HSG.CO.D.13: Constructing an Equilateral Triangle, Square and Regular Hexagon in a Circle
In plain English: HSG.CO.D.13 is the Common Core geometry standard that asks students to construct an equilateral triangle, a square and a regular hexagon inscribed in a circle with compass and straightedge. The key idea is that all radii are congruent, so each construction splits the circle into equal central angles of 120°, 90° or 60°. It is usually taught in high school Geometry.
Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.
Common Core State Standards for Mathematics · Domain: Congruence (CO) · Cluster: Make geometric constructions Also written as HSG-CO.D.13 or G-CO.13 · Official standard
Students construct the three regular polygons named in the standard inside a given circle, using only a compass and a straightedge, and explain why each construction works. The hexagon comes first because it needs a single compass setting: the radius fits around the circle exactly six times, since the center and two neighboring marks form an equilateral triangle with 60° at the center. Connecting every other mark of the hexagon gives the equilateral triangle, and a diameter with its perpendicular bisector gives the square.
Every construction is paired with a justification. Students name the congruent radii, the central angles (120°, 90° and 60°) and the congruent chords that follow, and they check their drawings by measuring. By the end, students can build each polygon from a blank circle and say which fact makes each side congruent.
Learning Objectives
By the end of this lesson, students will be able to:
Construct a regular hexagon inscribed in a circle by stepping a compass set to the radius around the circle
Construct an equilateral triangle inscribed in a circle, either from the hexagon marks or from a diameter and one arc
Construct a square inscribed in a circle by drawing a diameter and its perpendicular bisector
Justify each construction with congruent radii, equal central angles and congruent chords
Check a construction by measuring and explain what a mismatch says about the drawing
Prior Knowledge Required
Students should already be comfortable with:
Definitions of circle, radius, diameter, chord and perpendicular lines HSG.CO.A.1
Basic constructions: copying a segment and constructing a perpendicular bisector HSG.CO.D.12
Properties of equilateral and isosceles triangles, and the triangle angle sum 8.G.A.5
Properties of rectangles and squares, including their diagonals HSG.CO.C.11
The Pythagorean Theorem for checking side lengths 8.G.B.7
Each student draws a circle with a radius of 5 cm and marks one point on it. Without changing the compass opening, students place the compass point on the mark, swing a short arc across the circle, move to the new crossing and repeat.
Warm-Up Prompt
"Keep your compass at the radius of your circle and step it around the circle. How many steps bring you back to the starting mark? Why should the answer be the same for every circle?"
Collect answers. Careful drawings close up after six steps; drawings that overshoot or fall short usually have a compass that slipped. Ask students to draw the two radii to one pair of neighboring marks and describe the triangle they see. Leave the question "why exactly six?" on the board for Direct Instruction.
Direct Instruction20 minutes
Vocabulary. A polygon is inscribed in a circle when all of its vertices lie on the circle. A regular polygon has congruent sides and congruent angles. In an inscribed polygon, each side is a chord, and chords with equal central angles are congruent. Model each construction on the board with Diagram 1 and Diagram 2, and write the reason next to each step:
Regular hexagon: set the compass to the radius OA. From A, mark B on the circle; from B, mark C; continue to F. Connect the six marks in order. Reason: OA = OB = AB = r, so triangle OAB is equilateral and angle AOB = 60°. Six 60° central angles fill 360°, so the sixth arc lands back on A and all six chords equal r.
Equilateral triangle from the hexagon: after marking A through F, connect A, C and E. Reason: each side skips one mark, so it spans two 60° central angles. All three central angles are 120°, so the three chords are congruent.
Equilateral triangle from a diameter: draw diameter AD. With the compass at the radius, draw one arc centered at D; it crosses the circle at B and C. Connect A, B and C. Reason: triangles OBD and OCD are equilateral, so angle BOC = 120°, and angles AOB and AOC are each 180° - 60° = 120°.
Square: draw diameter AC, construct its perpendicular bisector (it passes through O), and let it meet the circle at B and D. Connect A, B, C and D. Reason: the diagonals AC and BD are congruent diameters that bisect each other at right angles, so the four central angles are 90° and the four sides are congruent.
Check by measuring: for radius r, the hexagon side is r, the triangle side is r√3 and the square side is r√2. A measured side that is off by more than a millimeter or two points to a slipped compass or a misplaced mark.
Work through the examples. Before each one ask: "Which segments are radii, and what central angle does each side use?"
Regular hexagon
A circle has radius 4 cm. A student keeps the compass at 4 cm, marks six points around the circle and connects neighbors. Explain why the hexagon is regular and find its perimeter.
Equation: OA = OB = AB = 4, so each central angle is 60° and 6 · 60° = 360°; each side is 4 cm and the perimeter is 24 cm
Equilateral triangle from the hexagon marks
On the same 4 cm circle, connect every other mark, A, C and E. Find the central angle of each side and the side length.
Equation: central angle 2 · 60° = 120°; AD is a diameter, so angle ACD = 90° and AC = √(8² - 4²) = 4√3 ≈ 6.93 cm
Square from perpendicular diameters
A circle has radius 5 cm. Diameter AC and its perpendicular bisector BD are drawn, and A, B, C, D are connected. Why is ABCD a square, and how long is each side?
Equation: OA = OB = OC = OD = 5 and AC ⊥ BD, so the four right triangles at O are congruent; side = √(5² + 5²) = 5√2 ≈ 7.07 cm
Equilateral triangle from one arc
A circle with center O has radius 3 in. Diameter AD is drawn, and an arc centered at D with radius 3 in crosses the circle at B and C. Show that triangle ABC is equilateral.
Equation: OB = OD = BD = 3, so angle BOD = 60°; likewise angle COD = 60°, so angle BOC = 120° and angles AOB = AOC = 180° - 60° = 120°; equal central angles give AB = BC = CA
Point out that the hexagon and triangle constructions use one compass setting for the whole drawing, while the square needs a second setting for the perpendicular bisector. In both diagrams every claim rests on the same fact: radii of one circle are congruent.
Guided Practice10-15 minutes
Pairs draw one circle with a radius of 6 cm and construct all three polygons on it, using different colors. Before measuring, each pair predicts the side lengths: hexagon 6 cm, triangle 6√3 ≈ 10.4 cm, square 6√2 ≈ 8.5 cm. Then they measure and record the difference. Circulate and check three habits: the compass point sits exactly on each new crossing, the marks are connected in order around the circle, and the perpendicular bisector is constructed rather than drawn by eye.
Guided practice predictions for a 6 cm circle
Polygon
Central angle
Predicted side
Measured side
Regular hexagon
60°
6 cm
_____
Square
90°
6√2 ≈ 8.5 cm
_____
Equilateral triangle
120°
6√3 ≈ 10.4 cm
_____
Independent Practice10 minutes
Students work alone on a fresh circle of any size. (a) Construct an inscribed equilateral triangle using the diameter-and-one-arc method, and write the reason each side is congruent. (b) Construct an inscribed square when only the circle and its center are given, with no diameter drawn yet, and explain why the two diameters you draw are perpendicular. (c) Explain in one sentence why the compass must stay at the radius for the whole hexagon construction.
Closure5 minutes
Exit ticket: "Point P is on circle O. List the construction steps for a regular hexagon inscribed in the circle with one vertex at P, and then say which points you would connect to turn it into an inscribed equilateral triangle. Give one reason the triangle is equilateral." Collect and sort the tickets by whether the reason names the radii or the 120° central angles.
Differentiation Strategies
For Struggling Students
Give a circle with the center already marked and a step-by-step card with a small sketch after each step
Use a compass with a locking wheel so the radius does not slip during the hexagon construction
Let students fold a paper circle in half twice to see the square before they construct it
For Advanced Students
Construct a regular dodecagon by bisecting the six central angles of the hexagon, and explain why it is regular
Construct an inscribed square when the center of the circle is not given, by first locating the center with two perpendicular bisectors of chords
Prove that the side of the inscribed equilateral triangle is √3 times the radius, using the right angle inscribed in a semicircle
Assessment Guidance
What to Look For
Check the drawings for visible construction marks: the six arcs of the hexagon, the two crossing arcs of the perpendicular bisector, and a straightedge line through them. A polygon drawn by eye does not count as a construction, even when it looks right. In written reasons, look for the congruent radii and the central angle of each side (60°, 90° or 120°), not only "it looks equal". When a measured side misses the predicted length, ask students which step could have caused it.
02
Classroom Activities
3 Activities
1
Hexagon and Triangle Construction Relay
20 minPairs
Partners take turns doing one step of the hexagon construction while the other states the reason for it. The finished hexagon then becomes the starting point for the equilateral triangle, so students see that both polygons come from the same six marks.
Procedure
Partner 1 draws a circle with a radius of 7 cm, labels the center O and marks a point A on the circle
Partners alternate: one steps the compass (still at 7 cm) to the next mark, the other says why the new chord is 7 cm long. Continue until the sixth arc returns to A
Label the marks A to F, connect them in order, and measure two sides and one interior angle (sample result: 7 cm and 120°)
On the same drawing, connect A, C and E in a second color. Predict the side length (7√3 ≈ 12.1 cm) and measure it
Discussion Questions
Why does the sixth step land on A and not somewhere near it?
What equilateral triangle is hidden in each step of the construction?
Which other points could you connect to get a second inscribed equilateral triangle?
Modification for Distance Learning
Use a free dynamic geometry tool: construct the circle with its radius as a segment, use the compass tool with that segment six times, and drag the original point to show that the hexagon stays regular. Students share a screenshot with one reason written under each step.
2
Two Ways to Make a Square
15 minPairs
Pairs build an inscribed square twice: once with compass and straightedge, and once by folding a paper circle. Comparing the two methods shows that both rely on two perpendicular diameters.
Procedure
Compass method: on a circle with a radius of 4.5 cm, draw a diameter AC. Construct its perpendicular bisector with two pairs of crossing arcs, label the new endpoints B and D, and connect A, B, C, D
Folding method: cut out a paper circle, fold it in half, then fold it in half again so the first crease lies on itself. Unfold and connect the four crease endpoints
For both squares, measure one side and both diagonals. Predicted side: 4.5√2 ≈ 6.4 cm for the compass square
Write one sentence for each method explaining why the creases or lines are perpendicular
Discussion Questions
If you draw two diameters that are not perpendicular and connect their endpoints, what shape do you get, and why?
In the folding method, which fold makes the second crease perpendicular to the first?
Why must the perpendicular bisector of a diameter pass through the center of the circle?
3
Justification Card Sort
20 minGroups of 3-4
Groups receive 6 cards, each with one reason. They build all three constructions on one large circle and place each card next to the construction step it justifies. Some cards justify more than one construction.
The 6 Cards
Card 1: All radii of a circle are congruent
Card 2: A triangle with three congruent sides is equilateral, so each of its angles is 60°
Card 3: Chords with congruent central angles are congruent
Card 4: The perpendicular bisector of a chord passes through the center
Card 5: A quadrilateral whose diagonals are congruent, bisect each other and are perpendicular is a square
Card 6: The central angles around the center add up to 360°
Procedure
Construct the hexagon, the triangle ACE and the square on one circle with a radius of 8 cm, in three colors
Place every card next to at least one step. Record which card or cards justify each polygon
Write one complete justification for one polygon of the group's choice, citing card numbers
Challenge Variation
Going further, beyond the standard: bisect each 60° central angle of the hexagon and construct a regular dodecagon. Groups decide which cards still apply and write the central angle of each side (30°).
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Regular Hexagon and Equilateral Triangle
Left: with the compass fixed at the radius r, six arcs step around the circle. Each arc is centered at one mark and passes through the next, so triangle OAB is equilateral and every central angle is 60°. Right: connecting every other mark gives triangle ACE, whose sides each span 120° and measure r√3. Points are computed exactly at 60° intervals.
Diagram 2: Square from a Diameter and Its Perpendicular Bisector
Arcs of equal radius centered at A and C cross above and below the center. The line through the crossings is the perpendicular bisector of diameter AC and gives the second diameter BD. The four half-diagonals are radii (tick marks), and the right angle at O makes ABCD a square with side r√2. Drawn to scale with r = 120 units.
Directions: Use a compass and straightedge for every construction and leave all construction marks visible. Measure with a centimeter ruler to the nearest tenth, and give a written reason for each construction.
Part 1: Constructions (Problems 1-3)
Draw a circle with a radius of 3.5 cm. Construct a regular hexagon inscribed in it. Measure two sides, give the perimeter you expect, and explain why the compass returns to the starting point after six arcs.
Draw a circle with a radius of 5.5 cm and draw one diameter. Construct an inscribed equilateral triangle using one arc, with the compass at the radius, centered at an endpoint of the diameter. Predict the side length to the nearest tenth, then measure it.
Draw a circle with a radius of 6.5 cm. Construct an inscribed square using a diameter and its perpendicular bisector. Predict the side length to the nearest tenth and measure it. Name the construction you used to make the second diameter.
Part 2: Reasons and Errors (Problems 4-6)
Jordan draws two diameters of a circle that look perpendicular, but he draws the second one by eye, and connects the four endpoints. Explain what kind of quadrilateral he is guaranteed to get, why it might not be a square, and which construction step would fix it.
Regular hexagon PQRSTU is inscribed in a circle with center O and a radius of 10 cm. Find the central angle POQ, the interior angle PQR and the length of PR, and explain why triangle PRT is equilateral.
Marisol tries to construct a regular hexagon in a circle with a radius of 5 cm, but her compass opens to 5.2 cm while she works. Is each central angle she steps off larger or smaller than 60°? Will her sixth arc land before or after the starting point? Explain.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Construction Accuracy
Measured sides within 0.2 cm of the prediction
One polygon off by more than 0.2 cm
Polygons drawn by eye or not closed
Construction Marks
All arcs and lines visible and correct
Some marks missing
No construction marks
Justification
Names congruent radii and central angles for each polygon
Reasons given for some polygons
No valid reasons
Error Analysis
Problems 4 and 6 identify the cause and the fix
Cause identified without a fix
No analysis
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
To mark the vertices of a regular hexagon inscribed in a circle, what compass opening do you keep for every step?
Answer: B
The radius works because each step makes an equilateral triangle with the center, so each central angle is 60° and six steps fill 360°. Choice C is a common error: a fixed opening that is not the radius makes the arcs miss the starting point. Choice A would only reach the opposite point of the circle.
Question 2 of 20 · Multiple Choice
In the hexagon construction, why is triangle OAB equilateral, where O is the center and A and B are neighboring marks?
Answer: A
OA = OB because they are radii, and AB = OA because the compass opening was the radius, so all three sides are equal. Choice C uses measurement, which checks a construction but does not justify it. Choice D is false: AB is a chord much shorter than a diameter.
Question 3 of 20 · Multiple Choice
Six marks A, B, C, D, E, F are stepped off in order around a circle for a regular hexagon. Which marks do you connect to get an inscribed equilateral triangle?
Answer: C
Connecting every other mark gives sides that each span two 60° central angles, so all three central angles are 120° and the sides are congruent. Choice A uses neighboring marks, which gives a triangle with a 120° angle at B. Choice B contains the diameter AD, so it is a right triangle.
Question 4 of 20 · Multiple Choice
What is the central angle of each side of an equilateral triangle inscribed in a circle?
Answer: D
The three congruent sides split the circle into three equal arcs: 360° ÷ 3 = 120°. Choice A is the interior angle of the triangle, not the central angle. Choice B is the central angle of a side of the inscribed square.
Question 5 of 20 · Multiple Choice
What is the central angle of each side of a square inscribed in a circle?
Answer: A
Four congruent sides split the circle into four equal arcs: 360° ÷ 4 = 90°, which is why the diagonals of the inscribed square are perpendicular diameters. Choice B is the angle between a side and a diagonal. Choice C belongs to the hexagon.
Question 6 of 20 · Multiple Choice
A diameter AC of a circle is drawn. Which construction gives the other two vertices of an inscribed square?
Answer: B
The perpendicular bisector of AC passes through the center O and is perpendicular to AC, so it is the second diameter BD and gives four 90° central angles. Choice A gives a line that never enters the circle except at A. Choice C would need a chord as long as the diameter, which only reaches C again. Choice D describes the line AC itself.
Question 7 of 20 · Multiple Choice
A circle has a radius of 9 cm. How long is each side of the regular hexagon inscribed in it?
Answer: D
Each side is the base of an equilateral triangle whose other sides are radii, so the side equals the radius: 9 cm. Choice A is the diameter. Choice B is the side of the inscribed equilateral triangle, not the hexagon.
Question 8 of 20 · Multiple Choice
A circle has a radius of 10 cm. About how long is each side of the square inscribed in it?
Answer: A
Two radii meet at a right angle at the center, so the side is √(10² + 10²) = 10√2 ≈ 14.1 cm. Choice C is the diagonal (a diameter), not the side. Choice D is 5√2, which uses half the radius.
Question 9 of 20 · Multiple Choice
A circle has a radius of 2 in. What is the exact side length of the equilateral triangle inscribed in it?
Answer: C
The side spans a 120° central angle. From the diameter through one vertex, the triangle side is a leg of a right triangle with hypotenuse 4 and other leg 2, so it is √(4² - 2²) = √12 = 2√3 ≈ 3.46 in. Choice A is the hexagon side. Choice B is the diameter, which is longer than any side of the triangle.
Question 10 of 20 · Multiple Choice
A student draws two diameters of a circle at an angle of about 70° to each other and connects their four endpoints. Which shape is guaranteed?
Answer: A
The diagonals are two diameters, so they are congruent and bisect each other, which makes the quadrilateral a rectangle. It is a square only if the diameters are perpendicular, so choice B is not guaranteed. Choice C would need perpendicular diagonals too.
Question 11 of 20 · Multiple Choice
In circle O, diameter AD is drawn, and an arc centered at D with radius OD crosses the circle at B and C. What is angle BOC?
Answer: B
Triangles OBD and OCD are equilateral, so angles BOD and COD are each 60°, and angle BOC = 60° + 60° = 120°. Choice A is only angle BOD. Then triangle ABC has three 120° central angles and is equilateral.
Question 12 of 20 · Multiple Choice
Why does connecting every other vertex of an inscribed regular hexagon produce an equilateral triangle?
Answer: D
Chords with congruent central angles are congruent, and each side of triangle ACE spans 120°. Choice A notices a pattern but gives no geometric reason. Choice B is false: a diameter spans 180°, and joining every third vertex would give one.
Question 13 of 20 · Multiple Choice
A paper circle is folded in half, and then folded in half again so that the first crease lands on itself. When it is unfolded, what do the creases show?
Answer: B
The first fold is a line of symmetry, so it is a diameter. Folding the diameter onto itself makes the second crease its perpendicular bisector through the center, so the creases are perpendicular diameters. Choice A would give a rectangle that is not a square.
Question 14 of 20 · Multiple Choice
What is the measure of each interior angle of a regular hexagon inscribed in a circle, found from the equilateral triangles of the construction?
Answer: D
Each interior angle, such as angle ABC, is made of two 60° angles from the neighboring equilateral triangles OAB and OBC: 60° + 60° = 120°. Choice A is the central angle. Choice B is the interior angle of a regular pentagon, and choice C belongs to a regular octagon.
Question 15 of 20 · Short Answer
Write the construction steps for a regular hexagon inscribed in a circle with center O, and give the reason the hexagon is regular.
Set the compass to the radius. Mark a point A on the circle, then with the compass point at A, mark B where the arc crosses the circle. Move to B and mark C, and continue to D, E and F. Connect A to F in order. Reason: OA = OB = AB, so each triangle such as OAB is equilateral with a 60° central angle. Six 60° angles make 360°, so all six chords are congruent (each equals the radius), and each interior angle is 60° + 60° = 120°.
Question 16 of 20 · Short Answer
A circle has a radius of 2.5 cm. Find the side lengths of the inscribed equilateral triangle, square and regular hexagon, exactly and to the nearest tenth.
Triangle: 2.5√3 ≈ 4.3 cm. Square: 2.5√2 ≈ 3.5 cm. Hexagon: 2.5 cm, equal to the radius. The more sides the polygon has, the shorter each side, because each side spans a smaller central angle (120°, 90°, 60°).
Question 17 of 20 · Short Answer
Explain why the quadrilateral formed by connecting the endpoints of two perpendicular diameters is a square, and not only a rectangle.
The four half-diagonals are radii, so they are congruent, and they meet at 90° angles at the center. The four right triangles at the center are therefore congruent by SAS, so all four sides are congruent. Each angle of the quadrilateral is made of two 45° base angles of these isosceles right triangles, so it is 90°. Four congruent sides and four right angles make a square. Congruent diagonals alone give only a rectangle; the right angle at the center makes the sides equal.
Question 18 of 20 · Short Answer
A student's hexagon construction does not close: the sixth arc lands a few millimeters past the starting point. What went wrong, and how should the student fix it?
The compass opening became larger than the radius during the construction (it slipped open or was set wrong). A chord longer than the radius has a central angle larger than 60°, so six steps pass 360°. The fix is to reset the compass exactly on a radius, lock it if possible, and start again from the first mark rather than adjusting the last step.
Question 19 of 20 · Short Answer
Point A is on circle O. Describe how to construct an equilateral triangle with vertex A inscribed in the circle using only one diameter and one arc, and explain why it works.
Draw the diameter from A through O to the opposite point D. With the compass set to the radius, draw an arc centered at D; it meets the circle at B and C. Connect A, B and C. OB = OD = BD, so triangle OBD is equilateral and angle BOD = 60°; the same holds for angle COD. So angle BOC = 120°, and angles AOB and AOC are 180° - 60° = 120°. Equal central angles give congruent chords AB, BC and CA.
Question 20 of 20 · Short Answer
Square WXYZ is inscribed in a circle with a radius of 4 cm. Find the exact side length and the perimeter to the nearest tenth.
Two radii meet at a right angle at the center, so the side is √(4² + 4²) = 4√2 cm ≈ 5.66 cm. The perimeter is 4 · 4√2 = 16√2 ≈ 22.6 cm.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.CO.D.13 mean?
HSG.CO.D.13 asks students to construct an equilateral triangle, a square and a regular hexagon inscribed in a circle. Inscribed means every vertex lies on the circle. Students build each polygon with construction tools and explain why its sides and angles are congruent.
Is HSG.CO.D.13 taught in Geometry?
Yes, it is usually taught in high school Geometry, in the constructions unit next to HSG.CO.D.12. It often returns in the circles unit, where students use central angles and inscribed angles to explain the same constructions.
Why does the radius fit exactly six times around a circle?
Because each step makes an equilateral triangle with the center. Two radii and one chord equal to the radius form a triangle with three equal sides, so its angle at the center is 60°, and 360° ÷ 60° = 6.
How do you construct an equilateral triangle inscribed in a circle?
Step off the six hexagon marks with the compass at the radius and connect every other mark. A shorter method: draw a diameter from the chosen vertex, draw one arc with radius r centered at the far end of the diameter, and connect the two crossings to the chosen vertex. In both methods each side spans a 120° central angle.
How do you construct a square inscribed in a circle?
Draw a diameter, construct its perpendicular bisector, and connect the four points where the two lines meet the circle. The two diameters are congruent, bisect each other and are perpendicular, so the quadrilateral is a square.
Do students have to use a compass and straightedge for HSG.CO.D.13?
Compass and straightedge are the usual tools. The related standard HSG.CO.D.12 also lists string, reflective devices, paper folding and dynamic geometry software, and many teachers accept these methods as long as the student can justify each step. Paper folding is a natural way to make the square.
What mistakes do students make with these constructions?
Common errors include:
letting the compass slip during the hexagon, so the sixth arc misses the start
drawing the second diameter of the square by eye instead of constructing it
connecting neighboring marks instead of every other mark for the triangle
erasing construction marks, which removes the evidence of the construction
How is HSG.CO.D.13 usually assessed?
Students may be asked to carry out a construction with visible marks, to put given steps in order, to name the step that makes a figure a square, or to explain why a construction works. Computational items may ask for central angles or for side lengths in terms of the radius.
What is the difference between an inscribed and a circumscribed polygon?
A polygon inscribed in a circle has all its vertices on the circle, so the circle is outside the polygon. A polygon circumscribed about a circle has every side tangent to the circle, so the circle is inside. HSG.CO.D.13 is about inscribed polygons.
How does HSG.CO.D.13 connect to other standards?
It applies the basic constructions of HSG.CO.D.12 and the symmetry of regular polygons from HSG.CO.A.3. The same inscribed polygons appear in HSG.C.A.3, where students construct circles through the vertices of a triangle, and in HSG.GMD.A.1, where polygons with more and more sides are used to argue about the circumference and area of a circle.
07
Related Standards
6 standards
These standards connect to HSG.CO.D.13: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.CO.A.1Prerequisite
Know precise definitions of angle, circle, perpendicular and parallel lines, segment
