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Interior & exterior angles of polygons

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visibility 120update 8 months agobookmarkshare

🎯 In this topic you will

Derive and use a formula for the sum of the interior angles of an $n$-sided polygon.
Calculate the interior angle of a regular polygon and solve related problems.
Recognise that the sum of the exterior angles of any polygon is $360^\circ$.
Use exterior angles of regular polygons to find unknown angles and justify your reasoning.

🧠 Key Words

regular polygon
exterior angle of a polygon

Show Definitions

regular polygon: A polygon in which all sides are the same length and all interior angles are equal.
exterior angle of a polygon: The angle formed between one side of a polygon and the extension of an adjacent side.

This is a pentagon. It has five sides and five angles. By joining vertices, you can split the pentagon into three triangles as shown. You can see that the angles of the triangles make the interior angles of the pentagon.

The sum of the interior angles of the pentagon = the sum of the angles of the three triangles
$= 3 \times 180^\circ = 540^\circ

The sum of the interior angles of any pentagon is $540^\circ$.

In a regular polygon, all the sides are the same length and all the angles are the same size. This is a regular pentagon. The sum of the five angles is $540^\circ$, so each angle of a regular pentagon is $540^\circ \div 5 = 108^\circ$.

🧠 PROBLEM-SOLVING Strategy

Interior angles of polygons

Use these steps for pentagons, hexagons, octagons, decagons, and “regular” cases.

Identify the number of sides $n$ and list all given angles (mark equal ones like $x^\circ$).
Use the interior-angle sum formula: $(n-2)\times180^\circ$.
For a missing angle in a non-regular polygon: add known angles, then subtract from the sum.
Missing$= (n-2)\times180^\circ - (\text{sum of known})$.
For a regular polygon: each interior angle is $\dfrac{(n-2)\times180^\circ}{n}$.
If angles involve variables (e.g., $x^\circ$, $(x 25)^\circ$, $(y 10)^\circ$):
write an equation “sum of the five/six… angles $=(n-2)\times180^\circ$”, solve for the variable, then evaluate any requested largest/smallest angles.
For parts that mention exterior or meeting-around-a-point facts: remember $\text{exterior angle of a regular } n\text{-gon}=\dfrac{360^\circ}{n}$ and angles around a point sum to $360^\circ$. Also, $\text{interior}=180^\circ-\text{exterior}$.
Always check your result: totals should match $(n-2)\times180^\circ$; in “regular” cases all angles should be equal.

Polygon	Sides $n$	Sum of interior angles	Each (regular)
triangle	$3$	$180^\circ$	$60^\circ$
quadrilateral	$4$	$360^\circ$	$90^\circ$
pentagon	$5$	$540^\circ$	$108^\circ$
hexagon	$6$	$720^\circ$	$120^\circ$
octagon	$8$	$1080^\circ$	$135^\circ$
decagon	$10$	$1440^\circ$	$144^\circ$

Quick patterns:$\text{pentagon sum}=540^\circ$, $\text{hexagon sum}=720^\circ$, $\text{octagon sum}=1080^\circ$, $\text{decagon sum}=1440^\circ$.

❓ EXERCISES

1. Work out the missing interior angle of this pentagon.

👀 Show answer

Sum of interior angles of any pentagon is $540^\circ$.
Given angles: $120^\circ 125^\circ 100^\circ 85^\circ = 430^\circ$.
Missing angle $= 540^\circ - 430^\circ = 110^\circ$. Answer:$110^\circ$.

2. Four angles of a pentagon are $125^\circ$ each. Work out the size of the fifth angle.

👀 Show answer

Sum is $540^\circ$. Four angles total $4\times125^\circ=500^\circ$.
Fifth angle $=540^\circ-500^\circ=40^\circ$. Answer:$40^\circ$.

3. Two angles of a pentagon are $112^\circ$ each and two angles are $90^\circ$ each. Calculate the fifth angle.

👀 Show answer

Given total $=2\times112^\circ 2\times90^\circ=224^\circ 180^\circ=404^\circ$.
Fifth angle $=540^\circ-404^\circ=136^\circ$. Answer:$136^\circ$.

a. Work out the value of $x$.

b. Work out the largest angle of the pentagon.

👀 Show answer

From the diagram, angles are $x^\circ, x^\circ, x^\circ, (x 25)^\circ$ and $90^\circ$. Sum: $x x x (x 25) 90=540$ ⇒ $4x 115=540$ ⇒ $4x=425$ ⇒ $x=106.25^\circ$.
Largest angle is $x 25=131.25^\circ$. Answers:$x=106.25^\circ$; largest angle $=131.25^\circ$.

5. The angles of a pentagon are $y^\circ$, $(y 10)^\circ$, $(y 20)^\circ$, $(y 30)^\circ$ and $(y 40)^\circ$.

a. Work out the value of $y$.

b. Work out the largest angle of the pentagon.

👀 Show answer

Sum: $y (y 10) (y 20) (y 30) (y 40)=540$ ⇒ $5y 100=540$ ⇒ $5y=440$ ⇒ $y=88^\circ$.
Largest angle $=y 40=88^\circ 40^\circ=128^\circ$. Answers:$y=88^\circ$; largest angle $=128^\circ$.

a. A hexagon has $6$ sides. Draw a hexagon.

b. By joining vertices, split the hexagon into triangles.

c. Show that the sum of the interior angles of a hexagon is $720^\circ$.

d. How big is each angle of a regular hexagon?

👀 Show answer

a. Any sketch of a $6$-sided polygon (hexagon) is acceptable.

b. From one vertex, join to all non-adjacent vertices to form $6-2=4$ triangles.

c. Using $(n-2)\times180^\circ$ with $n=6$ gives $(6-2)\times180^\circ=720^\circ$.

d. For a regular hexagon, each interior angle $= \dfrac{720^\circ}{6}=120^\circ$.

❓ EXERCISES

a. Calculate the missing angle of this hexagon.

b. Calculate the value of $x$.

👀 Show answer

a. For a hexagon, sum of interior angles is $(6-2)\times180^\circ=720^\circ$.
Given total: $132^\circ 136^\circ 125^\circ 108^\circ 110^\circ=611^\circ$.
Missing angle $=720^\circ-611^\circ=109^\circ$.

b. For a pentagon, sum is $540^\circ$. Known total: $150^\circ 150^\circ 130^\circ=430^\circ$.
Remaining for two equal angles: $540^\circ-430^\circ=110^\circ$, so $2x=110^\circ \Rightarrow x=55^\circ$.

💡 Tip

An octagon has $8$ sides. A decagon has $10$ sides.

8. Work out the sum of the interior angles of

a. an octagon b. a decagon.
Justify your answer in each case.

👀 Show answer

a. Use $(n-2)\times180^\circ$ with $n=8$: $(8-2)\times180^\circ=6\times180^\circ=1080^\circ$.

b. With $n=10$: $(10-2)\times180^\circ=8\times180^\circ=1440^\circ$.
Justification: Draw diagonals from one vertex to form $(n-2)$ triangles, each contributing $180^\circ$.

a. Copy and complete this table.

Polygon	Number of sides	Sum of interior angles
triangle	$3$	$180^\circ$
quadrilateral	$4$	$360^\circ$
pentagon	$5$	$540^\circ$
hexagon	$6$	$720^\circ$
octagon	$8$	$1080^\circ$
decagon	$10$	$1440^\circ$

b. Derive a formula for the sum of the interior angles of a polygon with $n$ sides.

c. A nonagon is a polygon with $9$ sides. Show that your formula from part $b$ gives the correct answer for the sum of the angles of a nonagon.

👀 Show answer

b. From one vertex, draw diagonals to form $(n-2)$ triangles. Each triangle totals $180^\circ$, so sum $=(n-2)\times180^\circ$.

c. With $n=9$: $(9-2)\times180^\circ=7\times180^\circ=1260^\circ$. This matches the nonagon’s interior-angle sum.

10.

a. $7$ of the interior angles of an octagon are $140^\circ$ each. Work out the eighth angle.

b. Work out the interior angle of a regular octagon.

👀 Show answer

a. Octagon sum $=1080^\circ$. Seven angles total $7\times140^\circ=980^\circ$. Eighth angle $=1080^\circ-980^\circ=100^\circ$.

b. Regular octagon interior angle $=\dfrac{1080^\circ}{8}=135^\circ$.

11. This is a regular decagon. How big is each interior angle?

👀 Show answer

Decagon sum $=(10-2)\times180^\circ=1440^\circ$.
Each interior angle (regular) $=\dfrac{1440^\circ}{10}=144^\circ$.

❓ EXERCISES

12.

a. Show that it is possible for $2$ squares and $3$ equilateral triangles to meet at one point.

👀 Show answer

Each square contributes $90^\circ$ and each equilateral triangle contributes $60^\circ$. Total around the point: $2\times90^\circ 3\times60^\circ=180^\circ 180^\circ=360^\circ$, which exactly fills the full turn, so it is possible (with equal side lengths so the edges meet).

b. Draw a different way for $2$ squares and $3$ equilateral triangles with sides the same length to meet at one point.

👀 Show answer

One arrangement is to place the polygons in the cyclic order S–T–S–T–T around the point (S = square, T = equilateral triangle). This still sums to $360^\circ$ and uses the same side length.

c. Can you work out a third way for $2$ squares and $3$ equilateral triangles to meet at one point? Give a reason for your answer.

👀 Show answer

No (up to rotation and reflection). With $2$ identical squares (S) and $3$ identical triangles (T), there are only two essentially different orders around the point: S–S–T–T–T (squares adjacent) and S–T–S–T–T (squares separated by one triangle). You cannot separate all three triangles because there are only $2$ squares to go between them, so a third essentially different arrangement is impossible.

🧠 Think like a Mathematician

A tessellation is an arrangement of shapes that completely covers a space. Squared paper is a tessellation of squares.

Tasks:

a. Draw a tessellation of equilateral triangles.
b. Draw a tessellation of regular hexagons.
c. Explain why it is not possible to draw a tessellation of regular pentagons.
d. Draw a tessellation of squares and equilateral triangles. Is there more than one way to do this?
e. Draw a tessellation using regular octagons and squares.
f. What other tessellations can you draw using regular polygons?

Tip: For part d, look at your answer to Question $12$.

Equipment: Square or triangular grid paper, ruler, pencil, eraser, colored pencils.

Method:

Choose the required regular polygon(s). Sketch one accurately on grid paper.
Translate (slide) the shape so copies fit edge-to-edge with no gaps or overlaps. Keep sides the same length.
Repeat to extend the pattern across the page. Use colors to highlight repeating units.
For mixed tilings, check the angles around each meeting point add to $360^\circ$ before repeating.

👀 Show answer

a. Yes. Each equilateral triangle has angle $60^\circ$; six meet at a point: $6\times60^\circ=360^\circ$.
b. Yes. A regular hexagon has interior angle $120^\circ$; three meet at a point: $3\times120^\circ=360^\circ$.
c. Not possible with regular pentagons alone. Interior angle is $108^\circ$; multiples are $3\times108^\circ=324^\circ$ and $4\times108^\circ=432^\circ$, neither equals $360^\circ$, so gaps/overlaps occur.
d. Yes. Squares ($90^\circ$) and equilateral triangles ($60^\circ$) can meet as $2$ squares $3$ triangles around a vertex: $2\times90^\circ 3\times60^\circ=360^\circ$. There is more than one arrangement (e.g., squares adjacent: S–S–T–T–T, or separated: S–T–S–T–T); both tessellate when repeated.
e. Use the semi-regular pattern with one square and two octagons at each vertex: $90^\circ 135^\circ 135^\circ=360^\circ$ (often denoted $4.8.8$).
f. Examples with regular polygons include the three regular tilings: triangles ($3^6$), squares ($4^4$), hexagons ($6^3$); and semi-regular tilings such as $3.3.3.4.4$ (triangles & squares), $3.6.3.6$ (triangles & hexagons), $4.6.12$, $3.12.12$, and $4.8.8$.

Here is a pentagon $ABCDE$. The exterior angles are shown. Imagine you are walking round the pentagon in an anticlockwise direction, starting and finishing at $P$. At $A$ you turn through $a^\circ$. At $B$ you turn through $b^\circ$, and so on. When you get back to $P$, you have turned through one whole turn or $360^\circ$. This shows that $a^\circ b^\circ c^\circ d^\circ e^\circ = 360^\circ$ and so the sum of the exterior angles of a pentagon is $360^\circ$.

There is nothing special about a pentagon. You can do the same for any polygon.

The sum of the exterior angles of any polygon is $360^\circ$.

📘 Worked example

The exterior angle of a regular polygon is $36^\circ$. How many sides does it have?

Answer:

All exterior angles in a regular polygon are equal and the sum of the exterior angles is $360^\circ$.
Number of sides $=\dfrac{360^\circ}{36^\circ}=10$. So the polygon has $10$ sides (a decagon).

Reasoning: Walking once around any polygon turns you through a full turn $360^\circ$. For a regular $n$-gon, each exterior angle is the same, so $n \times \text{exterior} = 360^\circ$. Hence $n = \dfrac{360^\circ}{\text{exterior}}$.

🧠 PROBLEM-SOLVING Strategy

Exterior angles & number of sides

Use these steps to find interior/exterior angles and the number of sides of a regular polygon, and to justify sums like $360^\circ$.

Identify what is given. Is it an exterior angle, an interior angle, or a piece of a diagram that reveals one of them?
Core facts.
- Sum of all exterior angles (taken in one direction) is $360^\circ$.
- For a regular $n$-gon: exterior angle $=\dfrac{360^\circ}{n}$, interior angle $=180^\circ-\dfrac{360^\circ}{n}$.
- Equivalently, if interior is known: $n=\dfrac{360^\circ}{\,180^\circ-\text{interior}\,}$.
Given an exterior angle (e.g., parts $10$, $7$): compute $n=\dfrac{360^\circ}{\text{exterior}}$, then interior $=180^\circ-\text{exterior}$.
Given an interior angle (e.g., part $8$a): first get exterior $=180^\circ-\text{interior}$, then use step 3 for $n$ if needed.
From a partial diagram (e.g., the $24^\circ$ picture): identify the marked angle as an exterior turn; then use $n=\dfrac{360^\circ}{24^\circ}=15$.
Vertices meeting around a point (tilings/arrangements like octagons & squares): the angles around the point must sum to $360^\circ$; set up an equation and solve.
Always check. Verify that $n$ is an integer and that your interior/exterior pair adds to $180^\circ$.

Exterior	Number of sides $n$	Interior (regular)
$45^\circ$	$8$	$135^\circ$
$36^\circ$	$10$	$144^\circ$
$30^\circ$	$12$	$150^\circ$
$24^\circ$	$15$	$156^\circ$
$20^\circ$	$18$	$160^\circ$
$18^\circ$	$20$	$162^\circ$
$15^\circ$	$24$	$165^\circ$

Common slips to avoid: mixing interior with exterior, or adding mixed-direction exterior angles that don’t represent the consistent “walk-around” turn.

❓ EXERCISES

a. Draw a hexagon.

b. Draw the exterior angles on your hexagon.

c. Explain why the sum of the exterior angles is $360^\circ$.

👀 Show answer

a. Any sketch of a polygon with $6$ sides (a hexagon) is correct.

b. At each vertex, extend one side and mark the outside “turning” angle between the extension and the next side, keeping the same direction around the polygon. You should have $6$ exterior angles, one at each vertex.

c. As you walk once around the polygon, the total turning is one full turn, which is $360^\circ$. Therefore the sum of the exterior angles is $360^\circ$. Equivalent argument: at each vertex, interior $ $ exterior $=180^\circ$; summing over $n$ vertices gives $n\cdot180^\circ - (n-2)\cdot180^\circ = 360^\circ$.

❓ EXERCISES

2. Work out the lettered angles in these diagrams.

👀 Show answer

The screenshot does not state all necessary relationships (e.g., which lines are parallel or straight) clearly enough to determine unique values for the lettered angles. Not enough information provided.Method hint: use straight-line sums ($180^\circ$), angles at a point ($360^\circ$), vertically opposite angles, and (if marked) parallel-line facts (corresponding/alternate/co-interior).

a. There are two exterior angles at one vertex in this diagram. Are these two angles equal? Give a reason for your answer.

b. Do these four angles add up to $360^\circ$? Give a reason for your answer.

👀 Show answer

a.No. At a vertex you can draw two different exterior angles (extending either adjacent side). They are generally not equal; only one is used when you measure the “turning” in a fixed direction. Each exterior angle is supplementary to the interior angle ($\text{interior} \text{exterior}=180^\circ$), but the two exterior angles are not equal unless the interior angle were $180^\circ$ (which cannot occur in a polygon).

b.Not necessarily. Exterior angles sum to $360^\circ$ only when they are the turning angles taken consistently in the same direction around a closed polygon. The four angles shown are not guaranteed to be such a consistent set, so their sum need not be $360^\circ$.

4. What are the exterior angles of

a. an equilateral triangle b. a square c. a regular pentagon?

👀 Show answer

Use $\text{exterior}=\dfrac{360^\circ}{n}$ for a regular $n$-gon. a.$\dfrac{360^\circ}{3}=120^\circ$. b.$\dfrac{360^\circ}{4}=90^\circ$. c.$\dfrac{360^\circ}{5}=72^\circ$.

a. What is the sum of the exterior angles of an octagon?

b. Work out the exterior angle of a regular octagon.

👀 Show answer

a. The sum of exterior angles of any polygon is $360^\circ$. b. Regular octagon exterior angle $=\dfrac{360^\circ}{8}=45^\circ$.

a. Copy and complete this table of regular polygons.

Regular polygon	Number of sides	Exterior angle
equilateral triangle	$3$	$120^\circ$
square	$4$
regular pentagon	$5$	$72^\circ$
regular hexagon
regular octagon	$8$
regular decagon

b. Derive a formula for the exterior angle of a regular polygon with $n$ sides.

c. Use your formula from part $b$ to work out the exterior angle of a regular polygon with i.$12$ sides ii.$20$ sides.

👀 Show answer

Regular polygon	Number of sides	Exterior angle
equilateral triangle	$3$	$120^\circ$
square	$4$	$90^\circ$
regular pentagon	$5$	$72^\circ$
regular hexagon	$6$	$60^\circ$
regular octagon	$8$	$45^\circ$
regular decagon	$10$	$36^\circ$

b. For a regular $n$-gon, the exterior angle is $\dfrac{360^\circ}{n}$.

c.i.$\dfrac{360^\circ}{12}=30^\circ$ ii.$\dfrac{360^\circ}{20}=18^\circ$.

7. The exterior angle of a regular polygon is $40^\circ$.

a. Work out the number of sides.

b. Write the interior angle.

👀 Show answer

a.$n=\dfrac{360^\circ}{40^\circ}=9$ sides (a nonagon).

b. Interior angle $=180^\circ-40^\circ=140^\circ$ (or $\dfrac{(n-2)\times180^\circ}{n}=\dfrac{7\times180^\circ}{9}=140^\circ$).

❓ EXERCISES

a. Work out the interior angle of a regular polygon when the exterior angle is i$30^\circ$ii$20^\circ$iii$10^\circ$.

b. Work out the number of sides of each of the regular polygons in part a.

👀 Show answer

a. Interior $=180^\circ-\text{exterior}$ → i. $180^\circ-30^\circ=150^\circ$ ii. $180^\circ-20^\circ=160^\circ$ iii. $180^\circ-10^\circ=170^\circ$.

b. For a regular $n$-gon, exterior $=\dfrac{360^\circ}{n}$ → i. $n=\dfrac{360^\circ}{30^\circ}=12$ ii. $n=\dfrac{360^\circ}{20^\circ}=18$ iii. $n=\dfrac{360^\circ}{10^\circ}=36$.

9. Here is part of a regular polygon. How many sides does the polygon have?

👀 Show answer

Exterior angle shown is $24^\circ$, so $n=\dfrac{360^\circ}{24^\circ}=15$. Answer:$15$ sides.

10. Work out the number of sides in a regular polygon when the exterior angle is

a. $45^\circ$ b. $30^\circ$ c. $18^\circ$ d. $15^\circ$.

👀 Show answer

a.$n=\dfrac{360^\circ}{45^\circ}=8$.

b.$n=\dfrac{360^\circ}{30^\circ}=12$.

c.$n=\dfrac{360^\circ}{18^\circ}=20$.

d.$n=\dfrac{360^\circ}{15^\circ}=24$.

11. Here are two regular octagons.

a. Show that the angle $ABC$ is a right angle.

b. Draw a diagram to show how four regular octagons can be arranged around a square.

👀 Show answer

a. In a regular octagon, exterior angle $=\dfrac{360^\circ}{8}=45^\circ$. At $B$, moving from side $BA$ to side $BC$ turns through two exterior angles: $45^\circ 45^\circ=90^\circ$. Hence $\angle ABC=90^\circ$ (a right angle).

b. Place a square in the centre and attach a regular octagon to each side so that they share a full edge. At each square corner the angles are $90^\circ 135^\circ 135^\circ=360^\circ$, so the arrangement fits with no gaps (this is the semi-regular tiling $4.8.8$).

12. This diagram shows part of three identical regular polygons joined together around an equilateral triangle. Work out the number of sides of each regular polygon. Show how you get your answer.

👀 Show answer

At each vertex of the central equilateral triangle the interior angle is $60^\circ$. Two identical polygons meet there as well, so around that point $2\times \text{(polygon interior)} 60^\circ = 360^\circ$ ⇒ polygon interior $=150^\circ$. For a regular $n$-gon, interior $=\dfrac{(n-2)180^\circ}{n}$. Solve $\dfrac{(n-2)180^\circ}{n}=150^\circ$ → $180n-360=150n$ → $30n=360$ → $n=12$. Each polygon is a regular dodecagon ($12$ sides).

13. Show that it is possible for the interior angle of a regular polygon to be $168^\circ$ or $170^\circ$ but it is not possible for the interior angle of a regular polygon to be $169^\circ$.

👀 Show answer

For a regular polygon, exterior angle $=180^\circ-\text{interior}$ and must equal $\dfrac{360^\circ}{n}$ for some integer $n$.

$168^\circ$ → exterior $=12^\circ$ → $n=\dfrac{360^\circ}{12^\circ}=30$ ✓
$170^\circ$ → exterior $=10^\circ$ → $n=\dfrac{360^\circ}{10^\circ}=36$ ✓
$169^\circ$ → exterior $=11^\circ$ → $n=\dfrac{360^\circ}{11^\circ}$ is not an integer ✗

Therefore $168^\circ$ and $170^\circ$ are possible interior angles of regular polygons, but $169^\circ$ is not.

📘 What we've learned

The sum of the interior angles of an $n$-gon is $(n-2)\times180^\circ$.
For a regular$n$-gon, each interior angle is $\dfrac{(n-2)\times180^\circ}{n}$.
The sum of the exterior angles of any polygon is $360^\circ$.
For a regular $n$-gon, each exterior angle is $\dfrac{360^\circ}{n}$ and it satisfies $\text{interior} \text{exterior}=180^\circ$.
To find the number of sides from a given exterior angle $E$, use $n=\dfrac{360^\circ}{E}$ (must be an integer).
We solved problems with pentagons, hexagons, octagons, decagons and used these formulas to find missing angles and largest/smallest angles.
We used “angles around a point = $360^\circ$” to justify vertex arrangements and tessellations (e.g., squares with equilateral triangles, octagons with squares).

Interior & exterior angles of polygons

🎯 In this topic you will

🧠 Key Words

🧠 PROBLEM-SOLVING Strategy

Interior angles of polygons

❓ EXERCISES

❓ EXERCISES

💡 Tip

❓ EXERCISES

🧠 Think like a Mathematician

🧠 PROBLEM-SOLVING Strategy

Exterior angles & number of sides

❓ EXERCISES

❓ EXERCISES

❓ EXERCISES

📘 What we've learned

Related Past Papers

Related Tutorials

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