Polygons and Their Interior Angles

The sum of the interior angles of any convex polygon, $(n - 2) \times 180^{\circ}$ , and how to use it: unknown angles, the number of sides, and each interior angle of a regular polygon.

Before you start

A polygon is a closed plane figure with three or more straight sides that meet only at their ends: a pentagon has 5 sides, a hexagon 6, a heptagon 7, an octagon 8, a nonagon 9 and a decagon 10, and a polygon with $n$ sides is an $n$ -sided polygon. It is convex if every interior angle is less than 180°, and regular if all its sides are equal and all its interior angles are equal. A regular polygon with $n$ sides has $n$ lines of symmetry; when $n$ is even, half of them join opposite vertices (from the lesson "Polygons and Regular Polygons").
The angles of a triangle add up to 180° (∠ sum of △) (from the lesson "Triangles and Their Angles"). The angles of a quadrilateral add up to 360°, because a diagonal cuts it into two triangles (∠ sum of quad.) (from the lesson "Properties of Special Quadrilaterals"): if three angles of a quadrilateral are 80°, 95° and 110°, the fourth is $360^{\circ} - 80^{\circ} - 95^{\circ} - 110^{\circ} = 75^{\circ}$ .
The base angles of an isosceles triangle are equal (base ∠s of isos. △), and a triangle with two equal angles has the sides opposite them equal (isos. △) (from the lesson "Isosceles and Equilateral Triangles").
An equation such as $5x + 40 = 540$ is solved by doing the same to both sides: $5x = 500$ , so $x = 100$ (from the lesson "Solving Linear Equations").

The Sum of the Interior Angles

Every polygon in this lesson is convex.

The angles of a quadrilateral add up to 360° because one diagonal cuts it into two triangles. The same idea works for any convex polygon. Choose one vertex, $A$ , and draw every diagonal from it.

4 sides: 2 triangles

5 sides: 3 triangles

6 sides: 4 triangles

From $A$ a diagonal goes to every other vertex except the two next to $A$ . So a polygon with $n$ sides has $n - 3$ diagonals from one vertex, and they cut it into $n - 2$ triangles: the pentagon has 2 diagonals from $A$ and 3 triangles, the hexagon 3 diagonals and 4 triangles.

Another way to see the $n - 2$ : two of the $n$ sides meet at $A$ , and each of the other $n - 2$ sides is the far side of exactly one triangle.

The angles of these triangles fit together to make exactly the interior angles of the polygon. At $A$ the angles of all the triangles together make the interior angle at $A$ ; at each other vertex, the interior angle is one angle of a triangle, or two angles side by side. Nothing is left over, and nothing is counted twice.

Each triangle's angles add up to 180° (∠ sum of △), so the interior angles of the polygon add up to $(n - 2) \times 180^{\circ}$ .

Sum of the interior angles of a polygon

The sum of the interior angles of a convex polygon with $n$ sides is $(n - 2) \times 180^{\circ}$ .

The reason is written (∠ sum of polygon), short for "the angle sum of a polygon".

Number of sides, $n$	Number of triangles, $n - 2$	Sum of the interior angles
3	1	$1 \times 180^{\circ} = 180^{\circ}$
4	2	$2 \times 180^{\circ} = 360^{\circ}$
5	3	$3 \times 180^{\circ} = 540^{\circ}$
6	4	$4 \times 180^{\circ} = 720^{\circ}$
7	5	$5 \times 180^{\circ} = 900^{\circ}$
8	6	$6 \times 180^{\circ} = 1080^{\circ}$
9	7	$7 \times 180^{\circ} = 1260^{\circ}$
10	8	$8 \times 180^{\circ} = 1440^{\circ}$

Each extra side adds one more triangle, so the angle sum goes up by 180° each time. The first two rows are the facts you already knew: 180° for a triangle and 360° for a quadrilateral.

Example 1: The angle sum of a 12-sided polygon

A polygon has 12 sides.

(a) How many diagonals can be drawn from one of its vertices, and into how many triangles do they cut it?

(b) Find the sum of its interior angles.

Solution

(a) From one vertex there are diagonals to every other vertex except the two next to it: $12 - 3 = 9$ diagonals.

They cut the polygon into $12 - 2 = 10$ triangles.

(b) Sum of the interior angles $= (12 - 2) \times 180^{\circ} = 10 \times 180^{\circ} = 1800^{\circ}$ (∠ sum of polygon)

(a) 9 diagonals, cutting it into 10 triangles. (b) The interior angles add up to 1800°.

Example 2: The number of sides from the angle sum

(a) The interior angles of a polygon add up to 1980°. How many sides does it have?

(b) Explain why no polygon has interior angles that add up to 1000°.

Solution

(a) Let the polygon have $n$ sides. Then

$(n - 2) \times 180 = 1980$ (∠ sum of polygon)

$n - 2 = 1980 \div 180 = 11$

$n = 13$

Check: $(13 - 2) \times 180^{\circ} = 11 \times 180^{\circ} = 1980^{\circ}$ .

(b) If $(n - 2) \times 180 = 1000$ , then $n - 2 = 1000 \div 180 = 5\frac{5}{9}$ , which is not a whole number. A polygon has a whole number of sides, so $n - 2$ must be a whole number.

Another way to see it: an angle sum is always a multiple of 180°. A polygon with 7 sides has angle sum 900° and one with 8 sides has 1080°; 1000° lies between them, and no polygon has a number of sides between 7 and 8.

(a) The polygon has 13 sides. (b) $1000 \div 180$ is not a whole number, so 1000° is not $(n - 2) \times 180^{\circ}$ for any whole number $n$ .

Finding Unknown Angles

To find an unknown interior angle: count the sides, work out the angle sum $(n - 2) \times 180^{\circ}$ , and form an equation with all the interior angles. Give the reason (∠ sum of polygon), and check at the end that the angles add up to the angle sum.

Example 3: One unknown angle

Find $x$ in pentagon $ABCDE$ .

Solution

A pentagon has 5 sides, so its interior angles add up to $(5 - 2) \times 180^{\circ} = 540^{\circ}$ .

$100 + 115 + 98 + 120 + x = 540$ (∠ sum of polygon)

$433 + x = 540$

$x = 107$

Check: $100^{\circ} + 115^{\circ} + 98^{\circ} + 120^{\circ} + 107^{\circ} = 540^{\circ}$ .

$x = 107$ , so $\angle AED = 107^{\circ}$ .

Example 4: Angles given in terms of x

Find $x$ in hexagon $ABCDEF$ , and find its largest interior angle.

Solution

A hexagon has 6 sides, so its interior angles add up to $(6 - 2) \times 180^{\circ} = 720^{\circ}$ .

$(x + 30) + 2x + 125 + (x + 25) + 135 + (2x - 15) = 720$ (∠ sum of polygon)

Collect the terms in $x$ and the numbers: $6x + 300 = 720$

$6x = 420$

$x = 70$

The angles are $\angle FAB = 70^{\circ} + 30^{\circ} = 100^{\circ}$ , $\angle ABC = 2 \times 70^{\circ} = 140^{\circ}$ , $\angle BCD = 125^{\circ}$ , $\angle CDE = 70^{\circ} + 25^{\circ} = 95^{\circ}$ , $\angle DEF = 135^{\circ}$ and $\angle EFA = 2 \times 70^{\circ} - 15^{\circ} = 125^{\circ}$ .

Check: $100^{\circ} + 140^{\circ} + 125^{\circ} + 95^{\circ} + 135^{\circ} + 125^{\circ} = 720^{\circ}$ , and every angle is less than 180°, as in a convex hexagon.

$x = 70$ . The largest interior angle is $\angle ABC = 140^{\circ}$ .

Example 5: Some angles given and the rest equal

The frame of an arched window is an octagon $ABCDEFGH$ . The angles at $A$ and $B$ , on the window sill, are right angles, and the other six interior angles are all equal, each $x^{\circ}$ . Find $x$ .

Solution

An octagon has 8 sides, so its interior angles add up to $(8 - 2) \times 180^{\circ} = 1080^{\circ}$ .

$90 + 90 + 6x = 1080$ (∠ sum of polygon)

$6x = 900$

$x = 150$

Check: $90^{\circ} + 90^{\circ} + 6 \times 150^{\circ} = 180^{\circ} + 900^{\circ} = 1080^{\circ}$ .

$x = 150$ : each of the six equal angles is 150°.

Each Interior Angle of a Regular Polygon

The interior angles of a regular polygon are all equal. So to find one of them, share the angle sum equally among the $n$ angles.

The sharing needs only the angles to be equal. A rectangle that is not a square is not regular, but its four angles are equal, and each is $\frac{(4 - 2) \times 180^{\circ}}{4} = 90^{\circ}$ .

Each interior angle of a regular polygon

Each interior angle of a regular polygon with $n$ sides is $\frac{(n - 2) \times 180^{\circ}}{n}$ .

This needs all the interior angles to be equal, as they are in a regular polygon. If the angles are not all equal, the angle sum cannot be shared out equally.

Example 6: Each interior angle of a regular hexagon and a regular decagon

Find each interior angle of (a) a regular hexagon and (b) a regular decagon.

Solution

(a) A hexagon has 6 sides. Its interior angles add up to $(6 - 2) \times 180^{\circ} = 720^{\circ}$ (∠ sum of polygon).

The six angles are equal, so each is $720^{\circ} \div 6 = 120^{\circ}$ .

(b) A decagon has 10 sides. Its interior angles add up to $(10 - 2) \times 180^{\circ} = 1440^{\circ}$ (∠ sum of polygon).

The ten angles are equal, so each is $1440^{\circ} \div 10 = 144^{\circ}$ .

(a) Each interior angle of a regular hexagon is 120°. (b) Each interior angle of a regular decagon is 144°.

Worked out in the same way, each interior angle of the regular polygons you will meet most often:

Regular polygon	Sum of the interior angles	Each interior angle
Equilateral triangle	180°	$180^{\circ} \div 3 = 60^{\circ}$
Square	360°	$360^{\circ} \div 4 = 90^{\circ}$
Regular pentagon	540°	$540^{\circ} \div 5 = 108^{\circ}$
Regular hexagon	720°	$720^{\circ} \div 6 = 120^{\circ}$
Regular octagon	1080°	$1080^{\circ} \div 8 = 135^{\circ}$
Regular decagon	1440°	$1440^{\circ} \div 10 = 144^{\circ}$

In a regular polygon, a diagonal that joins two vertices with one vertex between them makes a triangle with two sides of the polygon. These sides are equal, so the triangle is isosceles, and its base angles are equal (base ∠s of isos. △).

Example 7: A diagonal of a regular pentagon

$ABCDE$ is a regular pentagon, and the diagonal $AC$ is drawn.

Find (a) $\angle ABC$ , (b) $\angle BCA$ and (c) $\angle ACD$ .

Solution

(a) Each interior angle of a regular pentagon is $\frac{(5 - 2) \times 180^{\circ}}{5} = \frac{540^{\circ}}{5} = 108^{\circ}$ (∠ sum of polygon), so $\angle ABC = 108^{\circ}$ .

(b) The sides of a regular pentagon are equal, so $BA = BC$ , and $\triangle ABC$ is isosceles with apex $B$ . Its base angles are equal: $\angle BCA = \angle BAC$ (base ∠s of isos. △).

$\angle BCA = (180^{\circ} - 108^{\circ}) \div 2 = 36^{\circ}$ (∠ sum of △)

$\angle ACD = 108^{\circ} - 36^{\circ} = 72^{\circ}$

Check: in $\triangle ABC$ , $108^{\circ} + 36^{\circ} + 36^{\circ} = 180^{\circ}$ ; at $C$ , $36^{\circ} + 72^{\circ} = 108^{\circ}$ .

(a) $\angle ABC = 108^{\circ}$ (b) $\angle BCA = 36^{\circ}$ (c) $\angle ACD = 72^{\circ}$ .

Common Mistakes

Multiplying by $n$ instead of $n - 2$ . The angle sum of a hexagon is $(6 - 2) \times 180^{\circ} = 720^{\circ}$ , not $6 \times 180^{\circ} = 1080^{\circ}$ . Test the rule on a triangle: $(3 - 2) \times 180^{\circ} = 180^{\circ}$ is right.
Forgetting the 2 when finding $n$ . If the angle sum is 1980°, then $1980 \div 180 = 11$ is $n - 2$ , so the polygon has 13 sides, not 11.
Dividing by $n$ when the angles are not all equal. Sharing the angle sum equally among the angles gives each angle only when all the angles are equal. When they are not, form an equation with the angles you are given.

Practice

Try each question, then open its solution to check your working. The figures are drawn to scale.

Practice 1: Names and angle sums

Name each polygon and find the sum of its interior angles: (a) 7 sides, (b) 9 sides, (c) 11 sides, (d) 16 sides.

ShowHide solution

(a) A heptagon: $(7 - 2) \times 180^{\circ} = 5 \times 180^{\circ} = 900^{\circ}$ (∠ sum of polygon)

(b) A nonagon: $(9 - 2) \times 180^{\circ} = 7 \times 180^{\circ} = 1260^{\circ}$

(d) A 16-sided polygon: $(16 - 2) \times 180^{\circ} = 14 \times 180^{\circ} = 2520^{\circ}$

(a) Heptagon, 900° (b) nonagon, 1260° (c) 11-sided polygon, 1620° (d) 16-sided polygon, 2520°.

Practice 2: One unknown angle in a hexagon

Find $x$ in hexagon $ABCDEF$ .

ShowHide solution

A hexagon's interior angles add up to $(6 - 2) \times 180^{\circ} = 720^{\circ}$ .

$115 + 140 + 98 + 132 + 125 + x = 720$ (∠ sum of polygon)

$610 + x = 720$

$x = 110$

Check: $115^{\circ} + 140^{\circ} + 98^{\circ} + 132^{\circ} + 125^{\circ} + 110^{\circ} = 720^{\circ}$ .

$x = 110$ , so $\angle AFE = 110^{\circ}$ .

Practice 3: Angles in terms of x in a pentagon

Find $x$ in pentagon $ABCDE$ , and find its smallest interior angle.

ShowHide solution

A pentagon's interior angles add up to $(5 - 2) \times 180^{\circ} = 540^{\circ}$ .

$2x + (x + 15) + (2x - 5) + (x + 20) + 120 = 540$ (∠ sum of polygon)

$6x + 150 = 540$

$6x = 390$

$x = 65$

The angles are $\angle EAB = 2 \times 65^{\circ} = 130^{\circ}$ , $\angle ABC = 65^{\circ} + 15^{\circ} = 80^{\circ}$ , $\angle BCD = 2 \times 65^{\circ} - 5^{\circ} = 125^{\circ}$ , $\angle CDE = 65^{\circ} + 20^{\circ} = 85^{\circ}$ and $\angle DEA = 120^{\circ}$ .

Check: $130^{\circ} + 80^{\circ} + 125^{\circ} + 85^{\circ} + 120^{\circ} = 540^{\circ}$ .

$x = 65$ . The smallest interior angle is $\angle ABC = 80^{\circ}$ .

Practice 4: The number of sides

(a) The interior angles of a polygon add up to 2160°. How many sides does the polygon have?

(b) Siti says she has drawn a polygon whose interior angles add up to 1500°. Explain why she must be wrong.

ShowHide solution

(a) $(n - 2) \times 180 = 2160$ (∠ sum of polygon)

$n - 2 = 2160 \div 180 = 12$ , so $n = 14$ .

Check: $(14 - 2) \times 180^{\circ} = 12 \times 180^{\circ} = 2160^{\circ}$ .

(b) If $(n - 2) \times 180 = 1500$ , then $n - 2 = 1500 \div 180 = 8\frac{1}{3}$ , which is not a whole number. The angle sums of polygons go 1440° (10 sides), then 1620° (11 sides): 1500° is not one of them.

(a) 14 sides. (b) $1500 \div 180$ is not a whole number, so no polygon has an angle sum of 1500°.

Practice 5: Each interior angle of a regular polygon

Find each interior angle of (a) a regular 12-sided polygon, (b) a regular 9-sided polygon, (c) a regular 60-sided polygon.

ShowHide solution

(a) $\frac{(12 - 2) \times 180^{\circ}}{12} = \frac{1800^{\circ}}{12} = 150^{\circ}$

(b) $\frac{(9 - 2) \times 180^{\circ}}{9} = \frac{1260^{\circ}}{9} = 140^{\circ}$

Check (a): $12 \times 150^{\circ} = 1800^{\circ} = 10 \times 180^{\circ}$ .

(a) 150° (b) 140° (c) 174°.

Practice 6: A stop sign

Mr Tan notices that a stop sign at a road junction is a regular octagon.

(a) Find the sum of its interior angles.

(b) Find each of its interior angles.

(c) Mr Tan imagines the amber line $AC$ , joining two corners of the sign that have the corner $B$ between them. Find $\angle BAC$ and $\angle BCA$ .

ShowHide solution

(a) $(8 - 2) \times 180^{\circ} = 6 \times 180^{\circ} = 1080^{\circ}$ (∠ sum of polygon)

(b) The 8 interior angles of a regular octagon are equal: $1080^{\circ} \div 8 = 135^{\circ}$ .

(c) In $\triangle ABC$ , the sides $BA$ and $BC$ are sides of the regular octagon, so $BA = BC$ and the triangle is isosceles, with apex $B$ . The angle between them, $\angle ABC$ , is an interior angle of the octagon, 135°.

$\angle BAC = \angle BCA$ (base ∠s of isos. △)

$\angle BAC = (180^{\circ} - 135^{\circ}) \div 2 = 22.5^{\circ}$ (∠ sum of △)

Check: $135^{\circ} + 22.5^{\circ} + 22.5^{\circ} = 180^{\circ}$ .

(a) 1080° (b) 135° (c) $\angle BAC = \angle BCA = 22.5^{\circ}$ .

Practice 7: A tile with a corner cut off

Mdm Lim cuts one corner off a square tile with a straight cut. The tile is now a pentagon $ABCDE$ with right angles at $A$ , $B$ and $E$ , and $\angle BCD = \angle CDE$ . Find $\angle BCD$ .

ShowHide solution

A pentagon's interior angles add up to $(5 - 2) \times 180^{\circ} = 540^{\circ}$ .

Let $\angle BCD = \angle CDE = x^{\circ}$ .

$90 + 90 + 90 + x + x = 540$ (∠ sum of polygon)

$270 + 2x = 540$

$2x = 270$ , so $x = 135$ .

Check: $3 \times 90^{\circ} + 2 \times 135^{\circ} = 270^{\circ} + 270^{\circ} = 540^{\circ}$ .

$\angle BCD = 135^{\circ}$ (and $\angle CDE = 135^{\circ}$ ).

Practice 8: Two diagonals of a regular pentagon

$ABCDE$ is a regular pentagon, and the diagonals $AC$ and $AD$ are drawn.

(a) Find $\angle BAC$ and $\angle EAD$ .

(b) Find $\angle CAD$ .

ShowHide solution

(a) Each interior angle of a regular pentagon is $540^{\circ} \div 5 = 108^{\circ}$ (∠ sum of polygon).

In $\triangle ABC$ , $BA = BC$ (sides of the regular pentagon), so $\angle BAC = \angle BCA = (180^{\circ} - 108^{\circ}) \div 2 = 36^{\circ}$ (base ∠s of isos. △, ∠ sum of △).

In $\triangle AED$ , $EA = ED$ , so in the same way $\angle EAD = \angle EDA = (180^{\circ} - 108^{\circ}) \div 2 = 36^{\circ}$ (base ∠s of isos. △, ∠ sum of △).

(b) The interior angle at $A$ , $\angle BAE = 108^{\circ}$ , is cut into three angles by the two diagonals:

$\angle CAD = 108^{\circ} - 36^{\circ} - 36^{\circ} = 36^{\circ}$

In the same way $\angle ACD = \angle BCD - \angle BCA = 108^{\circ} - 36^{\circ} = 72^{\circ}$ (as in Example 7).

So $\angle ACD = \angle ADC$ , and the sides opposite them are equal: $AD = AC$ (isos. △).

Check in $\triangle ACD$ : $36^{\circ} + 72^{\circ} + 72^{\circ} = 180^{\circ}$ .

(a) $\angle BAC = \angle EAD = 36^{\circ}$ (b) $\angle CAD = 36^{\circ}$ (c) $\angle ADC = 72^{\circ} = \angle ACD$ , so $AC = AD$ and $\triangle ACD$ is isosceles.

Practice 9: A honeycomb cell

Each cell of a honeycomb is a regular hexagon. Priya draws one cell, $ABCDEF$ , and the diagonal $AD$ , which joins two opposite vertices.

(a) Find each interior angle of the hexagon.

(b) $AD$ is a line of symmetry of the hexagon. Use this to find $\angle BAD$ .

ShowHide solution

(a) The angles of a hexagon add up to $(6 - 2) \times 180^{\circ} = 720^{\circ}$ (∠ sum of polygon), and in a regular hexagon they are equal: each is $720^{\circ} \div 6 = 120^{\circ}$ .

(b) $AD$ joins opposite vertices, so it is a line of symmetry. Folding along $AD$ puts $AB$ onto $AF$ , so $AD$ cuts $\angle BAF$ into two equal angles:

$\angle BAD = 120^{\circ} \div 2 = 60^{\circ}$ ( $AD$ is a line of symmetry)

(c) In the same way $AD$ cuts $\angle CDE$ into two equal angles: $\angle ADC = 120^{\circ} \div 2 = 60^{\circ}$ ( $AD$ is a line of symmetry).

Check with quadrilateral $ABCD$ : $\angle BAD + \angle ABC + \angle BCD + \angle CDA = 60^{\circ} + 120^{\circ} + 120^{\circ} + 60^{\circ} = 360^{\circ}$ (∠ sum of quad.).

(a) 120° (b) $\angle BAD = 60^{\circ}$ (c) $\angle ADC = 60^{\circ}$ ; the angles of $ABCD$ add up to 360°.

Practice 10: How many sides?

Ali draws a convex polygon. Three of its interior angles are 100° each, and all its other interior angles are 160° each. How many sides does the polygon have?

ShowHide solution

Let the polygon have $n$ sides. It has $n$ interior angles: 3 of them are 100° and the other $n - 3$ are 160°.

$3 \times 100 + 160(n - 3) = (n - 2) \times 180$ (∠ sum of polygon)

$300 + 160n - 480 = 180n - 360$

$160n - 180 = 180n - 360$

$360 - 180 = 180n - 160n$

$180 = 20n$ , so $n = 9$ .

Check: 3 angles of 100° and 6 angles of 160° give $300^{\circ} + 960^{\circ} = 1260^{\circ}$ , and $(9 - 2) \times 180^{\circ} = 1260^{\circ}$ .

The polygon has 9 sides.

Key Points

The diagonals from one vertex cut a polygon with $n$ sides into $n - 2$ triangles, so the sum of its interior angles is $(n - 2) \times 180^{\circ}$ (∠ sum of polygon).
To find $n$ from an angle sum, divide by 180° and add 2; if the division does not give a whole number, there is no such polygon.
Each interior angle of a regular polygon with $n$ sides is $\frac{(n - 2) \times 180^{\circ}}{n}$ ; sharing the angle sum equally needs all the angles to be equal.
In a regular polygon, a diagonal joining two vertices with one vertex between them makes an isosceles triangle with two sides of the polygon.
Form an equation with all the interior angles, solve it, and check that the angles add up to the angle sum.