Inscribed Regular-Polygon Constructions

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Learning goals

Construct a regular hexagon in a given circle by retaining a fixed compass opening
Justify closure and regularity using equal chords and central angles
Derive an inscribed equilateral triangle from alternating hexagon vertices
Build an inscribed square using compass-constructed perpendicular diameters
Calculate exact lengths and areas from the justified constructions

Start with the given circle

An inscribed polygon has all its vertices on a circle. Its sides are straight chords; the curved arcs between the vertices are not polygon sides. A regular polygon has equal sides and equal interior angles.

Throughout this lesson, the given circle has a known center $O$ and positive radius $r$ . Choose any point $A$ on its boundary. The construction tools let us retain a compass opening and join distinct points with an unmarked straightedge. A measured angle or a shape that merely looks symmetric does not supply a compass construction.

Step the radius to construct a hexagon

Set the compass opening to $OA$ . Keep that opening throughout the following steps, and keep the compass arcs visible.

First crossing. With center $A$ and opening $OA$ , draw arcs cutting the given circle at $B$ and $F$ . Choose $B$ on one side of line $OA$ to begin traveling around the circle. The two circles have equal positive radius $r$ and their centers are $r$ apart, so they intersect in two distinct points, one on each side of $OA$ .

The first pair of compass crossings gives a choice of direction. Keep that direction for the remaining steps.

Continue the same way. With center $B$ and the same opening $OA$ , mark $C$ at the intersection with the given circle other than $A$ .

With center $C$ and opening $OA$ , mark $D$ at the intersection other than $B$ .

With center $D$ and opening $OA$ , mark $E$ at the intersection other than $C$ .

With center $E$ and opening $OA$ , select the intersection other than $D$ . It is the point $F$ already found from $A$ .

Finally, an arc centered at $F$ with opening $OA$ returns to $A$ at its intersection other than $E$ . Join consecutive vertices $A,B,C,D,E,F$ , then join $F$ to $A$ .

The claims that the fifth step reaches the original $F$ and the sixth returns to $A$ need reasons. Six marks that happen to look evenly spaced are not enough.

Why the steps close and produce a regular hexagon#

For a circle of radius $4$ cm, the first compass crossing gives $AB=4$ cm. Also $OA=4$ cm and $OB=4$ cm, so triangle $OAB$ is equilateral. Its angle at $O$ is $60^\circ$ .

The same argument works for every $r>0$ . The given circle supplies the two radii, and the compass arc supplies the new chord:

\begin{gathered} OA=r\\ OB=r\\ AB=r. \end{gathered}

Thus every successive center triangle is equilateral and each step advances $60^\circ$ around $O$ . At each vertex the two possible intersections lie $60^\circ$ in opposite directions from its radius. Excluding the previous vertex selects the forward intersection.

The first five forward turns are less than a full turn, so their points are distinct. After five turns, the radius points toward the other first crossing $F$ . After six turns,

6\cdot60^\circ=360^\circ.

The ray has returned to $OA$ , which meets the given circle at the unique point $A$ . This proves closure. The six equal chords have length $r$ . At each vertex the two adjacent equilateral center triangles contribute two $60^\circ$ angles, giving a $120^\circ$ interior angle. The vertices occur in order around the circle, so the joined polygon is convex and has six equal sides and equal angles. It is a regular hexagon.

Completed hexagon construction

Stage 6 of 6. From F, keep the same compass opening and choose A rather than returning to E. Join A, B, C, D, E and F in order. Each radius step makes a 60° central angle, so six steps close the regular hexagon. The given circle has radius r. Every compass arc shown has radius r.

Six equal radius steps close the construction. The retained compass arcs record the equal chord lengths.

Use First stage in the construction below, then advance through its stages. Before each next crossing appears, predict which earlier vertex must be excluded. Watch the retained arcs, and explain why keeping the compass opening fixes a chord length rather than a curved arc length.

Construct an inscribed hexagon

Stage Construction

Use First stage, then advance through the construction. Earlier compass arcs remain visible so you can justify each new point.

Worked example 1 Find the perimeter of the constructed hexagon

Problem. A given circle has radius $5.5$ cm. Construct its inscribed regular hexagon with a compass and unmarked straightedge. State the opening used for every forward step and find the hexagon’s perimeter.

Solution. Choose $A$ on the circle and set the opening to $OA$ , or $5.5$ cm. Move the compass point successively to $A,B,C,D,E,F$ , each time choosing the forward circle intersection as described above. Join the six consecutive vertices. The closure proof makes this a regular hexagon, and each side has the retained opening as its length.

\begin{aligned} s&=5.5\text{ cm}\\ P&=6(5.5)\\ P&=33\text{ cm}. \end{aligned}

Check your understanding

A compass has advanced four successive radius-length chords around a circle in one direction. What total turn have their four center angles made?

120^\circ

180^\circ

240^\circ

360^\circ

Use alternate hexagon vertices for a triangle

Keep the justified hexagon $ABCDEF$ . Join $A$ to $C$ , $C$ to $E$ , and $E$ to $A$ . Every new side skips exactly one hexagon vertex. No new compass opening is needed. The retained hexagon arcs show how all three selected vertices were located on the given circle.

Why alternate vertices give an equilateral triangle#

Two neighboring $60^\circ$ center angles combine to give

60^\circ+60^\circ=120^\circ.

The three center angles $AOC$ , $COE$ , and $EOA$ are therefore equal. In a circle of radius $6$ cm, each of these center triangles has two sides of $6$ cm and an included $120^\circ$ angle, so SAS makes them congruent.

For any positive circle radius, the triangles $AOC$ , $COE$ , and $EOA$ likewise have equal pairs of radii and the same included $120^\circ$ angle. SAS makes them congruent, so their opposite chords $AC$ , $CE$ , and $EA$ are equal. These three distinct vertices lie on the given circle, making $ACE$ an inscribed equilateral triangle. Its three equal angles are $60^\circ$ by the triangle angle sum.

Completed equilateral triangle construction

Stage 3 of 3. The three central angles are 120°. Their equal chords make ACE an inscribed equilateral triangle. The given circle has radius r. Every compass arc shown has radius r.

Joining A, C and E skips one hexagon vertex on each side. The three equal center angles justify the equal chords.

Use First stage in the construction below and advance until all six hexagon vertices are marked before joining the triangle. Before advancing to Stage 2, predict which three vertices the triangle will use. Watch which hexagon vertex each triangle side skips, and explain why three equal center angles justify three equal chords.

Construct an inscribed equilateral triangle

Stage Construction

Stage 3 of 3. The three central angles are 120°. Their equal chords make ACE an inscribed equilateral triangle. The given circle has radius r. Every compass arc shown has radius r.

Use First stage, then advance through the construction. Earlier compass arcs remain visible so you can justify each new point.

Worked example 2 Recover the triangle side from the circle

Problem. Construct an equilateral triangle inside a given circle of radius $7$ cm by using alternate hexagon vertices. Find its exact side length and perimeter.

Solution. Step the $7$ cm radius around the given circle to make $ABCDEF$ , then join $AC$ , $CE$ , and $EA$ . The proof above establishes the triangle’s equal sides.

Let $M$ be the midpoint of $AC$ . In isosceles triangle $AOC$ , the segment $OM$ bisects the $120^\circ$ center angle and is perpendicular to $AC$ . Each right half is a $30$ - $60$ - $90$ triangle with hypotenuse $7$ cm. Its shorter leg is $OM$ , while its longer leg is half of $AC$ .

OM meets AC at its midpoint and at a right angle. Use the right half of center triangle AOC.

\begin{aligned} OM&=\frac72\text{ cm}\\ AM&=\frac{7\sqrt3}{2}\text{ cm}\\ AC&=7\sqrt3\text{ cm}\\ P&=21\sqrt3\text{ cm}. \end{aligned}

Check your understanding

Triangle ACE uses alternate vertices of constructed hexagon ABCDEF. Which fact establishes that AC and CE have equal lengths?

A Their two center triangles have equal radii and equal included center angles B Every chord of a circle is a diameter C The triangle was drawn with a straightedge D The radius equals every triangle chord

Construct a square using perpendicular diameters

Begin a fresh construction in the given circle. Draw a line through $O$ and call its circle intersections $A$ and $C$ . Thus $AC$ is a diameter, with $O$ its midpoint.

Set the compass opening to the full length $AC$ . Draw arcs centered at $A$ above and below $AC$ . With center $C$ and the same opening $AC$ , draw arcs crossing them at $U$ and $V$ . These are genuine compass-circle intersections; $U$ and $V$ are not points estimated above and below the center.

The opening $AC$ is greater than half of $AC$ , so the two equal circles intersect in two distinct points. Draw line $UV$ . Call its intersections with the given circle $B$ and $D$ . Finally join $AB$ , $BC$ , $CD$ , and $DA$ .

Both helper circles use opening AC. Their two exterior crossings determine line UV; the given circle supplies B and D.

This uses the earlier perpendicular-bisector construction. Retaining its arcs makes the reason for the second diameter visible.

Why the constructed quadrilateral is a square#

If the given circle has radius $3$ cm, then $AC=6$ cm. Each helper-circle intersection is $6$ cm from both $A$ and $C$ , so both lie on the perpendicular bisector of $AC$ .

For any positive radius the same equal-distance argument applies. Points $U$ and $V$ each lie on both helper circles, giving

\begin{gathered} UA=UC\\ VA=VC. \end{gathered}

The perpendicular-bisector converse puts both points on that bisector. Since $U$ and $V$ are distinct, their line is the bisector. It passes through the known midpoint $O$ , so $BD$ is a diameter perpendicular to $AC$ .

The four center angles are $90^\circ$ . Their center triangles have equal radii and equal included angles, so SAS gives four equal chords. Each interior angle of $ABCD$ subtends a diameter: for example, angle $ABC$ subtends $AC$ . The angle in a semicircle is a right angle, so all four interior angles are right angles. The four vertices occur in order around the circle. Thus $ABCD$ is an inscribed square.

Completed square construction

Stage 4 of 4. Join A, B, C and D in order. Perpendicular diameters give equal chords, and each corner subtends a semicircle, so ABCD is a square. The given circle has radius r. The two compass circles each have radius 2r.

The perpendicular-bisector line passes through O. The two perpendicular diameters divide the circle into four equal central angles.

Use First stage in the construction below, then advance one step at a time. Predict where the two equal compass circles will meet. Watch those arcs remain after $UV$ is drawn, and explain why the new diameter must pass through $O$ even though neither helper circle is centered there.

Construct an inscribed square

Stage Construction

Use First stage, then advance through the construction. Earlier compass arcs remain visible so you can justify each new point.

Worked example 3 Construct a square from a given diameter

Problem. A circle has diameter $15$ cm and known center $O$ . Construct its inscribed square using a compass and unmarked straightedge, and find its exact side length and perimeter.

Solution. Draw diameter $AC$ . Use opening $AC$ , or $15$ cm, at $A$ and then $C$ to obtain the two helper intersections $U,V$ . Draw $UV$ , mark $B,D$ on the given circle, and join consecutive vertices. The equal-distance proof establishes perpendicular diameters and hence a square.

Its diagonal $AC$ is the hypotenuse of right isosceles triangle $ABC$ . For positive square side $s$ ,

\begin{aligned} 2s^2&=15^2\\ s&=\frac{15\sqrt2}{2}\text{ cm}\\ P&=30\sqrt2\text{ cm}. \end{aligned}

The $15$ cm compass opening locates the helper intersections. It is the diameter, not the square’s side length.

Check your understanding

Equal compass arcs centered at the endpoints of a diameter meet at U and V. What establishes that UV passes through the circle center?

A The helper arcs have the same color B U and V lie on the original circle C Every line through U is a diameter D UV is the perpendicular bisector of the diameter

Connect the constructions to length and area

Construction proves what the shapes are. The earlier regular-polygon area formula then applies: the apothem $a$ is the perpendicular distance from $O$ to a side, and $A=\tfrac12 Pa$ . The apothem is denoted by $a$ here; it was $h$ in the finite polygon bounds. It is not the slanted radius to a vertex.

The center triangles above give these exact dimensions for a circle of positive radius $r$ :

Constructed polygon	Side length	Apothem	Area
Regular hexagon	$r$	$\dfrac{r\sqrt3}{2}$	$\dfrac{3\sqrt3}{2}r^2$
Equilateral triangle	$r\sqrt3$	$\dfrac r2$	$\dfrac{3\sqrt3}{4}r^2$
Square	$r\sqrt2$	$\dfrac{r\sqrt2}{2}$	$2r^2$

For the hexagon, halve one equilateral center triangle. Its half-side is $r/2$ , so its $30$ - $60$ - $90$ apothem is $r\sqrt3/2$ . For the alternate-vertex triangle, halving a $120^\circ$ center triangle gives apothem $r/2$ and half-side $r\sqrt3/2$ , as in Worked Example 2. For the square, a $90^\circ$ center triangle has hypotenuse $r\sqrt2$ as its polygon side; its altitude from $O$ is half that side. These use the earlier special right triangles, rather than measuring the diagrams.

Substituting each perimeter and apothem into the proved area formula gives the last column. For example, for the hexagon,

\begin{aligned} P&=6r\\ A&=\frac12(6r)\frac{r\sqrt3}{2}\\ A&=\frac{3\sqrt3}{2}r^2. \end{aligned}

Worked example 4 Use the height supplied by the construction

Problem. A regular hexagon is constructed inside a given circle of radius $14$ cm. Find the hexagon’s apothem and exact area.

Solution. Its side is $14$ cm by the compass construction. Halving an equilateral center triangle gives a right triangle with hypotenuse $14$ cm and shorter leg $7$ cm. Its perpendicular height is therefore $7\sqrt3$ cm.

The apothem reaches a side midpoint. The radius and half-side determine this perpendicular height.

\begin{aligned} a&=7\sqrt3\text{ cm}\\ P&=6(14)\\ P&=84\text{ cm}\\ A&=\frac12(84)(7\sqrt3)\\ A&=294\sqrt3\text{ cm}^2. \end{aligned}

Check your understanding

A square is constructed in a given circle of radius 13 cm. Which calculation gives its exact area in square centimeters?

13^2

2(13^2)

4(13^2)

13^2/2

Worked example 5 Compare two constructions in the same circle

Problem. An equilateral triangle and a square are constructed in the same given circle of radius $3\sqrt5$ cm. Find each exact area and the ratio of triangle area to square area.

Solution. Both constructions use the same radius, whose square is

r^2=(3\sqrt5)^2.

r^2=45\text{ cm}^2.

Use the established shape-specific area relationships separately.

\begin{aligned} A_{\triangle}&=\frac{3\sqrt3}{4}(45)\\ A_{\triangle}&=\frac{135\sqrt3}{4}\text{ cm}^2\\ A_{\square}&=2(45)\\ A_{\square}&=90\text{ cm}^2. \end{aligned}

The ratio divides the two areas, so the square-centimeter units cancel:

\frac{A_{\triangle}}{A_{\square}}=\frac{3\sqrt3}{8}.

The square has the greater area. The common circle does not make the polygons’ side lengths or areas equal.

Check your understanding

A constructed inscribed triangle and regular hexagon use the same circle of radius 10 cm. What is the triangle area divided by the hexagon area?

1/4

1/3

1/2

2

Inscribed Regular-Polygon Constructions

Learning goals

Start with the given circle

Step the radius to construct a hexagon

Use alternate hexagon vertices for a triangle

Construct a square using perpendicular diameters

Connect the constructions to length and area

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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