MathematicsClass 8ICSE

Walk Once Around Any Polygon and You Turn Full Circle

Learn why the exterior angles of every convex polygon total 360 degrees, calculate each angle of a regular polygon, find the number of sides from an angle, and solve ratio problems.

Why do the exterior angles of a polygon always add to 360 degrees?

Because walking once around the outside of the figure turns you through one complete revolution.

Imagine walking along the edge of a hexagonal park. At each corner you turn by a small amount and set off along the next side. After six corners you are back where you started, facing the direction you began in — so the turns must have added up to a full circle,

36 0^{\circ}

.

That is precisely what the exterior angles measure: the turn at each corner. And nothing in the argument mentioned six. Walk around a triangle, an octagon or a fifty-sided figure and you still end up facing your original direction, so the exterior angles still total

36 0^{\circ}

.

This is the strangest and most useful fact in the chapter. The interior angle sum grows without limit as sides are added, while the exterior sum never budges. This page covers the second part of the ICSE Class 8 Mathematics chapter on understanding shapes.

What exactly is an exterior angle, and how is it linked to the interior one?

An exterior angle is formed by extending one side of the polygon, and it makes a straight line with the interior angle at that vertex.

Because they sit on a straight line, at every vertex

interior angle + exterior angle = 18 0^{\circ}

That relationship is the bridge between everything in this page and everything in the previous one. Give me either angle and I can produce the other by subtracting from

18 0^{\circ}

.

Worked example 1 — verifying the total on a triangle. Take a triangle with interior angles

5 0^{\circ}

6 0^{\circ}

and

7 0^{\circ}

. The exterior angles are

180 - 50 = 13 0^{\circ}, 180 - 60 = 12 0^{\circ}, 180 - 70 = 11 0^{\circ}

Adding them:

130 + 120 + 110 = 36 0^{\circ}

. The rule holds.

Worked example 2 — verifying on a quadrilateral. Interior angles

8 0^{\circ}

10 0^{\circ}

7 0^{\circ}

and

11 0^{\circ}

total

36 0^{\circ}

, as they must. The exterior angles are

10 0^{\circ}

8 0^{\circ}

11 0^{\circ}

and

7 0^{\circ}

, which also total

36 0^{\circ}

.

**Why both totals are

36 0^{\circ}

for a quadrilateral, and never again.** For

n = 4

the interior sum happens to be

(4 - 2) \times 180 = 36 0^{\circ}

, matching the exterior sum by coincidence. For a pentagon the interior sum is

54 0^{\circ}

while the exterior stays

36 0^{\circ}

, and the gap only widens. Treating the quadrilateral case as the general rule is a mistake this topic is designed to expose.

Worked example 3 — proving the total algebraically. At each of

n

vertices the two angles sum to

18 0^{\circ}

, so

all interior + all exterior = 180 n

The interior angles total

(n - 2) \times 180 = 180 n - 360

. Subtracting:

all exterior = 180 n - (180 n - 360) = 36 0^{\circ}

The

n

cancels completely — which is the algebraic version of the walking argument, and the reason the answer cannot depend on the number of sides.

Formula

How do you find each angle of a regular polygon?

For a regular polygon with

n

sides, all the angles are equal, so each one is the total shared out evenly:

each exterior angle = \frac{36 0 ^{\circ}}{n}

each interior angle = 18 0^{\circ} - \frac{36 0 ^{\circ}}{n} = \frac{( n - 2 ) \times 18 0 ^{\circ}}{n}

Always start with the exterior angle. Dividing

360

n

is easier than dividing

(n - 2) \times 180

n

, and one subtraction then gives the interior angle. The two formulas agree, but one is far quicker to use under pressure.

Worked example 1 — a regular hexagon.

exterior = \frac{360}{6} = 6 0^{\circ}, interior = 180 - 60 = 12 0^{\circ}

Check with the other formula:

\frac{( 6 - 2 ) \times 180}{6} = \frac{720}{6} = 12 0^{\circ}

. They agree.

Worked example 2 — a regular octagon.

exterior = \frac{360}{8} = 4 5^{\circ}, interior = 13 5^{\circ}

Worked example 3 — a regular polygon with twelve sides.

exterior = \frac{360}{12} = 3 0^{\circ}, interior = 15 0^{\circ}

Its interior angles sum to

12 \times 150 = 180 0^{\circ}

, which matches

(12 - 2) \times 180 = 180 0^{\circ}

. Correct.

The values worth recognising instantly:

- Equilateral triangle: exterior

12 0^{\circ}

, interior

6 0^{\circ}

- Square: exterior

9 0^{\circ}

, interior

9 0^{\circ}

- Regular pentagon: exterior

7 2^{\circ}

, interior

10 8^{\circ}

- Regular hexagon: exterior

6 0^{\circ}

, interior

12 0^{\circ}

- Regular octagon: exterior

4 5^{\circ}

, interior

13 5^{\circ}

- Regular decagon: exterior

3 6^{\circ}

, interior

14 4^{\circ}

The square is the one case where both angles are equal, because

90 + 90 = 180

. For fewer sides the exterior angle is larger; for more sides the interior angle is larger, and the crossover happens exactly at four sides.

These formulas are for regular polygons only. An irregular hexagon has exterior angles totalling

36 0^{\circ}

, but individually they can be anything —

\frac{360}{n}

describes each one only when every corner is identical.

How do you find the number of sides when an angle is given?

**Convert to the exterior angle, then divide

36 0^{\circ}

by it.

Worked example 1 — from the exterior angle.** *Each exterior angle of a regular polygon is

2 4^{\circ}

. How many sides?*

n = \frac{360}{24} = 15

Fifteen sides. Its interior angle is

180 - 24 = 15 6^{\circ}

, and checking with the other formula:

\frac{( 15 - 2 ) \times 180}{15} = \frac{2340}{15} = 15 6^{\circ}

. Correct.

Worked example 2 — from the interior angle. *Each interior angle of a regular polygon is

16 5^{\circ}

. How many sides?*

First the exterior angle:

180 - 165 = 1 5^{\circ} \Rightarrow n = \frac{360}{15} = 24

Twenty-four sides. Check:

\frac{( 24 - 2 ) \times 180}{24} = \frac{3960}{24} = 16 5^{\circ}

. Correct.

Worked example 3 — from the angle sum. *The interior angles of a regular polygon sum to

144 0^{\circ}

, and each interior angle is asked for.*

(n - 2) \times 180 = 1440 \Rightarrow n = 10

So each interior angle is

\frac{1440}{10} = 14 4^{\circ}

, and each exterior angle is

3 6^{\circ}

.

Worked example 4 — a question with no answer. *Can a regular polygon have an interior angle of

14 5^{\circ}

?*

The exterior angle would be

180 - 145 = 3 5^{\circ}

, so

n = \frac{360}{35} = \frac{72}{7}

which is not a whole number. No such regular polygon exists. A polygon cannot have a fractional number of sides, and this is the standard trap in the topic: the arithmetic runs perfectly and the answer is still impossible.

The test to apply every time. The exterior angle of a regular polygon must divide

36 0^{\circ}

exactly. So

3 6^{\circ}

4 0^{\circ}

4 5^{\circ}

and

2 4^{\circ}

are all possible;

3 5^{\circ}

5 0^{\circ}

and

7 0^{\circ}

are not. Running that check before you commit to an answer takes one division and saves a whole question.

How do you handle ratio questions on interior and exterior angles?

**Use the fact that the two angles add to

18 0^{\circ}

**, so the ratio simply divides

18 0^{\circ}

into parts.

Worked example 1. *The interior and exterior angles of a regular polygon are in the ratio

5 : 1

. Find the number of sides.*

There are

5 + 1 = 6

parts sharing

18 0^{\circ}

, so one part is

3 0^{\circ}

interior = 5 \times 30 = 15 0^{\circ}, exterior = 3 0^{\circ}

n = \frac{360}{30} = 12

A regular twelve-sided polygon. Check:

\frac{( 12 - 2 ) \times 180}{12} = 15 0^{\circ}

, and

150 : 30

does reduce to

5 : 1

. Correct.

Worked example 2. *The ratio is

3 : 2

.*

Five parts share

18 0^{\circ}

, so one part is

3 6^{\circ}

interior = 10 8^{\circ}, exterior = 7 2^{\circ}, n = \frac{360}{72} = 5

A regular pentagon, whose angles are indeed

10 8^{\circ}

and

7 2^{\circ}

. Correct.

Worked example 3. *The ratio is

7 : 2

.*

Nine parts share

18 0^{\circ}

, so one part is

2 0^{\circ}

interior = 14 0^{\circ}, exterior = 4 0^{\circ}, n = \frac{360}{40} = 9

A regular nonagon. Check:

\frac{( 9 - 2 ) \times 180}{9} = \frac{1260}{9} = 14 0^{\circ}

. Correct.

Worked example 4 — the interior angle as a multiple. An interior angle of a regular polygon is four times its exterior angle.

This is the ratio

4 : 1

, so five parts share

18 0^{\circ}

and one part is

3 6^{\circ}

. The exterior angle is

3 6^{\circ}

and

n = 10

— a regular decagon.

**Why the ratio must always be divided into

18 0^{\circ}

, never

36 0^{\circ}

. The ratio compares the two angles at one vertex**, and at one vertex they lie on a straight line. Dividing

36 0^{\circ}

instead is the error to guard against, and it doubles every answer — giving an exterior angle of

6 0^{\circ}

in the first example, and an impossible

n = 6

for a polygon whose interior angle was supposed to be five times its exterior.

A quick plausibility test. If the ratio puts the interior angle above the exterior, the polygon has more than four sides. If they are equal, it is a square. If the exterior angle is larger, it is a triangle — and no other polygon is possible, since only

n = 3

gives an exterior angle above

9 0^{\circ}

Exam tip

Exam tip: always go through the exterior angle

**Exterior angles of any convex polygon total

36 0^{\circ}

, whatever

n

is.** Interior angles total

(n - 2) \times 18 0^{\circ}

, which grows with

n

.

At every vertex, **interior

+

exterior

= 18 0^{\circ}

.

For a regular** polygon, find the exterior angle first:

\frac{360}{n}

is one easy division, and the interior angle follows by subtraction.

**Check that the exterior angle divides

36 0^{\circ}

exactly.** An interior angle of

14 5^{\circ}

needs an exterior angle of

3 5^{\circ}

, and

\frac{360}{35}

is not a whole number — so no such regular polygon exists.

For ratio questions, divide

18 0^{\circ}

into the parts, never

36 0^{\circ}

. A ratio of

5 : 1

gives

15 0^{\circ}

and

3 0^{\circ}

, so

n = 12

.

**Do not apply

\frac{360}{n}

to an irregular polygon** — the total is still

36 0^{\circ}

, but individual angles vary.

Remember that the quadrilateral is a coincidence: both its totals are

36 0^{\circ}

, and no other polygon shares that.

Learn the common values: exterior angles

12 0^{\circ}

9 0^{\circ}

7 2^{\circ}

6 0^{\circ}

4 5^{\circ}

3 6^{\circ}

for

3

4

5

6

8

10

sides.

And verify with the second formula —

\frac{( n - 2 ) \times 180}{n}

should reproduce your interior angle exactly.

Did you know

The one angle sum that refuses to grow

Set the two sums side by side and the contrast is striking.

The interior angles of a triangle total

18 0^{\circ}

, of a hexagon

72 0^{\circ}

, of a hundred-sided figure

1764 0^{\circ}

— a number with no upper limit. The exterior angles of all three total

36 0^{\circ}

.

The walking argument explains it. However many corners you round, you finish the circuit pointing the way you set off, so the turning adds to one revolution and no more. Adding sides means more corners, but each turn is correspondingly smaller.

And that is where the formula becomes genuinely practical. A regular polygon's exterior angle is

\frac{360}{n}

, so more sides means a gentler turn at each corner:

12 0^{\circ}

for a triangle,

6 0^{\circ}

for a hexagon,

3. 6^{\circ}

for a hundred-sided figure. The turn never quite reaches zero, which is why the figure never quite becomes a circle — but it gets close enough that the difference stops being visible.

The same fact makes tiling work. Regular hexagons fit around a point with no gaps because three interior angles of

12 0^{\circ}

make exactly

36 0^{\circ}

. Regular pentagons cannot, since

10 8^{\circ}

does not divide

36 0^{\circ}

— three of them leave

3 6^{\circ}

over. Every honeycomb and every tiled floor is that arithmetic, worked out in advance.

Key takeaways

Exterior angles and regular polygons: quick revision

- The **exterior angles of any convex polygon total

36 0^{\circ}

**, independent of

n

— one full turn on a walk around the figure.
- At each vertex, **interior

+

exterior

= 18 0^{\circ}

.
- Algebraic proof**: all angles together make

180 n

, the interior ones make

180 n - 360

, so the exterior ones make

36 0^{\circ}

and

n

cancels.
- Triangle with interior angles

5 0^{\circ}, 6 0^{\circ}, 7 0^{\circ}

has exterior angles

13 0^{\circ}, 12 0^{\circ}, 11 0^{\circ}

, totalling

36 0^{\circ}

.
- For a regular polygon: each exterior angle is

\frac{36 0 ^{\circ}}{n}

and each interior angle is

18 0^{\circ} - \frac{36 0 ^{\circ}}{n} = \frac{( n - 2 ) \times 18 0 ^{\circ}}{n}

.
- Values to know: triangle

12 0^{\circ} /6 0^{\circ}

, square

9 0^{\circ} /9 0^{\circ}

, pentagon

7 2^{\circ} /10 8^{\circ}

, hexagon

6 0^{\circ} /12 0^{\circ}

, octagon

4 5^{\circ} /13 5^{\circ}

, decagon

3 6^{\circ} /14 4^{\circ}

.
- A regular twelve-sided polygon has

3 0^{\circ}

and

15 0^{\circ}

, with an interior total of

180 0^{\circ}

.
- **Finding

n

**: exterior

2 4^{\circ}

gives

n = 15

and interior

15 6^{\circ}

. Interior

16 5^{\circ}

gives exterior

1 5^{\circ}

, so

n = 24

.
- From a sum:

(n - 2) \times 180 = 1440

gives

n = 10

, interior

14 4^{\circ}

, exterior

3 6^{\circ}

.
- **The exterior angle must divide

36 0^{\circ}

exactly.** An interior angle of

14 5^{\circ}

needs

\frac{360}{35}

sides, so no such regular polygon exists.
- **Ratio questions divide

18 0^{\circ}

**, not

36 0^{\circ}

. Ratio

5 : 1

gives

15 0^{\circ} /3 0^{\circ}

and

n = 12

;

3 : 2

gives

10 8^{\circ} /7 2^{\circ}

and

n = 5

;

7 : 2

gives

14 0^{\circ} /4 0^{\circ}

and

n = 9

; interior four times exterior gives

n = 10

.
-

\frac{360}{n}

applies to regular polygons only; irregular ones still total

36 0^{\circ}

but vary vertex to vertex.
- The quadrilateral is a coincidence — both its sums are

36 0^{\circ}

, and no other polygon repeats it.
- Three hexagons meet at a point because

3 \times 12 0^{\circ} = 36 0^{\circ}

; pentagons cannot tile, since

10 8^{\circ}

does not divide

36 0^{\circ}

.

Work out the interior angle of regular polygons with

5

8

9

10

and

12

sides from memory, then list which exterior angles between

1 0^{\circ}

and

9 0^{\circ}

actually divide

36 0^{\circ}

— that short list is the answer to half the questions in this topic.

Why do the exterior angles of a polygon always add to 360 degrees?

What exactly is an exterior angle, and how is it linked to the interior one?

How do you find each angle of a regular polygon?

How do you find the number of sides when an angle is given?

How do you handle ratio questions on interior and exterior angles?

Exam tip: always go through the exterior angle

The one angle sum that refuses to grow

Exterior angles and regular polygons: quick revision

Related study notes

Ready to put this into practice?