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Inscribed Regular-Polygon Constructions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Complete the fixed-opening construction

A given circle has center O and radius 6.4 cm. Point A lies on it. An arc centered at A with opening OA selects B on the circle, as shown.

On paper, copy the given configuration at any convenient scale. Complete regular hexagon ABCDEF with an unmarked straightedge and compass. Retain every compass arc. At each next step, name the center, opening and circle intersection to exclude.

Using the stated radius, find the perimeter of triangle OAB and its angle at O. Explain which circle supplies each side length.

The initial compass arc centered at A selects the next circle point B.

Text description of this figure

The given circle has center O and radius 6.4 cm. A radius-opening compass arc centered at A crosses the circle at the selected point B. Chord AB and radii OA and OB are shown. The remaining hexagon vertices have not been constructed.
1. Hint 1
  Keep every compass opening equal to the given radius; the two circle conditions determine each next point.
2. Hint 2
  At a new center, one crossing is the point just left. Which crossing continues forward?
3. Hint 3
  Separate the original circle radii from the first compass circle radius to identify triangle OAB.
Answer

With fixed opening OA, centers B, C, D, E, F produce C, D, E, F, A, excluding A, B, C, D, E respectively.
$\begin{gathered}P_{OAB}=19.2\text{ cm},\\\angle AOB=60^\circ.\end{gathered}$

Full solution

Reproduce the first step with center A and opening OA, choosing one crossing as B.
Keep that opening for every remaining compass arc.

Center B: choose C at the given-circle intersection other than A.

Center C: choose D at the given-circle intersection other than B.

Center D: choose E at the given-circle intersection other than C.

Center E: choose F at the given-circle intersection other than D.

Center F: choose A at the given-circle intersection other than E.
Retain all six arcs and join A, B, C, D, E, F, A in order.

Every constructed chord equals the given radius.
Its two center-to-endpoint radii have that same length, so each center triangle is equilateral with a 60-degree center angle.
Six forward center angles make one full turn; the equal sides and equal 120-degree interior angles give the regular hexagon.

OA and OB are radii of the original circle.
AB is a radius of the compass circle centered at A, whose opening was set to OA.

$\begin{gathered}OA=6.4\text{ cm}\\OB=6.4\text{ cm}\\AB=6.4\text{ cm}.\end{gathered}$

Add the three sides of equilateral triangle OAB and divide its angle sum equally.

$\begin{aligned}P_{OAB}&=3(6.4)\\P_{OAB}&=19.2\text{ cm}\\\angle AOB&=180^\circ\div3\\\angle AOB&=60^\circ.\end{aligned}$

Answer

With fixed opening OA, centers B, C, D, E, F produce C, D, E, F, A, excluding A, B, C, D, E respectively.
$\begin{gathered}P_{OAB}=19.2\text{ cm},\\\angle AOB=60^\circ.\end{gathered}$

Key idea

A fixed compass opening and the forward intersection make the six equal-radius chords close.
Problem 2 Construct and measure the inscribed triangle

A given circle has center O, radius 5 cm and boundary point A.

On paper, copy the given circle at any convenient scale. Construct an inscribed equilateral triangle with an unmarked straightedge and compass. Retain the helper arcs and label its other vertices C and E in circle order. Explain which helper vertices are skipped and why the three sides are equal.

Find the smaller central angle AOC, the exact length AC and the perimeter of triangle ACE.

Text description of this figure

The given circle has center O, radius 5 cm, and boundary point A. The given radius OA is drawn.
1. Hint 1
  First locate six equal radius-length chords around the given circle.
2. Hint 2
  Choose alternate helper vertices, so each triangle side spans two hexagon center angles.
3. Hint 3
  Drop the perpendicular from O to the midpoint of AC. Use the resulting special right triangle to find half of AC.
Answer

Construct triangle ACE from alternate hexagon vertices, skipping B, D and F.
$\begin{gathered}\angle AOC=120^\circ,\\AC=5\sqrt3\text{ cm},\\P_{ACE}=15\sqrt3\text{ cm}.\end{gathered}$

Full solution

Set the opening to OA.
With center A mark B on the given circle.
With the same opening and centers B, C, D, E and F in turn, always exclude the preceding vertex and mark C, D, E, F and A respectively.
Retain the six compass arcs.

Join alternate vertices A, C and E, skipping B, D and F.
Each selected central angle spans two of the hexagon center angles.

$\angle AOC=2(60^\circ).$

$\angle AOC=120^\circ.$

The three center triangles have equal radii and equal included 120-degree angles.
SAS gives equal chords AC, CE and EA, so the constructed triangle is equilateral.

Let M be the midpoint of AC.
The perpendicular OM bisects isosceles triangle OAC into two 30-60-90 right triangles.
OA is 5 cm, so the longer leg AM is five times the square root of 3 divided by 2 cm.

$\begin{aligned}AC&=2\left(\frac{5\sqrt3}{2}\right)\\AC&=5\sqrt3\text{ cm}\\P_{ACE}&=3(5\sqrt3)\\P_{ACE}&=15\sqrt3\text{ cm}.\end{aligned}$

Answer

Construct triangle ACE from alternate hexagon vertices, skipping B, D and F.
$\begin{gathered}\angle AOC=120^\circ,\\AC=5\sqrt3\text{ cm},\\P_{ACE}=15\sqrt3\text{ cm}.\end{gathered}$

Key idea

Alternate vertices retain the justified hexagon marks while doubling each selected center angle.
Problem 3 A square from a measured circle

A given circle has known center O and diameter 18.4 cm. Construct its inscribed square using a compass and unmarked straightedge.

State the centers and opening of the two helper circles, identify the line determined by their intersections, and find the exact square side and perimeter.

Construct the square from this circle.

Text description of this figure

The given circle has center O and diameter AC of length 18.4 cm.
1. Hint 1
  Begin with a diameter AC. Its endpoints are the helper-circle centers.
2. Hint 2
  An opening equal to the whole diameter gives two distinct helper intersections.
3. Hint 3
  A square diagonal has length 18.4 cm; it is the hypotenuse of a right isosceles triangle.
Answer

Helper centers A and C, common opening 18.4 cm; line UV is the perpendicular bisector of AC. Side $9.2\sqrt2$ cm; perimeter $36.8\sqrt2$ cm.

Full solution

Draw diameter AC through O.
Draw equal arcs with center A and center C, both using opening AC, or 18.4 cm.
Name their two distinct intersections U,V and join them.
The common radius exceeds half the separation AC, ensuring two intersections.

Each intersection has equal distances to A and C.
Their line UV is therefore the perpendicular bisector of AC, passing through its midpoint O.
Mark its circle intersections B,D and join AB, BC, CD, DA.
The perpendicular diameters give four equal chords, and each square vertex subtends a diameter and is a right angle.

The square diagonal is AC.
For positive side s, Pythagoras gives the side and then the perimeter.

$\begin{aligned}2s^2&=18.4^2\\s&=\frac{18.4}{\sqrt2}\\s&=9.2\sqrt2\text{ cm}\\P&=4(9.2\sqrt2)\\P&=36.8\sqrt2\text{ cm}.\end{aligned}$

Retained compass arcs justify the perpendicular diameter.

Answer

Helper centers A and C, common opening 18.4 cm; line UV is the perpendicular bisector of AC. Side $9.2\sqrt2$ cm; perimeter $36.8\sqrt2$ cm.

Key idea

The helper-circle opening locates a perpendicular diameter; it is not the square side.
Problem 4 An apothem from a compass opening

A regular hexagon is constructed in a given circle of radius $4\sqrt5$ cm. Find its exact apothem, perimeter, and area.

M is a side midpoint and OM is perpendicular to that side.

Text description of this figure

Circle O contains the constructed regular hexagon ABCDEF. The circle radius is labeled 4√5 cm. OM is the perpendicular to the midpoint M of the first polygon side.
1. Hint 1
  Every hexagon side equals the compass opening.
2. Hint 2
  Half of one equilateral center triangle has shorter leg half the radius.
3. Hint 3
  Use the perpendicular apothem with the entire six-side perimeter.
Answer

Apothem $2\sqrt{15}$ cm; perimeter $24\sqrt5$ cm; area $120\sqrt3\text{ cm}^2$ .

Full solution

The constructed hexagon has side four times the square root of 5 cm.
Halving an equilateral center triangle gives half-side two times the square root of 5 cm and apothem the square root of 3 times that half-side.

$\begin{aligned}a&=(2\sqrt5)\sqrt3\\a&=2\sqrt{15}\text{ cm}\\P&=6(4\sqrt5)\\P&=24\sqrt5\text{ cm}.\end{aligned}$

The apothem is the perpendicular height for each center triangle.
Use all six bases through the perimeter.

$\begin{aligned}A&=\frac12(24\sqrt5)(2\sqrt{15})\\A&=24\sqrt{75}\\A&=120\sqrt3\text{ cm}^2.\end{aligned}$

Answer

Apothem $2\sqrt{15}$ cm; perimeter $24\sqrt5$ cm; area $120\sqrt3\text{ cm}^2$ .

Key idea

A fixed compass opening determines the side first; a special right triangle supplies the perpendicular height.
Problem 5 Recover the original opening

An equilateral triangle was constructed by joining alternate vertices of a radius-stepped regular hexagon. The triangle perimeter is $63\sqrt3$ cm.

Find the original compass opening, the diameter of the given circle, and the hexagon perimeter.

The triangle and hexagon have the same supporting circle.

Text description of this figure

Circle O contains the alternate-vertex triangle ACE and hexagon ABCDEF. The circle radius is labeled r.
1. Hint 1
  Divide the triangle perimeter into three equal sides.
2. Hint 2
  The triangle side is the circle radius times the square root of 3.
3. Hint 3
  The hexagon uses six radius-length sides.
Answer

Compass opening 21 cm; diameter 42 cm; hexagon perimeter 126 cm.

Full solution

The alternate-vertex triangle has three equal sides, each subtending a 120-degree center angle.
Its side is the circle radius times the square root of 3.

$\begin{aligned}s_T&=\frac{63\sqrt3}{3}\\s_T&=21\sqrt3\text{ cm}\\r&=\frac{21\sqrt3}{\sqrt3}\\r&=21\text{ cm}.\end{aligned}$

The radius was the compass opening for the hexagon.
Double it for the given circle's diameter and multiply it by six for the hexagon perimeter.

$\begin{aligned}d&=2(21)\\d&=42\text{ cm}\\P_H&=6(21)\\P_H&=126\text{ cm}.\end{aligned}$

Answer

Compass opening 21 cm; diameter 42 cm; hexagon perimeter 126 cm.

Key idea

Reverse the triangle's side-to-radius relationship before recovering the underlying construction lengths.
Problem 6 Use a different helper opening

A given circle has diameter AC = 24 cm and center O. To construct a perpendicular diameter, draw equal helper circles centered at A and C with opening 15 cm. They intersect at U and V.

Find UV. Explain why joining U,V still locates a diameter of the given circle, and find the area of the square obtained from the two perpendicular diameters.

The original and helper circles have different radii.

Text description of this figure

The original circle has center O and diameter AC of 24 cm. Equal compass arcs of radius 15 cm centered at A and C intersect at U and V, inside the original circle. Their line meets the original circle at B and D.
1. Hint 1
  The 15 cm opening is greater than the 12 cm half-diameter, so two helper crossings exist.
2. Hint 2
  U and V lie on the perpendicular bisector of AC. In right triangle AOU, AO is 12 cm and AU is 15 cm.
3. Hint 3
  Add OU and OV, then use the original radius, not the helper opening, for the square's area.
Answer

$UV=18$ cm; line UV passes through O perpendicular to AC; square area $288\text{ cm}^2$ .

Full solution

The original radius is 12 cm.
Since each helper radius is 15 cm and the center separation is 24 cm, the helper circles cross twice.
Each crossing has equal distances to A and C, so both lie on the perpendicular bisector of AC.
Their line passes through the known midpoint O and cuts the original circle in a perpendicular diameter.

Pythagoras in right triangle AOU gives the distance to the upper crossing.
The lower crossing is equally far from O, on the opposite side of AC.

$\begin{aligned}OU^2&=15^2-12^2\\OU^2&=81\\OU&=9\text{ cm}\\UV&=9+9\\UV&=18\text{ cm}.\end{aligned}$

The two original diameters determine an inscribed square.
Its supporting circle still has radius 12 cm.

$\begin{aligned}A_S&=2(12^2)\\A_S&=288\text{ cm}^2.\end{aligned}$

Answer

$UV=18$ cm; line UV passes through O perpendicular to AC; square area $288\text{ cm}^2$ .

Key idea

Equal helper circles can use another sufficiently large opening; the original circle fixes the finished square.
Problem 7 Why the finished square is independent of the opening

Let a given circle have center O and radius r > 0, with diameter AC. Use equal helper circles centered at A and C with any common opening k > r. Let their two intersections be U,V. Their line cuts the given circle at B,D.

Prove that changing k, while keeping k > r, does not change the points B,D, and that joining A,B,C,D in circle order gives a square. Include both equal-side and right-angle reasons.

One permitted helper opening is shown; the proof covers every k greater than r.

Text description of this figure

A schematic given circle with center O and diameter AC. Equal arcs at A and C with one permitted common opening meet at U and V. The line through their intersections meets the circle at B and D.
1. Hint 1
  Since AC is 2r, the condition k > r guarantees two distinct helper crossings.
2. Hint 2
  Apply the perpendicular-bisector converse to both crossings. Is there more than one perpendicular bisector of AC?
3. Hint 3
  Use the resulting equal center angles for the sides and the semicircle theorem for the interior angles.
Answer

Every permitted opening gives the same perpendicular-bisector line through O, hence the same B,D as an unordered pair; ABCD has four equal sides and four right angles.

Full solution

The endpoints A,C are distinct since r is positive.
Their separation is 2r, so k greater than r makes the equal helper circles intersect in two distinct points U,V.
Their circle memberships give

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

The perpendicular-bisector converse puts both intersections on the perpendicular bisector of AC.
Two distinct points determine their joining line, so UV is that unique bisector.
It passes through the midpoint O of diameter AC and is perpendicular to AC.
This line depends on A,C, not on k.
Its two intersections with the fixed original circle are consequently the same unordered pair B,D for every permitted opening.
Label B on one chosen side of AC to fix the names as well.

The two perpendicular diameters make four 90-degree center angles.
Their center triangles have two radii of length r and an equal included angle, so SAS gives four equal chords.
At each vertex, the interior angle subtends the diameter joining the two neighboring vertices.
The semicircle theorem makes that angle a right angle.
Thus the quadrilateral has four equal sides and four right angles and is a square.

The strict condition on k matters: at k equal to r the helper circles would touch only at O, leaving no pair of distinct crossings to determine UV.

Answer

Every permitted opening gives the same perpendicular-bisector line through O, hence the same B,D as an unordered pair; ABCD has four equal sides and four right angles.

Key idea

A construction can allow different helper points while forcing one unique final line and polygon.
Problem 8 Transfer from a square area

A square is constructed using perpendicular diameters of a given circle. Its area is $46\text{ cm}^2$ . In that same circle, a regular hexagon is constructed by radius stepping and an equilateral triangle is formed using alternate hexagon vertices.

Find the hexagon perimeter and the exact triangle area.

The square diagonal passes through the supporting circle center.

Text description of this figure

Circle O contains the constructed square ABCD. The circle radius is labeled r.
1. Hint 1
  The square area is twice the squared supporting-circle radius.
2. Hint 2
  Keep r squared for the triangle's area, but take the positive square root for the hexagon's perimeter.
3. Hint 3
  The common circle gives the same radius to all three constructions, not the same side length.
Answer

Hexagon perimeter $6\sqrt{23}$ cm; triangle area $\frac{69\sqrt3}{4}\text{ cm}^2$ .

Full solution

The square's construction gives perpendicular radius legs in each center triangle, so its area is twice the squared circle radius.

$\begin{aligned}2r^2&=46\\r^2&=23\text{ cm}^2\\r&=\sqrt{23}\text{ cm}.\end{aligned}$

Each radius-stepped hexagon side has that radius as its length.
The alternate-vertex triangle has area three times the square root of 3 divided by 4 times the squared radius.

$\begin{aligned}P_H&=6\sqrt{23}\text{ cm}\\A_T&=\frac{3\sqrt3}{4}(23)\\A_T&=\frac{69\sqrt3}{4}\text{ cm}^2.\end{aligned}$

The same circle supports the newly constructed hexagon and triangle.

Answer

Hexagon perimeter $6\sqrt{23}$ cm; triangle area $\frac{69\sqrt3}{4}\text{ cm}^2$ .

Key idea

An area can determine the squared common radius directly, even when the radius is irrational.
Problem 9 Wire along constructed chords

A circular frame has radius $5\sqrt3$ cm. Six equally spaced attachment points are located by the compass hexagon construction. Wire runs once along each of the six hexagon sides and once along each side of the equilateral triangle formed by alternate points. The wire is straight between attachment points, and joints add no length.

Find the total wire length and the exact area enclosed by the triangle.

Wire follows the dashed outer hexagon and the solid inner triangle.

Text description of this figure

Circle O contains the alternate-vertex triangle ACE and hexagon ABCDEF. The circle radius is labeled 5√3 cm.
1. Hint 1
  The outer six wire pieces each have the original radius as their length.
2. Hint 2
  Each alternate-point triangle side is the radius times the square root of 3. The outer pieces and triangle pieces are distinct.
3. Hint 3
  For area, use the triangle side or the squared supporting-circle radius.
Answer

Total wire $(30\sqrt3+45)$ cm; triangle area $\frac{225\sqrt3}{4}\text{ cm}^2$ .

Full solution

The hexagon is justified by the radius-stepping construction, so the six outer pieces have total length six times the circle radius.
Each triangle side spans two hexagon steps and is the radius times the square root of 3.

$\begin{aligned}P_H&=6(5\sqrt3)\\P_H&=30\sqrt3\text{ cm}\\s_T&=(5\sqrt3)\sqrt3\\s_T&=15\text{ cm}\\P_T&=3(15)\\P_T&=45\text{ cm}.\end{aligned}$

The two sets of straight segments share vertices but no whole segment, so their lengths add without removing overlaps.
The triangle area follows from its 15 cm equilateral side.

$\begin{aligned}L&=30\sqrt3+45\text{ cm}\\A_T&=\frac{15^2\sqrt3}{4}\\A_T&=\frac{225\sqrt3}{4}\text{ cm}^2.\end{aligned}$

Answer

Total wire $(30\sqrt3+45)$ cm; triangle area $\frac{225\sqrt3}{4}\text{ cm}^2$ .

Key idea

Translate the actual straight pieces into polygon sides before adding lengths or calculating an enclosed area.
Problem 10 A new construction inside the first one

A regular hexagon ABCDEF is constructed in a given circle with center O and radius $2\sqrt{13}$ cm. Join alternate vertices to form equilateral triangle ACE.

Draw the largest circle centered at O that fits inside triangle ACE, then construct a square inscribed in this smaller circle using perpendicular diameters.

Find the smaller circle radius, the square side, the square area, and the fraction of the original hexagon area occupied by this square.

First locate the largest centered circle inside the triangle.

Text description of this figure

Circle O contains the alternate-vertex triangle ACE and hexagon ABCDEF. The circle radius is labeled 2√13 cm.
1. Hint 1
  The largest centered circle reaches the triangle sides at their perpendicular feet. Its radius is the triangle's apothem.
2. Hint 2
  An alternate-vertex triangle has apothem half the original circle radius.
3. Hint 3
  Use the smaller radius for the square and the original radius for the hexagon. Keep the two radii separate.
4. Hint 4
  Divide the square area by the hexagon area and simplify the exact quotient.
Answer

Smaller radius $\sqrt{13}$ cm; square side $\sqrt{26}$ cm; square area $26\text{ cm}^2$ ; fraction $\frac{\sqrt3}{9}$ .

Full solution

The regular triangle has the same perpendicular distance from O to each of its three sides.
A centered circle with that distance as radius touches each side and stays in their interior half-planes.
Any larger centered circle crosses a side, so this is the largest one.
Its radius is the triangle apothem, half the original circumradius.

$\begin{aligned}r_{\mathrm{small}}&=\frac{2\sqrt{13}}{2}\\r_{\mathrm{small}}&=\sqrt{13}\text{ cm}.\end{aligned}$

To draw the smaller circle, use the already constructed diameter AD and let M be its intersection with CE.
Both O and D are equally far from C and E: O by equal radii, and D by equal hexagon sides.
Thus line OD, which is line AD, is the perpendicular bisector of CE.
Draw the new circle with center O and opening OM.

For its square, draw a diameter PR.
Arcs centered at P and R, both with opening PR, meet at X,Y.
Their joining line cuts the smaller circle at Q,S.
Join P,Q,R,S in circle order.
The perpendicular-diameter proof gives a square wholly inside the smaller circle and hence inside the original triangle and hexagon.

$\begin{aligned}s_S&=\sqrt{13}\sqrt2\\s_S&=\sqrt{26}\text{ cm}\\A_S&=2(\sqrt{13})^2\\A_S&=26\text{ cm}^2.\end{aligned}$

The outer hexagon uses the original radius.
Its area and the required area fraction are

$\begin{aligned}A_H&=\frac{3\sqrt3}{2}(2\sqrt{13})^2\\A_H&=78\sqrt3\text{ cm}^2\\\frac{A_S}{A_H}&=\frac{26}{78\sqrt3}\\\frac{A_S}{A_H}&=\frac{\sqrt3}{9}.\end{aligned}$

The smaller radius comes from the triangle apothem.

Answer

Smaller radius $\sqrt{13}$ cm; square side $\sqrt{26}$ cm; square area $26\text{ cm}^2$ ; fraction $\frac{\sqrt3}{9}$ .

Key idea

An apothem can become the radius of a new construction; use each circle's own radius for its polygon.

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