so use the algebraic version if you prefer it.

GEOM-8.1

Polygon Interior & Exterior Angle Sums

Master the polygon angle formulas: interior sum (n-2)·180°, exterior sum 360°, regular n-gon angles, and how to work backward from an angle to the number of sides.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Polygon Interior & Exterior Angle Sums, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Polygon Interior & Exterior Angle Sums free

What this lesson covers

A triangle's angles add to

180°

. A quadrilateral's add to

360°

. That pattern is not a coincidence — every convex polygon can be sliced into triangles from a single vertex, and each triangle contributes another

180°

. In this lesson you'll turn that idea into two formulas you'll use constantly: the interior angle sum

(n-2)\cdot 180°

and the surprising fact that the exterior angles of any convex polygon always add to exactly

360°

, no matter how many sides it has.

You'll also learn to run the formulas in reverse. If a regular polygon has an interior angle of

162°

, how many sides does it have? Questions like that show up on quizzes, in tiling and floor-design problems, and later whenever you need to know whether shapes fit together around a point with no gaps.

Where the Interior Angle Sum Formula Comes From

Pick one vertex of a convex polygon and draw every diagonal from that vertex. In a pentagon you get 3 triangles; in a hexagon, 4; in an octagon, 6. In general an

n

-gon splits into

n-2

triangles, because the two sides touching your chosen vertex can't be used for a diagonal. Since every triangle contributes

180°

and no angle is counted twice or left out, the interior angles of the polygon add to

S = (n-2)\cdot 180°

This is a derivation worth remembering, not just a formula to memorize. If you ever blank on the formula, sketch a hexagon, draw the diagonals from one vertex, count 4 triangles, and you have

4\cdot 180° = 720°

immediately.

Polygon	$n$	Triangles ( $n-2$ )	Interior sum
Triangle	3	1	$180°$
Quadrilateral	4	2	$360°$
Pentagon	5	3	$540°$
Hexagon	6	4	$720°$
Octagon	8	6	$1080°$
Decagon	10	8	$1440°$

Two cautions. First,

(n-2)\cdot 180°

is the sum of all the interior angles, not one angle — students go wrong by reporting

1080°

as "the angle" of an octagon. Second, the formula holds for any simple polygon, but in this course we apply it to convex polygons, where every interior angle measures less than

180°

.

The interior sum grows without bound: a 100-gon has an interior sum of

98\cdot 180° = 17{,}640°

. That is normal, not an error.

Regular Polygons: One Angle at a Time

A regular polygon is both equilateral (all sides congruent) and equiangular (all angles congruent). Because the angles are all equal, you can divide the total by

n

to get a single interior angle:

\text{each interior angle} = \frac{(n-2)\cdot 180°}{n}

For a regular pentagon that is

\frac{3\cdot 180°}{5} = 108°

. For a regular hexagon,

\frac{4\cdot 180°}{6} = 120°

— which is exactly why regular hexagons tile a honeycomb with no gaps: three of them meet at a point, and

3\cdot 120° = 360°

.

The most common error here is dividing at the wrong moment. You must find the full sum first, then divide by

n

. Writing

\frac{n-2}{n}\cdot 180°

is the same thing and is fine; writing

(n-2)\cdot \frac{180°}{n}

is also the same. What is not fine is dividing

n-2

by 180, or dividing the sum by

n-2

.

A second warning: this division only works when the polygon is regular. If a quadrilateral has angles

100°

80°

120°

, and

60°

, the sum is still

360°

, but no single angle equals

90°

. When a problem gives you a picture with different-looking angles, use the sum as an equation instead. For example, if a pentagon's angles are

x

x

x

2x

, and

2x

, then

7x = 540°

, so

x \approx 77.1°

.

Notice also that as

n

grows, each interior angle of a regular

n

-gon creeps toward

180°

but never reaches it — the polygon looks more and more like a circle.

Why Exterior Angles Always Total 360°

An exterior angle is formed by extending one side of the polygon; it is the supplement of the interior angle at that vertex. Take exactly one exterior angle at each vertex, all going the same direction around the polygon. Their sum is always

360°

for every convex polygon — triangle, hexagon, 47-gon, all the same.

Here are two ways to see it. Walking argument: imagine walking around the outside of the polygon. At each corner you turn by the exterior angle. When you return to your starting point facing your original direction, you have made exactly one full rotation,

360°

. Algebra argument: each interior/exterior pair is a linear pair summing to

180°

, so all

n

pairs total

180n

. Subtract the interior sum:

180n - (n-2)\cdot 180 = 180n - 180n + 360 = 360°

For a regular

n

-gon, all exterior angles are equal, so

\text{each exterior angle} = \frac{360°}{n}

This is usually the fastest route to a single interior angle: find

\frac{360°}{n}

, then subtract from

180°

. For a regular 12-gon, the exterior angle is

30°

, so the interior angle is

150°

— much quicker than computing

1800° \div 12

.

The biggest misconception: students assume the exterior sum grows with

n

the way the interior sum does. It does not. It is locked at

360°

forever. A related slip is taking two exterior angles at one vertex (they are vertical angles, both valid, but you count only one per vertex).

Working Backward From an Angle to the Number of Sides

Many problems hand you an angle and ask for

n

. Always convert to the exterior angle first — the arithmetic is far easier because

360

has many divisors.

If you know one exterior angle

e

of a regular polygon, then

n = \frac{360°}{e}

If you know one interior angle

i

, first find

e = 180° - i

, then divide.

Given	Step 1	Step 2	Result
Interior angle $= 162°$	$e = 180 - 162 = 18°$	$n = 360/18$	20 sides
Exterior angle $= 24°$	already have $e$	$n = 360/24$	15 sides
Interior sum $= 1620°$	$(n-2)180 = 1620$	$n - 2 = 9$	11 sides

You can also solve directly from the interior angle:

\frac{(n-2)180}{n} = 162

gives

180n - 360 = 162n

, so

18n = 360

and

n = 20

. Same answer, more algebra.

Use a reality check. The answer for

n

must be a whole number and at least 3. If a problem claims a regular polygon has an interior angle of

130°

, then

e = 50°

and

n = 7.2

— impossible, so no such regular polygon exists. Recognizing impossibility is a legitimate answer, and it's exactly the kind of question that catches students who plug in without thinking.

One more check on the interior side: if

(n-2)\cdot 180 = S

and

S

is not a multiple of

180

, no polygon has that interior sum. An interior sum of

1000°

is impossible because

1000/180

is not an integer.

Key terms

Convex polygon.: A polygon in which every interior angle measures less than $180°$ , so no vertex points inward and any segment between two interior points stays inside.
Interior angle.: An angle formed inside a polygon by two adjacent sides meeting at a vertex.
Interior angle sum.: The total of all interior angle measures of an $n$ -gon, equal to $(n-2)\cdot 180°$ .
Exterior angle.: The angle between one side of a polygon and the extension of an adjacent side; it forms a linear pair with the interior angle at that vertex.
Exterior angle sum.: The total of one exterior angle at each vertex of a convex polygon, always exactly $360°$ regardless of $n$ .
Regular polygon.: A polygon that is both equilateral and equiangular; each interior angle is $\frac{(n-2)180°}{n}$ and each exterior angle is $\frac{360°}{n}$ .
Equiangular.: Having all angles congruent. A rectangle is equiangular but not regular unless it is a square.
n-gon.: A polygon with $n$ sides; the shorthand used when the number of sides is a variable or is large enough that a special name is unhelpful.

Worked example

A regular polygon has an interior angle that measures

156°

. (a) How many sides does it have? (b) What is the sum of its interior angles? (c) A second regular polygon is placed edge to edge with the first at a shared vertex, along with one more copy of the first polygon. Could a regular polygon fill the remaining gap exactly? If so, which one?

Part (a). Convert to the exterior angle: it forms a linear pair with the interior angle, so

e = 180° - 156° = 24°

. For a regular polygon the exterior angles are all equal and total

360°

, so

n = \frac{360°}{24°} = 15.

The polygon is a regular 15-gon. Check:

\frac{(15-2)\cdot 180°}{15} = \frac{2340°}{15} = 156°

. ✓

Part (b). Use the interior sum formula with

n = 15

S = (15-2)\cdot 180° = 13\cdot 180° = 2340°.

Part (c). Two copies of the 15-gon meet at the vertex, using

2\cdot 156° = 312°

. Angles around a point total

360°

, so the gap measures

360° - 312° = 48°

. We need a regular polygon whose interior angle is

48°

. Its exterior angle would be

180° - 48° = 132°

, giving

n = \frac{360}{132} \approx 2.7

, which is not a whole number and is less than 3. So no regular polygon fits the gap exactly.

The reasoning pattern to carry forward: exterior angle first, then divide into

360

, then check that

n

is a whole number at least 3.

Practice questions

The sum of the interior angles of a convex polygon is

1440°

. How many sides does it have?

Answer: 10

Set

(n-2)\cdot 180 = 1440

. Divide both sides by

180

to get

n - 2 = 8

, so

n = 10

. A frequent mistake is stopping at

8

— that is the number of triangles the polygon splits into, not the number of sides. Always add the 2 back. Quick check: a decagon's exterior angles are

360/10 = 36°

each, so each interior angle of a regular decagon is

144°

, and

10\cdot 144° = 1440°

. ✓

Five of the six interior angles of a convex hexagon measure

110°

125°

140°

98°

, and

132°

. Find the sixth angle.

105°
115°
120°
135°

Answer: 115°

The interior sum of a hexagon is

(6-2)\cdot 180° = 720°

. The five known angles total

110 + 125 + 140 + 98 + 132 = 605

. So the missing angle is

720° - 605° = 115°

. Note that the hexagon is not regular — the angles differ — so you cannot divide

720

6

here. The sum formula works for any convex hexagon; only the per-angle division requires regularity. The answer is also under

180°

, consistent with the polygon being convex.

Elena claims she has drawn a regular polygon in which each interior angle measures

145°

. Explain, using the exterior angle sum, why her claim cannot be correct. Then find the two regular polygons whose interior angles come closest to

145°

Answer: An interior angle of 145° would require an exterior angle of 35° and n = 360/35 ≈ 10.29, which is not a whole number, so no such regular polygon exists. The nearest possibilities are the regular decagon (144°) and the regular 11-gon (about 147.27°).

Each interior angle and its exterior angle form a linear pair, so

e = 180° - 145° = 35°

. In any convex polygon the exterior angles sum to

360°

, and in a regular polygon they are all congruent, so

n\cdot 35 = 360

, giving

n = \frac{360}{35} \approx 10.29

. Because

n

counts sides, it must be a whole number, so no regular polygon has a

145°

interior angle. Testing whole numbers on either side: for

n = 10

e = 36°

and

i = 144°

; for

n = 11

e = \frac{360}{11} \approx 32.73°

and

i \approx 147.27°

. This is a good habit — whenever

360

divided by your exterior angle is not an integer, the described polygon does not exist.

FAQ

Does the exterior angle sum really stay 360° even for a polygon with 100 sides?: Yes. The interior sum grows with $n$ , but the exterior sum never changes. Algebraically, the $n$ interior/exterior linear pairs total $180n$ , and subtracting the interior sum $(n-2)180$ leaves exactly $360$ every time. Intuitively, walking once around any convex polygon turns you through one complete rotation no matter how many corners you round.
What's the difference between the interior angle sum and one interior angle?: The sum $(n-2)\cdot 180°$ counts every angle in the polygon added together. One interior angle of a regular polygon is that sum divided by $n$ . For a regular octagon the sum is $1080°$ but each angle is $135°$ . Read the question carefully — 'the sum of the interior angles' and 'the measure of an interior angle' are different requests.
Can I use these formulas on a concave polygon?: The interior angle sum $(n-2)\cdot 180°$ still works for any simple polygon, including concave ones, as long as you count the reflex angle (greater than $180°$ ) at each inward-pointing vertex. The clean $360°$ exterior sum, however, is stated for convex polygons in this course, because concave vertices require signed turns that partly cancel. Stick to convex polygons unless your teacher says otherwise.
Which formula should I start with when a problem gives me an angle and asks for the number of sides?: Start with the exterior angle. Compute $e = 180° - i$ if you were given an interior angle, then use $n = \frac{360°}{e}$ . The numbers are smaller and $360$ divides evenly by many values, so mistakes are less likely than solving $\frac{(n-2)180}{n} = i$ algebraically. Both methods give the same answer, so use the algebraic version if you prefer it.

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