Constructing Regular Polygons

Construct a regular pentagon, hexagon, octagon or decagon accurately: from its side with a protractor, inside a circle from the angles at its centre, or a regular hexagon with compasses. Then measure a diagonal or check an angle.

Before you start

A regular polygon has all its sides equal and all its angles equal. For a regular polygon with $n$ sides, each interior angle is $\frac{(n - 2) \times 180^{\circ}}{n}$ and each exterior angle is $\frac{360^{\circ}}{n}$ (ext. ∠ of regular polygon); at each vertex they add up to $180^{\circ}$ (adj. ∠s on a str. line) (Exterior Angles and Regular Polygons).
The base angles of an isosceles triangle are equal (base ∠s of isos. △), and each angle of an equilateral triangle is $60^{\circ}$ (equilateral △) (Isosceles and Equilateral Triangles).
To construct a figure is to draw it accurately with instruments: a ruler, compasses and a protractor. Sketch it first, freehand, with the given lengths and angles marked, and leave your construction arcs and circles showing (Constructing Triangles).
Measure lengths to the nearest 0.1 cm and angles to the nearest degree. Before you measure what you are asked for, check yourself: measure a given length or angle again. An expected answer here is given with a range: your measurement is accurate if it is within 0.2 cm of a length, or within $2^{\circ}$ of an angle, given here.
The drawings on this page are not to the real size: do not measure the screen. Construct on paper and measure your own drawing.

From One Side: Measure the Interior Angle

All the sides of a regular polygon are equal and all its angles are equal. So once you know its interior angle, you can construct it from one side: draw a side, measure the interior angle at its end with the protractor, draw the next side the same length along it, and go on until the polygon closes.

Regular polygon	Number of sides	Each interior angle	Each exterior angle
Pentagon	5	$\frac{(5 - 2) \times 180^{\circ}}{5} = 108^{\circ}$	$\frac{360^{\circ}}{5} = 72^{\circ}$
Hexagon	6	$\frac{(6 - 2) \times 180^{\circ}}{6} = 120^{\circ}$	$\frac{360^{\circ}}{6} = 60^{\circ}$
Octagon	8	$\frac{(8 - 2) \times 180^{\circ}}{8} = 135^{\circ}$	$\frac{360^{\circ}}{8} = 45^{\circ}$
Decagon	10	$\frac{(10 - 2) \times 180^{\circ}}{10} = 144^{\circ}$	$\frac{360^{\circ}}{10} = 36^{\circ}$

A regular polygon from its side, with a protractor

1. Find each interior angle, $\frac{(n - 2) \times 180^{\circ}}{n}$ .

2. Draw one side $AB$ the given length.

3. At $B$ , measure the interior angle $\angle ABC$ with the protractor, on the side where the polygon will be, and draw the line. Mark $C$ on it, the same length from $B$ .

4. Do the same at $C$ , then at each new vertex, always measuring the angle on the inside.

5. Join the last vertex to $A$ . This last side, and the angles at its ends, are your check: they should be the same as the others.

Example 1: A regular pentagon with a protractor

Construct a regular pentagon $ABCDE$ with sides of 4 cm. Measure the diagonal $AC$ .

Solution

Each interior angle of a regular pentagon is $108^{\circ}$ .

Step 1. Draw $AB = 4$ cm.
Step 2. At $B$ , measure $\angle ABC = 108^{\circ}$ and draw the line. Mark $C$ on it, 4 cm from $B$ .
Step 3. At $C$ , measure $\angle BCD = 108^{\circ}$ on the inside of the pentagon, and mark $D$ , 4 cm from $C$ .
Step 4. At $D$ , measure $\angle CDE = 108^{\circ}$ and mark $E$ , 4 cm from $D$ .
Step 5. Join $EA$ . Check yourself: $EA$ should measure 4 cm, and the angles at $E$ and $A$ should each measure $108^{\circ}$ . Then join $AC$ and measure it.

Small errors build up as you go round, so the last side $EA$ is your check. If it is far from 4 cm, construct the pentagon again more carefully.

$AC = 6.5$ cm (accurate if from 6.3 cm to 6.7 cm).

Example 2: A regular octagon with a protractor

Construct a regular octagon $ABCDEFGH$ with sides of 3 cm. Measure the diagonal $AE$ .

Solution

Each interior angle $= \frac{(8 - 2) \times 180^{\circ}}{8} = \frac{1080^{\circ}}{8} = 135^{\circ}$ .

Step 1. Draw $AB = 3$ cm.
Step 2. At $B$ , measure $\angle ABC = 135^{\circ}$ and draw the line. Mark $C$ on it, 3 cm from $B$ .
Step 3. At $C$ , measure $\angle BCD = 135^{\circ}$ on the inside, and mark $D$ , 3 cm from $C$ .
Step 4. Go on in the same way: measure $135^{\circ}$ inside the octagon at $D$ , $E$ , $F$ and $G$ , and mark $E$ , $F$ , $G$ and $H$ , each 3 cm from the vertex before.
Step 5. Join $HA$ . Check yourself: $HA$ should measure 3 cm, and the angles at $H$ and $A$ should each measure $135^{\circ}$ . Then join $AE$ and measure it.

A regular hexagon ( $120^{\circ}$ ) and a regular decagon ( $144^{\circ}$ ) are constructed in the same way (Practice 1 and 4).

$AE = 7.8$ cm (accurate if from 7.6 cm to 8.0 cm).

In a Circle: Angles at the Centre

Join the centre $O$ of a regular polygon to each vertex. Turned about $O$ through $\frac{1}{n}$ of a complete turn, the polygon fits onto itself, each vertex moving to where the next one was (Polygons and Regular Polygons). So $OA$ turns onto $OB$ , $OB$ onto the next line, and so on: the lines from $O$ to the vertices are all equal, and the angles between them at $O$ are all equal. The polygon is cut into $n$ isosceles triangles meeting at $O$ . Their angles at $O$ fill the whole turn round $O$ , so each is $\frac{360^{\circ}}{n}$ (∠s at a pt). (This is the same size as the exterior angle.)

So a regular polygon can be constructed inside a circle: mark angles of $\frac{360^{\circ}}{n}$ at the centre with a protractor, and the points where the lines meet the circle are the vertices.

Regular polygon	Number of sides	Each angle at the centre
Pentagon	5	$\frac{360^{\circ}}{5} = 72^{\circ}$
Hexagon	6	$\frac{360^{\circ}}{6} = 60^{\circ}$
Octagon	8	$\frac{360^{\circ}}{8} = 45^{\circ}$
Decagon	10	$\frac{360^{\circ}}{10} = 36^{\circ}$

A regular polygon in a circle, from the angles at the centre

1. Draw a circle with centre $O$ and the given radius. Draw a radius $OA$ .

2. Each angle at the centre is $\frac{360^{\circ}}{n}$ . Put the protractor's centre on $O$ and its base line along $OA$ , and mark the multiples of $\frac{360^{\circ}}{n}$ on one side of $OA$ . Then turn the protractor round to the other side of $OA$ , centre still on $O$ and base line along $OA$ , and mark them there, reading the scale that starts at 0 on $OA$ . Draw a line from $O$ through each mark to the circle. If $n$ is even, the vertex opposite $A$ is where $AO$ , carried on past $O$ , meets the circle.

3. The points where the lines meet the circle are the vertices. Join them in order round the circle.

4. Check yourself: each angle at $O$ that you did not mark from $OA$ should measure $\frac{360^{\circ}}{n}$ too, and the last side should be the same length as the others.

Example 3: A regular pentagon in a circle

Construct a regular pentagon $ABCDE$ inside a circle of radius 4 cm, using the angles at the centre. Measure the side $AB$ and $\angle ABC$ .

Solution

Each angle at the centre $= \frac{360^{\circ}}{5} = 72^{\circ}$ . Measured from $OA$ , the lines are at $72^{\circ}$ and $2 \times 72^{\circ} = 144^{\circ}$ on each side of $OA$ .

Step 1. Set the compasses to 4 cm and draw a circle with centre $O$ . Draw a radius $OA$ .
Step 2. Put the protractor on $O$ with its base line along $OA$ . Mark $72^{\circ}$ and $144^{\circ}$ above $OA$ , and draw lines from $O$ through the marks to the circle, at $B$ and $C$ .
Step 3. Turn the protractor round to the other side of $OA$ , centre still on $O$ and base line along $OA$ . Mark $72^{\circ}$ and $144^{\circ}$ below $OA$ , reading the scale that starts at 0 on $OA$ , and draw lines from $O$ to the circle, at $E$ and $D$ .
Step 4. Join $AB$ , $BC$ , $CD$ , $DE$ and $EA$ . Check yourself: $\angle COD$ , which you did not draw, should measure $72^{\circ}$ , and $EA$ should be the same length as $AB$ . Then measure $AB$ and $\angle ABC$ .

Measure every angle from $OA$ , not from the line before. Then an error of $1^{\circ}$ in one line is not carried on to the next.

$AB = 4.7$ cm (accurate if from 4.5 cm to 4.9 cm) and $\angle ABC = 108^{\circ}$ (accurate if from $106^{\circ}$ to $110^{\circ}$ ), the interior angle of a regular pentagon.

Why the pentagon is regular

$OA = OB = OC = OD = OE = 4$ cm (radii of the circle), and every angle at $O$ is $72^{\circ}$ : the four you drew, and $\angle COD = 360^{\circ} - 4 \times 72^{\circ} = 72^{\circ}$ (∠s at a pt).

Turn the figure through $72^{\circ}$ about $O$ . Every radius is 4 cm and every angle at $O$ is $72^{\circ}$ , so $OA$ turns onto $OB$ , $OB$ onto $OC$ , $OC$ onto $OD$ , $OD$ onto $OE$ and $OE$ onto $OA$ . Each vertex lands where the next one was, so each side turns onto the next, and a turn does not change a length: $AB = BC = CD = DE = EA$ .

Each of the five triangles $OAB$ , $OBC$ , $OCD$ , $ODE$ and $OEA$ is isosceles (two radii), so its base angles are equal: each is $\frac{180^{\circ} - 72^{\circ}}{2} = 54^{\circ}$ (base ∠s of isos. △, ∠ sum of △). At each vertex two base angles meet, so every angle of the pentagon is $54^{\circ} + 54^{\circ} = 108^{\circ}$ .

All the sides are equal and all the angles are equal, so the pentagon is regular. The same reasons hold for any number of sides.

A regular octagon is constructed in the same way with angles of $45^{\circ}$ at the centre (Practice 6), and a regular decagon with angles of $36^{\circ}$ (Practice 7).

The Regular Hexagon with Compasses

A regular hexagon can also be constructed without a protractor: with compasses, inside a circle whose radius is the length of a side, and a ruler to join the points.

Example 4: A regular hexagon in a circle

Construct a regular hexagon $ABCDEF$ with sides of 4 cm, using compasses and a ruler. Measure the diagonal $AC$ .

Solution

Step 1. Set the compasses to 4 cm, the length of a side. Draw a circle with centre $O$ and mark a point $A$ on it.
Step 2. Keep the compasses at 4 cm. With centre $A$ , draw an arc cutting the circle at $B$ .
Step 3. With centre $B$ , draw an arc cutting the circle at $C$ . Go on round the circle in the same way to mark $D$ , $E$ and $F$ .
Step 4. Check yourself: an arc with centre $F$ should pass through $A$ . If it misses, the compasses have slipped: start again.
Step 5. With the ruler, join $AB$ , $BC$ , $CD$ , $DE$ , $EF$ and $FA$ . Then join $AC$ and measure it.

$AC = 6.9$ cm (accurate if from 6.7 cm to 7.1 cm).

Why the hexagon closes, and why it is regular

$OA = OB = 4$ cm (radii of the circle) and $AB = 4$ cm (the arc), so $\triangle OAB$ is equilateral and $\angle AOB = 60^{\circ}$ (equilateral △).

Each step round the circle makes another equilateral triangle with a $60^{\circ}$ angle at $O$ , and $6 \times 60^{\circ} = 360^{\circ}$ . So exactly six steps take you round the circle and back to $A$ .

The hexagon is six equilateral triangles meeting at the centre. Every side is 4 cm, and each interior angle is made of two $60^{\circ}$ angles: $60^{\circ} + 60^{\circ} = 120^{\circ}$ . All sides and all angles are equal, so the hexagon is regular.

This works only for the hexagon: its angle at the centre is $60^{\circ}$ , so its triangles are equilateral and its side equals the radius. For any other regular polygon the side is not the radius, so use the angles at the centre.

Measuring and Checking Afterwards

Once a regular polygon is constructed, you can measure a diagonal, or measure an angle and check it by reasoning. If the measurement and the reasoning do not agree within $2^{\circ}$ , look again at your construction.

Which method to use: given the side, measure the interior angle at each vertex; given the circle (its radius), mark the angles at the centre; for a regular hexagon, the side equals the radius, so compasses and a ruler will do.

Example 5: Checking an angle

In the regular pentagon $ABCDE$ of Example 1, join the diagonal $AC$ . Measure $\angle ACD$ , then check your measurement by reasoning.

Solution

Measured: $\angle ACD = 72^{\circ}$ .

Check by reasoning. $AB = BC$ (sides of a regular pentagon), so $\triangle ABC$ is isosceles and $\angle BCA = \angle BAC$ (base ∠s of isos. △).

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

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

The measurement agrees with the reasoning.

$\angle ACD = 72^{\circ}$ (accurate if from $70^{\circ}$ to $74^{\circ}$ ), as the reasoning shows.

Common mistakes

Reading the wrong scale of the protractor. For a regular pentagon you need $108^{\circ}$ ; the other scale gives $72^{\circ}$ (as $180^{\circ} - 108^{\circ} = 72^{\circ}$ ), and the pentagon will not close. An obtuse angle must look obtuse.
Measuring the interior angle on the outside. At each vertex the angle goes on the side where the polygon is being built. On the wrong side the sides zigzag and never close.
Putting the interior angle at the centre. The angles at the centre are $\frac{360^{\circ}}{n}$ : $72^{\circ}$ for a pentagon, not $108^{\circ}$ .
Measuring each angle at the centre from the line before. Measure every one from $OA$ ( $72^{\circ}$ , $144^{\circ}$ , …), so that one small error is not carried round the circle.
Stepping round the circle with the radius for a polygon that is not a hexagon. Only a regular hexagon has its side equal to the radius. Six steps of the radius always give a hexagon.
Changing the compasses while stepping round the circle for a regular hexagon. The radius must stay equal to the side all the way round, and the last arc must land on $A$ .
Rubbing out the circle and the arcs. They show how each vertex was found. Leave them showing.

Practice

For each question: sketch the polygon with the given data, construct it on paper with your instruments, leaving the arcs and circles showing, check yourself as the question says, and then measure.

A measurement within 0.2 cm of the length, or within $2^{\circ}$ of the angle, in the answer is accurate.

Practice 1

Construct a regular hexagon $ABCDEF$ with sides of 3.5 cm, using a protractor. Check yourself: the last side $FA$ should measure 3.5 cm. Then measure the diagonal $AD$ .

ShowHide solution

Each interior angle $= \frac{(6 - 2) \times 180^{\circ}}{6} = \frac{720^{\circ}}{6} = 120^{\circ}$ .

Step 1. Draw $AB = 3.5$ cm.
Step 2. At $B$ measure $\angle ABC = 120^{\circ}$ inside the hexagon and mark $C$ , 3.5 cm from $B$ . Do the same at $C$ , $D$ and $E$ to mark $D$ , $E$ and $F$ .
Step 3. Join $FA$ . Check $FA = 3.5$ cm, then join $AD$ and measure it.

Each interior angle is $120^{\circ}$ , and $AD = 7.0$ cm (accurate if from 6.8 cm to 7.2 cm).

Practice 2

Construct a regular pentagon $ABCDE$ with sides of 5 cm. Check yourself: the last side $EA$ should measure 5 cm. Then measure the diagonal $BE$ .

ShowHide solution

Step 1. Each interior angle of a regular pentagon is $108^{\circ}$ . Draw $AB = 5$ cm.
Step 2. At $B$ measure $108^{\circ}$ inside the pentagon and mark $C$ , 5 cm from $B$ . Do the same at $C$ and $D$ to mark $D$ and $E$ .
Step 3. Join $EA$ . Check $EA = 5$ cm, then join $BE$ and measure it.

$BE = 8.1$ cm (accurate if from 7.9 cm to 8.3 cm).

Practice 3

Priya is making a model of a stop sign, a regular octagon $ABCDEFGH$ with sides of 2.5 cm, with $AB$ along the bottom. First find each interior angle. Construct the octagon. Check yourself: the last side $HA$ should measure 2.5 cm. Then measure $AF$ , the height of her model.

ShowHide solution

Each interior angle $= \frac{(8 - 2) \times 180^{\circ}}{8} = 135^{\circ}$ .

Step 1. Draw $AB = 2.5$ cm.
Step 2. At $B$ measure $135^{\circ}$ and mark $C$ , 2.5 cm from $B$ . Go on round in the same way, measuring $135^{\circ}$ inside the octagon at $C$ , $D$ , $E$ , $F$ and $G$ , to mark $D$ , $E$ , $F$ , $G$ and $H$ .
Step 3. Join $HA$ . Check $HA = 2.5$ cm, then join $AF$ and measure it.

Each interior angle is $135^{\circ}$ , and $AF = 6.0$ cm (accurate if from 5.8 cm to 6.2 cm).

Practice 4

Construct a regular decagon $ABCDEFGHIJ$ with sides of 2 cm. First find each interior angle. Check yourself: the last side $JA$ should measure 2 cm. Then measure the diagonal $AF$ .

ShowHide solution

Each interior angle $= \frac{(10 - 2) \times 180^{\circ}}{10} = \frac{1440^{\circ}}{10} = 144^{\circ}$ .

Step 1. Draw $AB = 2$ cm.
Step 2. At $B$ measure $144^{\circ}$ inside the decagon and mark $C$ , 2 cm from $B$ . Go on round in the same way at $C$ , $D$ , …, $I$ to mark $D$ , $E$ , …, $J$ .
Step 3. Join $JA$ . Check $JA = 2$ cm, then join $AF$ and measure it.

Each interior angle is $144^{\circ}$ , and $AF = 6.5$ cm (accurate if from 6.3 cm to 6.7 cm).

Practice 5

Construct a regular pentagon $ABCDE$ inside a circle of radius 5 cm, using the angles at the centre $O$ . Check yourself: $\angle COD$ , which you do not draw, should measure $72^{\circ}$ . Then measure the side $AB$ and $\angle ABC$ .

ShowHide solution

Step 1. Each angle at the centre is $\frac{360^{\circ}}{5} = 72^{\circ}$ . Draw a circle of radius 5 cm with centre $O$ , and a radius $OA$ .
Step 2. With the protractor on $O$ and its base line along $OA$ , mark $72^{\circ}$ and $144^{\circ}$ on each side of $OA$ . Draw lines from $O$ to the circle: $B$ and $C$ on one side, $E$ and $D$ on the other.
Step 3. Join $AB$ , $BC$ , $CD$ , $DE$ and $EA$ . Check $\angle COD = 72^{\circ}$ , then measure $AB$ and $\angle ABC$ .

$AB = 5.9$ cm (accurate if from 5.7 cm to 6.1 cm) and $\angle ABC = 108^{\circ}$ (accurate if from $106^{\circ}$ to $110^{\circ}$ ).

Practice 6

Construct a regular octagon $ABCDEFGH$ inside a circle of radius 4 cm, using the angles at the centre $O$ . Check yourself: $\angle DOE$ should measure $45^{\circ}$ . Then measure the side $AB$ and $\angle ABC$ .

ShowHide solution

Step 1. Each angle at the centre is $\frac{360^{\circ}}{8} = 45^{\circ}$ . Draw a circle of radius 4 cm with centre $O$ , and a radius $OA$ .
Step 2. With the protractor along $OA$ , mark $45^{\circ}$ , $90^{\circ}$ and $135^{\circ}$ on each side of $OA$ , and draw lines from $O$ to the circle: $B$ , $C$ , $D$ on one side and $H$ , $G$ , $F$ on the other. $E$ is where $AO$ , carried on past $O$ , meets the circle ( $180^{\circ}$ from $OA$ ).
Step 3. Join the eight points in order. Check $\angle DOE = 45^{\circ}$ , then measure $AB$ and $\angle ABC$ .

$AB = 3.1$ cm (accurate if from 2.9 cm to 3.3 cm) and $\angle ABC = 135^{\circ}$ (accurate if from $133^{\circ}$ to $137^{\circ}$ ).

Practice 7

Ali is designing a badge for his school's CCA: a regular decagon $ABCDEFGHIJ$ inside a circle of radius 3.5 cm. Construct it, using the angles at the centre $O$ . Check yourself: $\angle EOF$ should measure $36^{\circ}$ . Then measure the side $AB$ and the diagonal $AC$ .

ShowHide solution

Step 1. Each angle at the centre is $\frac{360^{\circ}}{10} = 36^{\circ}$ . Draw a circle of radius 3.5 cm with centre $O$ , and a radius $OA$ .
Step 2. With the protractor along $OA$ , mark $36^{\circ}$ , $72^{\circ}$ , $108^{\circ}$ and $144^{\circ}$ on each side of $OA$ , and draw lines from $O$ to the circle: $B$ , $C$ , $D$ , $E$ on one side and $J$ , $I$ , $H$ , $G$ on the other. $F$ is where $AO$ , carried on past $O$ , meets the circle.
Step 3. Join the ten points in order. Check $\angle EOF = 36^{\circ}$ , then measure $AB$ , and join and measure $AC$ .

$AB = 2.2$ cm (accurate if from 2.0 cm to 2.4 cm) and $AC = 4.1$ cm (accurate if from 3.9 cm to 4.3 cm).

Practice 8

Siti is drawing a honeycomb pattern whose cells are regular hexagons with sides of 5 cm.

(a) Construct one cell, a regular hexagon $ABCDEF$ , in a circle with compasses and a ruler.

(b) Without measuring, find $\angle ABC$ and give the reason.

ShowHide solution

(a)

Step 1. Set the compasses to 5 cm and draw a circle with centre $O$ . Mark $A$ on it.
Step 2. Keeping the compasses at 5 cm, step round the circle: an arc with centre $A$ marks $B$ , an arc with centre $B$ marks $C$ , and so on to $F$ . Join the six points in order with the ruler.

(b) $OA = OB = AB = 5$ cm, so $\triangle OAB$ is equilateral and $\angle OBA = 60^{\circ}$ (equilateral △). In the same way $\triangle OBC$ is equilateral and $\angle OBC = 60^{\circ}$ . So $\angle ABC = 60^{\circ} + 60^{\circ} = 120^{\circ}$ , the interior angle of a regular hexagon.

(b) $\angle ABC = 120^{\circ}$ . (c) $AC = 8.7$ cm (accurate if from 8.5 cm to 8.9 cm).

Practice 9

Wei Ming is drawing a six-pointed star for a class art project. He constructs a regular hexagon $ABCDEF$ in a circle of radius 3.5 cm with compasses and a ruler, then joins $A$ , $C$ and $E$ to make one triangle, and $B$ , $D$ and $F$ to make another.

(a) Check yourself: an arc with centre $F$ should pass through $A$ . Then measure $AC$ and $\angle CAE$ .

(b) What kind of triangle is $ACE$ ? Give the reason.

ShowHide solution

Step 1. Set the compasses to 3.5 cm, draw a circle with centre $O$ and mark $A$ on it. Step round the circle with the same radius to mark $B$ , $C$ , $D$ , $E$ and $F$ , and check that the arc with centre $F$ passes through $A$ .
Step 2. Join $AC$ , $CE$ and $EA$ , then $BD$ , $DF$ and $FB$ . Measure $AC$ and $\angle CAE$ .

(b) Each interior angle of a regular hexagon is $120^{\circ}$ . $AB = BC$ (sides of a regular hexagon), so in $\triangle ABC$ , $\angle BCA = \frac{180^{\circ} - 120^{\circ}}{2} = 30^{\circ}$ (base ∠s of isos. △, ∠ sum of △). In the same way, $CD = DE$ , so $\angle DCE = 30^{\circ}$ .

$\angle ACE = \angle BCD - \angle BCA - \angle DCE = 120^{\circ} - 30^{\circ} - 30^{\circ} = 60^{\circ}$ . In the same way, $\angle CEA = 60^{\circ}$ and $\angle EAC = 60^{\circ}$ .

All three angles are $60^{\circ}$ , so $\triangle ACE$ is equilateral (equilateral △), which agrees with the measured $\angle CAE$ .

(a) $AC = 6.1$ cm (accurate if from 5.9 cm to 6.3 cm) and $\angle CAE = 60^{\circ}$ (accurate if from $58^{\circ}$ to $62^{\circ}$ ). (b) $\triangle ACE$ is equilateral.

Practice 10

Construct a regular hexagon $ABCDEF$ with sides of 4.5 cm, by either method. Check yourself: the last side $FA$ should measure 4.5 cm. Join $AC$ and $AD$ . Measure $AD$ and $\angle ACD$ , then check $\angle ACD$ by reasoning.

ShowHide solution

Step 1. With compasses set to 4.5 cm, draw a circle and step round it to mark $A$ to $F$ (the side equals the radius), or draw $AB = 4.5$ cm and measure $120^{\circ}$ inside the hexagon at each vertex.
Step 2. Join the six points. Check $FA = 4.5$ cm, then join $AC$ and $AD$ , and measure $AD$ and $\angle ACD$ .

Check by reasoning. $AB = BC$ (sides of a regular hexagon) and $\angle ABC = 120^{\circ}$ , so $\angle BCA = \frac{180^{\circ} - 120^{\circ}}{2} = 30^{\circ}$ (base ∠s of isos. △, ∠ sum of △).

$\angle ACD = \angle BCD - \angle BCA = 120^{\circ} - 30^{\circ} = 90^{\circ}$

$AD = 9.0$ cm (accurate if from 8.8 cm to 9.2 cm) and $\angle ACD = 90^{\circ}$ (accurate if from $88^{\circ}$ to $92^{\circ}$ ), as the reasoning shows.

Practice 11

Construct a regular pentagon $ABCDE$ inside a circle of radius 4.5 cm, using the angles at the centre. Check yourself: $\angle COD$ should measure $72^{\circ}$ . Join $AC$ and $AD$ . Measure $AC$ and $\angle CAD$ , then check $\angle CAD$ by reasoning.

ShowHide solution

Step 1. Draw a circle of radius 4.5 cm with centre $O$ and a radius $OA$ . Mark $72^{\circ}$ and $144^{\circ}$ on each side of $OA$ , and draw lines from $O$ to the circle at $B$ , $C$ , $E$ and $D$ .
Step 2. Join the five points in order. Check $\angle COD = 72^{\circ}$ , then join $AC$ and $AD$ , and measure $AC$ and $\angle CAD$ .

Check by reasoning. Each interior angle of a regular pentagon is $108^{\circ}$ . $AB = BC$ , so in $\triangle ABC$ , $\angle BAC = \frac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}$ (base ∠s of isos. △, ∠ sum of △). In the same way, $AE = ED$ , so $\angle EAD = 36^{\circ}$ .

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

$AC = 8.6$ cm (accurate if from 8.4 cm to 8.8 cm) and $\angle CAD = 36^{\circ}$ (accurate if from $34^{\circ}$ to $38^{\circ}$ ), as the reasoning shows.

Practice 12

Each angle at the centre of a regular polygon is $40^{\circ}$ .

(a) How many sides has the polygon? What is each of its interior angles?

(b) Construct the polygon inside a circle of radius 4 cm, naming its vertices $A$ , $B$ , $C$ , … in order. Check yourself: the angle at the centre that you did not draw should measure $40^{\circ}$ . Then measure the side $AB$ and $\angle ABC$ , and compare $\angle ABC$ with your answer to (a).

ShowHide solution

(a) The angles at the centre fill the whole turn: $n = \frac{360^{\circ}}{40^{\circ}} = 9$ . It has 9 sides (a regular nonagon).

Each interior angle $= \frac{(9 - 2) \times 180^{\circ}}{9} = \frac{1260^{\circ}}{9} = 140^{\circ}$ .

(b)

Step 1. Draw a circle of radius 4 cm with centre $O$ , and a radius $OA$ .
Step 2. With the protractor along $OA$ , mark $40^{\circ}$ , $80^{\circ}$ , $120^{\circ}$ and $160^{\circ}$ on each side of $OA$ , and draw lines from $O$ to the circle: $B$ , $C$ , $D$ , $E$ on one side and $I$ , $H$ , $G$ , $F$ on the other.
Step 3. Join the nine points in order. Check that $\angle EOF$ , the angle you did not draw, is $40^{\circ}$ (as $360^{\circ} - 2 \times 160^{\circ} = 40^{\circ}$ ). Then measure $AB$ and $\angle ABC$ .

(a) 9 sides; each interior angle is $140^{\circ}$ . (b) $AB = 2.7$ cm (accurate if from 2.5 cm to 2.9 cm) and $\angle ABC = 140^{\circ}$ (accurate if from $138^{\circ}$ to $142^{\circ}$ ), as in (a).

Key points

Sketch first, construct with instruments, leave the arcs and circles showing, check yourself on a given length or angle, then measure (to 0.1 cm and to the nearest degree).
From the side, with a protractor: draw one side, then at each vertex measure the interior angle inside the polygon ( $108^{\circ}$ pentagon, $120^{\circ}$ hexagon, $135^{\circ}$ octagon, $144^{\circ}$ decagon) and draw the next side the same length. The last side is the check.
In a circle, from the centre: mark angles of $\frac{360^{\circ}}{n}$ at the centre $O$ ( $72^{\circ}$ pentagon, $60^{\circ}$ hexagon, $45^{\circ}$ octagon, $36^{\circ}$ decagon), each measured from $OA$ ; the points on the circle are the vertices.
Why it is regular: the radii are equal and the angles at the centre are equal, so a turn of $\frac{360^{\circ}}{n}$ about $O$ takes each side onto the next (the sides are equal); the triangles at the centre are isosceles, so each angle of the polygon is two equal base angles (the angles are equal).
A regular hexagon with compasses and a ruler: radius = side; step round the circle; it is six equilateral triangles with $60^{\circ}$ at the centre.
Afterwards, measure a diagonal, or measure an angle and check it by reasoning with isosceles triangles.