Constructions of Polygons | ICSE Class 9 Maths Notes

This note covers polygon terminology, ruler-and-compasses methods, construction of quadrilaterals, parallelograms, rectangles, rhombuses and squares, construction of a regular hexagon, and checks based on sides, angles and diagonals.

What must you know before constructing a polygon?

A polygon is a simple closed plane figure made of line segments. Simple means that its boundary does not cross itself; closed means that the boundary returns to its starting point. A line segment is the straight part between two endpoints.

The boundary segments are its sides. Their meeting points are its vertices, with vertex meaning one such point. A quadrilateral is a polygon with four sides. A hexagon is a polygon with six sides.

How are points and lengths named?

Capital letters name points. In quadrilateral ABCD, the letters A, B, C and D follow the boundary in order. Its sides are AB, BC, CD and DA. Here AB names the segment joining A to B and, in a length statement, its length.

Adjacent sides share a vertex; opposite sides do not. A diagonal joins non-adjacent vertices. Thus AC and BD are the diagonals of ABCD. Joining points in a different order can turn a diagonal into an unintended boundary line.

An interior angle is the angle inside the polygon at a vertex. The notation ∠ABC means the angle between BA and BC, with B as its vertex. The symbol ° means degrees, the unit used here for angles; a right angle measures 90°.

Angles at adjacent vertices are adjacent angles; angles at opposite vertices are opposite angles. The symbol = means equals; + denotes addition, − subtraction and ÷ division.

What do the special shape names mean?

Parallel lines lie in the same plane and do not meet when extended. A parallelogram has both pairs of opposite sides parallel. A rhombus has four equal sides. A rectangle has four right angles, and a square has four equal sides and four right angles.

Definition: A regular polygon has equal side lengths and equal interior angles. Equilateral means equal-sided; equiangular means equal-angled. A regular hexagon must satisfy both conditions.

A convex quadrilateral has every interior angle less than 180°. The general quadrilateral methods below specify convex figures, so the position of a new vertex must agree with this condition as well as the given lengths.

How do ruler and compasses produce the basic constructions?

The ruler draws straight lines and sets out given lengths. The compasses transfer lengths and draw circles or arcs. A circle consists of points at a fixed distance from a fixed point called its centre; that distance is its radius. An arc is part of a circle.

To mark a point at a given distance from A, draw an arc centred at A with that distance as radius. To satisfy distances from two known points, use the intersection of their arcs. An intersection is a point common to both curves or lines.

How do you construct a perpendicular bisector?

A midpoint divides a segment into equal lengths. To bisect means to divide into two equal parts. Perpendicular lines meet at a right angle. A perpendicular bisector passes through a segment's midpoint at right angles to it.

1. Draw the segment AB whose midpoint is required.
2. Set the compasses to a radius greater than half AB. Draw arcs on both sides of AB from A.
3. Without changing the radius, draw arcs from B to meet the first pair. Name the intersections P and Q.
4. Join P to Q. Its intersection O with AB is the midpoint, and PQ is the perpendicular bisector.

The letters P and Q name the two arc intersections; O names the midpoint just obtained. Equal radii matter: changing the opening between centres A and B would remove the equal-distance condition on which the construction depends.

How do you construct the required angles?

A triangle is a polygon with three sides. An equilateral triangle has three equal sides and three angles of 60°. To construct one on AB, intersect arcs centred at A and B, each with radius AB, at C. Join AC and BC.

An angle bisector divides an angle into two equal angles. A ray starts at one point and extends indefinitely in one direction.

Worked example 1. Construct an angle of 30° using ruler and compasses.

Answer: Construct equilateral triangle ABC, giving ∠BAC = 60°. Draw an arc centred at A meeting AB and AC at P and Q. From P and Q draw equal-radius arcs meeting inside the angle at R. Join AR. This angle bisector divides ∠BAC into two equal angles, so ∠BAR = 60° ÷ 2 = 30°.

In this example, P and Q are on the first arc and R is the intersection of the second pair. The constructed ray AR starts at A and continues through R.

Note: Use ruler and compasses only. Numerical angle labels describe the result; they do not authorise setting the angle with a protractor.

How can four sides and one diagonal determine a quadrilateral?

A useful plan is to split the required quadrilateral into two triangles. If the four sides and one diagonal of a convex quadrilateral are given, each triangle has all three side lengths known. Construct the shared diagonal first.

For convex quadrilateral ABCD, suppose the supplied lengths are AB, BC, CD, DA and AC. No numerical values are needed to explain the method: each compass opening below is the corresponding length supplied in the question.

What is the construction order?

1. Draw diagonal AC with its given length. It is the common side of triangles ABC and ADC.
2. With A as centre and radius AB, draw an arc on one side of AC.
3. With C as centre and radius CB, intersect that arc at B. Join AB and BC.
4. With A as centre and radius AD, draw an arc on the other side of AC.
5. With C as centre and radius CD, intersect that arc at D. Join CD and DA.
6. Check that the completed boundary ABCD is convex and that all four sides and diagonal AC match the supplied data.

Why does the method work?

Each arc fixes a distance from its centre. B satisfies the required distances from A and C simultaneously; D does the same for its two distances. The diagonal therefore acts as a fixed shared starting segment for the two triangles.

Congruent triangles have the same shape and size. The side-side-side, or SSS, congruence condition states that three corresponding equal side lengths determine congruent triangles. It explains why the triangle construction is fixed, apart from reflection across its starting segment.

A reflection gives the mirror-image position on the other side of a line. For this convex quadrilateral method, choose B and D on opposite sides of AC. Still check convexity: opposite-side placement alone does not guarantee that arbitrary supplied lengths produce a convex quadrilateral.

Draw convex ABCD with diagonal AC inside it. Show the arc intersections locating B and D on opposite sides of AC. Label the four given sides and the common diagonal.

If the arcs do not intersect, do not force a vertex into the sketch. Check the specified lengths and compass openings. A triangle requires the sum of any two side lengths to be greater than the third.

How do you choose a starting triangle from other given data?

The best starting figure contains enough information to construct it before any other unknown vertex is needed. A rough sketch is an approximate planning diagram: it records the intended order of vertices and the supplied measurements, rather than claiming to be the completed construction.

What if three sides and two diagonals are given?

Suppose a convex quadrilateral ABCD is required and the given lengths are AB, BC, CD, AC and BD. Begin with triangle ABC, whose three sides are known. The remaining vertex D must be at the given distance BD from B and CD from C.

1. Draw AC and construct B using arcs with radii AB and CB.
2. Join AB and BC, completing the known triangle ABC.
3. From B draw an arc of radius BD; from C draw an arc of radius CD.
4. Choose an intersection D on the opposite side of AC from B that gives the required convex boundary.
5. Join CD and DA. Retain the construction of diagonal BD so that its prescribed length can be checked.

Here DA is not a given length. It is obtained by joining the constructed endpoints. Setting it independently would introduce an extra condition that the question did not supply.

How do special properties reduce the required data?

The included angle is the angle between the two supplied adjacent sides. A surrounding circle passes through all the polygon's vertices. In each special construction, use the defining properties to obtain missing data rather than choosing further measurements freely.

How do you construct a parallelogram from adjacent sides and their angle?

Property: Opposite sides of a parallelogram are equal

If ABCD is a parallelogram, AB equals CD and AD equals BC. Conversely, a quadrilateral with both pairs of opposite sides equal is a parallelogram. These facts let us locate the fourth vertex with distance arcs after placing two adjacent sides.

Begin at the vertex common to the two given sides. Construct their included angle there, then mark each length on the correct arm. In the example below, the arm AB receives the shorter given side and the arm AD receives the longer one.

Worked example 2. Construct parallelogram ABCD with AB = 4 cm, AD = 5 cm and ∠DAB = 30°. Here cm means centimetres, a unit of length.

Answer: Draw AB = 4 cm. At A construct a 60° angle and bisect it to obtain the 30° ray for AD. Mark AD = 5 cm on that ray. From B draw an arc of radius 5 cm; from D draw an arc of radius 4 cm. Choose their intersection C on the opposite side of BD from A. Join BC and CD.

Now AB = CD = 4 cm and AD = BC = 5 cm. Both pairs of opposite sides are equal, so ABCD is the required parallelogram.

How can the result be checked?

The opposite angles of a parallelogram are equal, and adjacent angles add to 180°. Thus the example has angles 30°, 150°, 30° and 150° in boundary order from A. These are consequences of the completed construction, not extra measurements to set out.

The alternative plan is to construct a line through D parallel to AB and a line through B parallel to AD. Their meeting point is C. The distance-arc method above avoids needing a separate parallel-line construction while producing the same required side relations.

Do not confuse the included angle at A with an angle between diagonals. The given 30° belongs between AB and AD. Drawing it at the crossing of AC and BD would solve a different construction problem.

How can a rectangle be constructed from its diagonals?

Property: Rectangle diagonals are equal and bisect each other

The diagonals of a rectangle have the same length and share a midpoint. Thus, if O is the point where diagonals AC and BD meet, AO equals OC and BO equals OD. Equality of both complete diagonals also makes all four half-diagonals equal.

Equal diagonals alone are insufficient: their common midpoint is an essential part of this construction. Conversely, joining the endpoints of equal segments that bisect each other gives a rectangle, provided the segments lie on different intersecting lines.

Worked example 3. Construct rectangle ABCD with diagonal AC = 8 cm and angle AOB = 60°, where O is the intersection of the diagonals.

Answer: Draw AC = 8 cm and construct its midpoint O. Construct a 60° ray OB from OA and extend its line through O in the opposite direction. Mark B and D on opposite rays with OB = OD = 4 cm. Join A, B, C and D in that order, then join D to A.

The complete second diagonal BD is 8 cm. Both diagonals are equal and bisect each other, so ABCD is a rectangle. The angle between OA and OB is the specified 60°.

Why must the diagonal angle be distinguished from a corner angle?

In this example, ∠AOB is formed at the interior crossing point O. The corner angles of the rectangle remain 90°. A rectangle's diagonals do not need to meet at right angles; that extra condition gives a square.

The diagonal length by itself allows rectangles of different shapes. The stated angle between the diagonals selects the shape required here. Label O before describing that angle so that there is no ambiguity about where it is constructed.

How do you construct a rhombus when both diagonals are given?

Property: Rhombus diagonals bisect each other at right angles

A rhombus is a parallelogram, so its diagonals bisect each other. They also meet at 90°. These conditions fix a rhombus from its two diagonal lengths: construct one complete diagonal, then place half of the other on either side of its midpoint.

Worked example 4. Construct rhombus ABCD whose diagonals AC and BD are 5 cm and 4 cm respectively.

Answer: Draw AC = 5 cm. Construct its perpendicular bisector, meeting AC at O. On the perpendicular line mark B and D on opposite sides of O, each 2 cm from O. Join AB, BC, CD and DA. Then AO = OC = 2.5 cm and BO = OD = 2 cm.

The diagonals have the prescribed lengths and bisect one another at 90°. Therefore ABCD is the required rhombus.

Why are all four sides equal?

The side-angle-side, or SAS, congruence condition states that triangles are congruent when two corresponding sides and their included angles are equal. Each small triangle around O has the same two half-diagonal lengths enclosing a right angle.

Applying that condition around O shows that the outer sides are equal. This gives the defining property of a rhombus, rather than relying on its appearance. The unequal diagonals in this example do not conflict with the four equal outer sides.

Draw AC = 5 cm and its perpendicular bisector through O. Place B and D at distances of 2 cm on opposite sides of O. Join the boundary ABCD and retain both diagonals.

Do not mark the full 4 cm on each side of O. That would make BD twice its required length. Write the half-lengths beside the sketch before opening the compasses, and check the complete diagonal after joining the vertices.

Although a square is also a rhombus, this example is not a square: a square's diagonals must be equal. The lengths 5 cm and 4 cm deliberately remain the given unequal diagonal lengths throughout the construction.

How can equal-sided triangles produce a rhombus?

Joining two equilateral triangles along a common side gives a useful rhombus construction. Place the triangles on opposite sides of that common segment. Their remaining sides form the outer boundary, while the shared side lies inside the completed quadrilateral as a diagonal.

How are the triangles located?

Draw the common side and use its length as the compass radius from both endpoints. The two circles meet on opposite sides of the segment. Each meeting point is at the required equal distance from both endpoints, so it completes an equilateral triangle.

Worked example 5. Join two equilateral triangles of side 4 cm on opposite sides of their common side BD. Construct the outer quadrilateral ABCD and find its sides and angles.

Answer: Draw BD = 4 cm. With B and D as centres and radius 4 cm, draw arcs meeting at A and C on opposite sides of BD. Join AB, AD, CB and CD. Each outer side is 4 cm, so ABCD is a rhombus.

Each triangle angle is 60°. The outer angles at A and C are therefore 60°. At B and D, two triangle angles meet, giving 60° + 60° = 120°. The boundary angles are 60°, 120°, 60° and 120°.

What makes the common side a diagonal?

The outer boundary is A to B to C to D to A. Along that boundary, B and D are not adjacent vertices. Therefore BD is a diagonal, even though it was the starting side of each separate triangle.

This distinction prevents a frequent labelling error. Count the outer sides after combining the triangles, rather than counting every segment remaining on the page as a polygon side. A construction line can remain visible without becoming part of the boundary.

The completed figure has four equal sides but its angles are not all equal. It is therefore a rhombus but not a regular quadrilateral. The square satisfies both the equal-side and equal-angle requirements for regularity.

How do you construct a square from one diagonal?

Property: Square diagonals are equal perpendicular bisectors

The diagonals of a square have equal lengths and bisect each other at right angles. Therefore one complete diagonal supplies the length of the other. Its midpoint supplies their crossing point, and the perpendicular bisector supplies the direction of the second diagonal.

Notice the three conditions working together: equal length, mutual bisection and right-angle intersection. A rhombus needs the last two conditions but its diagonals need not be equal. The additional equality produces the square.

Worked example 6. Construct square ABCD whose diagonal AC is 8 cm.

Answer: Draw AC = 8 cm. Construct the perpendicular bisector of AC and call their intersection O. Using OA as the compass radius, mark B and D on opposite sides of O along the perpendicular. Join AB, BC, CD and DA.

Here OA = OC = OB = OD = 4 cm, so BD = 8 cm. The diagonals are equal and bisect each other at 90°. Thus ABCD is the required square.

How does the same method apply to another given diagonal?

Worked example 7. Construct a square with diagonal 6 cm, using ruler and compasses only.

Answer: Name the diagonal AC and draw AC = 6 cm. Construct its perpendicular bisector through midpoint O. Transfer OA onto this line in both directions, obtaining OB = OD = 3 cm. Join AB, BC, CD and DA. Both complete diagonals are 6 cm and bisect each other at right angles, so the boundary is a square.

These examples begin with a diagonal length, not a side length. Drawing a side of 8 cm in the first example would not meet the question. Check which segment the stated measurement belongs to before drawing any line.

Draw AC = 6 cm, with O at its midpoint. Draw BD through O perpendicular to AC, with OB = OD = 3 cm. Join consecutive boundary vertices and mark the right angle between the diagonals.

The perpendicular-bisector arcs should remain clear enough to explain how O and the perpendicular were obtained. The final boundary and the auxiliary lines, meaning lines added to help the construction, serve different purposes and should be distinguishable.

How do you construct and verify a regular hexagon?

A regular hexagon has six equal sides and six equal interior angles. It can be constructed by stepping the radius around a circle without changing the compass opening. Here the radius equals the required side length.

What are the construction steps?

1. Set the compasses to the supplied side length and draw a circle with centre O.
2. Choose a point A on the circle. With A as centre and the same radius, cut the circle at B.
3. From B mark the next point C on the circle, taking the intersection other than A.
4. Continue in the same direction, marking D from C, E from D and F from E, each time taking the next intersection rather than the preceding vertex.
5. Join AB, BC, CD, DE, EF and FA to form the closed boundary.
6. Join O to the vertices to check the equal triangles that explain the construction.

In these steps A, B, C, D, E and F name consecutive vertices. A central angle has its vertex at the centre of the circle. Each triangle formed by O and two consecutive vertices has all three sides equal to the radius.

Why does stepping the radius close the figure?

Each of these triangles is equilateral, so its angle at O is 60°. Six such central angles make 360°, one complete turn. The sixth step returns to the starting point, and all six boundary sides have the same length.

Worked example 8. Find the exterior and interior angles of a regular hexagon. An exterior angle here is formed by one side and the extension of the adjacent side.

Answer: The six equal exterior angles together make 360°. Each exterior angle is 360° ÷ 6 = 60°. An interior angle and its adjacent exterior angle form a straight angle of 180°. Therefore each interior angle is 180° − 60° = 120°.

The 60° central angle is not the 120° interior angle at a boundary vertex. Joining a vertex to O divides its interior angle into two 60° angles. This verifies equal angles as well as equal side lengths.

Mark centre O and successive vertices A, B, C, D, E and F on the circle. Join neighbouring vertices and draw all six radii. Show one central angle of 60° and one interior angle of 120°.

Keep the compass radius unchanged throughout the stepping procedure. If the last side does not close accurately, check the original radius and each marked intersection. Changing the final opening to force closure would break the equal-side construction.

Glossary

• Polygon — A simple closed plane figure whose boundary consists entirely of straight line segments.
• Vertex — A meeting point of two consecutive sides on a polygon's boundary.
• Diagonal — A line segment joining two vertices that are not adjacent on the polygon.
• Regular polygon — A polygon with all side lengths equal and all interior angles equal.
• Parallelogram — A quadrilateral in which both pairs of opposite sides are parallel.
• Rhombus — A quadrilateral whose four sides are equal in length.
• Midpoint — The point on a line segment that divides it into two equal lengths.
• Perpendicular bisector — A line passing through a segment's midpoint at right angles to that segment.
• Angle bisector — A ray dividing a given angle into two angles of equal measure.
• Radius — The distance from a circle's centre to any point on the circle.
• Arc — A portion of a circle used to locate points at a fixed distance from its centre.
• Central angle — An angle formed by two radii with its vertex at the centre of a circle.
• Congruent triangles — Triangles having the same shape and size, with equal corresponding sides and angles.

Common errors and misconceptions

• Misconception: A protractor can be used to set the required angles. Correct: Use ruler and compasses only; construct the angle using equal-sided triangles, bisection or a perpendicular as appropriate.
• Misconception: Any two equal diagonals give a rectangle. Correct: The diagonals must also bisect each other. Equality alone does not establish that the completed quadrilateral is a rectangle.
• Misconception: A rhombus has equal diagonals. Correct: Its diagonals bisect each other at right angles but need not be equal. Equal diagonals give the special case of a square.
• Misconception: Put the complete second diagonal length on each side of the midpoint. Correct: Put half its length on each side, so the two parts together give the prescribed diagonal.
• Misconception: The vertices can be joined in any order. Correct: Follow the polygon's boundary order. Joining a pair of non-adjacent vertices draws a diagonal rather than a boundary side.
• Misconception: Six equal sides alone prove a hexagon is regular. Correct: Regularity requires equal interior angles too. The circle construction verifies both through its six equilateral triangles.
• Misconception: A regular hexagon's interior angle is 60°. Correct: Its central and exterior angles are 60°, while each interior angle is 120°.

Exam-style questions with model answers

Q1. State the two conditions that make a hexagon regular. [2 marks]

1. All six sides of the hexagon must have the same length.
2. All six interior angles of the hexagon must have the same measure.

Q2. A rhombus has diagonals AC = 5 cm and BD = 4 cm meeting at O. State the half-diagonal lengths and the angle between the diagonals. [3 marks]

1. A rhombus has diagonals that bisect each other, so O is the midpoint of AC. Hence AO and OC are each 5 ÷ 2 = 2.5 cm.
2. O is also the midpoint of BD. Therefore BO and OD are each 4 ÷ 2 = 2 cm.
3. The diagonals of a rhombus are perpendicular, so the angle between AC and BD is 90°.

Q3. Describe a ruler-and-compasses construction of square ABCD with diagonal AC = 6 cm, and justify the result. [5 marks]

1. Draw the given diagonal AC = 6 cm. A and C are opposite vertices of the required square, so AC is not one of its boundary sides.
2. From A and C draw equal-radius arcs above and below AC, using a radius greater than 3 cm. Join their intersections to form the perpendicular bisector.
3. Call its intersection with AC point O. This gives AO = OC = 3 cm and a line through O perpendicular to AC.
4. With centre O and radius OA, mark B and D on opposite sides of O along this perpendicular. Join AB, BC, CD and DA.
5. AC and BD are both 6 cm, share midpoint O and meet at 90°. Equal diagonals bisecting each other at right angles establish the required square.

Q4. Construct parallelogram ABCD with AB = 4 cm, AD = 5 cm and ∠DAB = 30°, using ruler and compasses only. Explain why it is a parallelogram. [5 marks]

1. Draw AB = 4 cm. Construct an equilateral triangle on AB to obtain a 60° angle at A, on the chosen side of AB.
2. Bisect that angle using equal-distance arcs: intersect its arms with an arc centred at A, then intersect equal-radius arcs from those points. Join A to their intersection.
3. This bisector makes 30° with AB. On it mark D so that AD = 5 cm, setting out the second adjacent side.
4. Draw arcs centred at B with radius 5 cm and at D with radius 4 cm. Choose C opposite A across BD; join BC and CD.
5. AB = CD = 4 cm and AD = BC = 5 cm. Since both pairs of opposite sides are equal, ABCD is a parallelogram with the required included angle.

Q5. Describe how to construct a regular hexagon whose side equals a supplied line segment AB. Use ruler and compasses only, and justify regularity. [6 marks]

1. Set the compasses to the length of the supplied segment AB. Draw a circle with this radius and name its centre O.
2. Choose a point P on the circle. With P as centre and the unchanged radius, mark a point Q where an arc cuts the circle.
3. Continue in the same direction, stepping the same radius from Q to R, then S, T and U, each time selecting the next intersection.
4. Join PQ, QR, RS, ST, TU and UP. Each side has the compass radius as its length and therefore equals the supplied segment AB.
5. Join O to all six vertices. The resulting triangles have three sides equal to the radius, so each is equilateral with central angle 60°. Six such angles complete 360°.
6. Each boundary vertex has two adjacent triangle angles of 60°, making its interior angle 120°. Equal sides and equal interior angles prove that PQRSTU is regular.

Q6. Two equilateral triangles ABD and CBD, each with side 4 cm, lie on opposite sides of BD. Identify the outer quadrilateral ABCD and find all its sides and interior angles. [4 marks]

1. The outer sides AB, BC, CD and DA are each sides of the equilateral triangles, so each measures 4 cm.
2. The outer figure has four equal sides and is therefore a rhombus. Its shared internal segment BD is a diagonal.
3. The angles at A and C are single angles of equilateral triangles. Each is therefore 60°.
4. At B and D, two 60° triangle angles meet. Each interior angle is 120°, giving angles 60°, 120°, 60° and 120° in boundary order.

Q7. Convex quadrilateral ABCD is to be constructed from supplied segments giving AB, BC, CD, DA and AC. Describe the method, assuming these lengths permit the required convex figure. [4 marks]

1. Draw AC equal to its supplied length. It will be the shared diagonal of the two triangles used in the construction.
2. On one side of AC, intersect arcs centred at A and C with radii AB and CB respectively. Name their intersection B and join AB and BC.
3. On the opposite side of AC, intersect arcs centred at A and C with radii AD and CD respectively. Name the appropriate intersection D.
4. Join CD and DA. Check the convex boundary ABCD and all five supplied lengths; the arcs have fixed each vertex at the prescribed distances from A and C.

Key takeaways

• Use ruler and compasses to construct the required lengths, angles and intersections, retaining enough construction marks to explain the method.
• Name vertices in boundary order and distinguish the polygon's sides from the diagonals drawn inside it.
• A quadrilateral with four sides and a diagonal supplied can be approached as two triangles sharing that diagonal.
• Use a parallelogram's equal opposite sides to locate its fourth vertex after constructing the adjacent sides and included angle.
• For a rhombus, construct perpendicular diagonals that bisect each other; equal diagonal lengths are not required.
• For a square, the diagonals must be equal as well as perpendicular and mutually bisecting.
• A regular hexagon is constructed by stepping the unchanged circle radius six times around its boundary, then joining successive vertices.
• Verify a completed construction against every given measurement and the properties that establish the requested type of polygon.

Test yourself

What are the diagonals of quadrilateral ABCD, named in boundary order?

AC and BD are its diagonals because each joins two non-adjacent vertices.

Why must the arcs used to construct a perpendicular bisector have equal radii?

Their intersections must be equally distant from the segment's two endpoints. Equal compass radii establish those equal distances.

Why are B and D placed on opposite sides of AC in the convex quadrilateral method?

AC divides a convex quadrilateral into two triangles on opposite sides of the diagonal. The placement must reproduce that arrangement.

A rhombus has one diagonal of 4 cm. How much of that diagonal lies on each side of its intersection with the other diagonal?

Each half is 2 cm because the diagonals of a rhombus bisect each other.

What extra diagonal condition distinguishes the square construction from the general rhombus construction?

The square's diagonals must also be equal, in addition to bisecting each other at right angles.

What compass opening is used to step off a regular hexagon on a circle?

Use the circle's radius and keep that opening unchanged while marking successive vertices.

Why does each central triangle in the regular hexagon construction have a 60° angle at the centre?

Its two radii and its hexagon side are equal, making it an equilateral triangle with three 60° angles.

Why does an accurate-looking sketch not complete a construction argument?

The explanation must connect the compass settings, intersections and geometric properties to every required length, angle and shape condition.