Six Steps of the Compass Round a Circle Give a Hexagon
Construct a quadrilateral from four sides and a diagonal, build a parallelogram from sides, diagonals and angles, draw a rhombus from its diagonals, and step off a regular hexagon with the compass alone.
Why does stepping the compass six times round a circle close it exactly?
Open your compass to some radius, draw a circle, then without changing the setting put the point on the circle and step the same distance round the edge. The sixth mark lands exactly back on the first one — not approximately, exactly.
Here is why. Join the centre to two neighbouring marks and . Then and are radii, and was stepped off at the same compass setting, so
The triangle is equilateral, so . Six such angles make , a complete turn, so six steps close the circle and the six chords form a regular hexagon.
So the side of a regular hexagon equals the radius of its circle. That is the only construction in this chapter that needs no measuring at all.
And it tells you what a construction really is. You are not drawing a picture that looks right; you are using the compass as a statement that two lengths are equal, and the ruler as a statement that three points are in a line. Every step must be one of those two claims — which is why constructions are marked on the arcs you leave behind, not on the neatness of the final figure.
This page covers the ICSE Class 9 Mathematics chapter on the construction of polygons: quadrilaterals from four sides with a diagonal or an angle, parallelograms, rhombuses, and the regular hexagon.
Here is why. Join the centre to two neighbouring marks and . Then and are radii, and was stepped off at the same compass setting, so
The triangle is equilateral, so . Six such angles make , a complete turn, so six steps close the circle and the six chords form a regular hexagon.
So the side of a regular hexagon equals the radius of its circle. That is the only construction in this chapter that needs no measuring at all.
And it tells you what a construction really is. You are not drawing a picture that looks right; you are using the compass as a statement that two lengths are equal, and the ruler as a statement that three points are in a line. Every step must be one of those two claims — which is why constructions are marked on the arcs you leave behind, not on the neatness of the final figure.
This page covers the ICSE Class 9 Mathematics chapter on the construction of polygons: quadrilaterals from four sides with a diagonal or an angle, parallelograms, rhombuses, and the regular hexagon.
How do you construct a quadrilateral from four sides and a diagonal?
Draw the diagonal first — it splits the quadrilateral into two triangles, and each triangle is then an SSS construction.
A quadrilateral has four sides and two diagonals, and five measurements are needed to fix it. Four sides alone are not enough, because a four-bar figure can lean over without any side changing length.
Worked construction 1. Construct quadrilateral with cm, cm, cm, cm and diagonal cm.
- Draw cm
- With as centre and radius cm, draw an arc; with as centre and radius cm, draw another arc. They cross at
- With as centre and radius cm, and as centre with radius cm, draw arcs on the other side of . They cross at
- Join , , ,
Check before you start that the arcs can meet. In you need , and , so they do. In , , so they do as well.
Worked construction 2 — four sides and one angle. Construct with cm, cm, cm, cm and .
- Draw cm, construct at , and cut cm along
- Arcs of cm from and cm from meet at
Here the given angle replaced the diagonal. The first triangle was fixed by SAS instead of SSS, and the second by SSS as before — so two triangles, each fixed by a criterion from the congruence chapter, is what every quadrilateral construction reduces to.
The boundary case that makes a construction impossible. If the diagonal is too long — say cm with cm and cm — then and the two arcs never intersect. The triangle inequality decides whether a construction exists at all, and noticing that before you draw saves the whole question.
A quadrilateral has four sides and two diagonals, and five measurements are needed to fix it. Four sides alone are not enough, because a four-bar figure can lean over without any side changing length.
Worked construction 1. Construct quadrilateral with cm, cm, cm, cm and diagonal cm.
- Draw cm
- With as centre and radius cm, draw an arc; with as centre and radius cm, draw another arc. They cross at
- With as centre and radius cm, and as centre with radius cm, draw arcs on the other side of . They cross at
- Join , , ,
Check before you start that the arcs can meet. In you need , and , so they do. In , , so they do as well.
Worked construction 2 — four sides and one angle. Construct with cm, cm, cm, cm and .
- Draw cm, construct at , and cut cm along
- Arcs of cm from and cm from meet at
Here the given angle replaced the diagonal. The first triangle was fixed by SAS instead of SSS, and the second by SSS as before — so two triangles, each fixed by a criterion from the congruence chapter, is what every quadrilateral construction reduces to.
The boundary case that makes a construction impossible. If the diagonal is too long — say cm with cm and cm — then and the two arcs never intersect. The triangle inequality decides whether a construction exists at all, and noticing that before you draw saves the whole question.
What combinations let you construct a parallelogram?
Use the fact that opposite sides are equal, or that the diagonals bisect each other — whichever the given data matches.
A parallelogram needs only three measurements, because its opposite sides are already known to be equal.
Worked construction 1 — two sides and the included angle. Construct parallelogram with cm, cm and .
- Draw cm and construct
- Cut cm along
- Arcs of cm from and cm from meet at , since and
Check your drawing by measuring the diagonal. By the cosine rule,
so cm. If your diagonal measures anything near cm you are right; if it measures cm, your has become an obtuse angle.
Worked construction 2 — two diagonals and the angle between them. Construct a parallelogram whose diagonals are cm and cm, meeting at .
The diagonals bisect each other, so work from the centre outwards:
- Draw cm and mark its mid-point
- Construct a line through at to
- Cut cm on either side of along that line
- Join , , ,
Check by calculation. The half-diagonals are cm and cm, so
The two adjacent sides come out different, because the two angles at the centre are and . That is the check to make: measure both sides, and they should differ by about cm.
One thing you cannot do. Two diagonals alone, with no angle, do not fix a parallelogram — you can swing them to any opening you like. Count your given measurements: three for a parallelogram, five for a general quadrilateral, and a question with fewer has more than one answer.
A parallelogram needs only three measurements, because its opposite sides are already known to be equal.
Worked construction 1 — two sides and the included angle. Construct parallelogram with cm, cm and .
- Draw cm and construct
- Cut cm along
- Arcs of cm from and cm from meet at , since and
Check your drawing by measuring the diagonal. By the cosine rule,
so cm. If your diagonal measures anything near cm you are right; if it measures cm, your has become an obtuse angle.
Worked construction 2 — two diagonals and the angle between them. Construct a parallelogram whose diagonals are cm and cm, meeting at .
The diagonals bisect each other, so work from the centre outwards:
- Draw cm and mark its mid-point
- Construct a line through at to
- Cut cm on either side of along that line
- Join , , ,
Check by calculation. The half-diagonals are cm and cm, so
The two adjacent sides come out different, because the two angles at the centre are and . That is the check to make: measure both sides, and they should differ by about cm.
One thing you cannot do. Two diagonals alone, with no angle, do not fix a parallelogram — you can swing them to any opening you like. Count your given measurements: three for a parallelogram, five for a general quadrilateral, and a question with fewer has more than one answer.
How do you construct a rhombus from its side or its diagonals?
Use the property from the previous chapter: the diagonals of a rhombus bisect each other at right angles. That turns the construction into two perpendicular lines.
Worked construction 1 — both diagonals given. Construct a rhombus whose diagonals are cm and cm.
- Draw cm and construct the perpendicular bisector of , meeting it at
- Cut cm on the perpendicular, one on each side
- Join the four sides
Check by calculation. The half-diagonals are cm and cm at right angles, so every side is
Measure all four sides: each must be cm. The area is cm.
Worked construction 2 — side and one diagonal given. Construct a rhombus of side cm with one diagonal cm.
- Draw cm
- Arcs of cm from and from meet at above the line and at below it
- Join , , ,
Check: the other diagonal should be
Notice that this is just the four-sides-and-a-diagonal construction with all four sides the same. The rhombus needs only two measurements because three of its sides are already determined.
Worked construction 3 — a square. Construct a square of side cm.
A square is a rhombus with a right angle, so construct at , cut cm, and complete with arcs of cm from and . Its diagonal must measure cm.
The feasibility limit for a rhombus. A diagonal must be shorter than twice the side, since the two half-sides have to reach across it: with side cm, a diagonal of cm would flatten the figure into a straight line and anything longer is impossible. Every construction in this chapter has such a limit, and it is always the triangle inequality in disguise.
Worked construction 1 — both diagonals given. Construct a rhombus whose diagonals are cm and cm.
- Draw cm and construct the perpendicular bisector of , meeting it at
- Cut cm on the perpendicular, one on each side
- Join the four sides
Check by calculation. The half-diagonals are cm and cm at right angles, so every side is
Measure all four sides: each must be cm. The area is cm.
Worked construction 2 — side and one diagonal given. Construct a rhombus of side cm with one diagonal cm.
- Draw cm
- Arcs of cm from and from meet at above the line and at below it
- Join , , ,
Check: the other diagonal should be
Notice that this is just the four-sides-and-a-diagonal construction with all four sides the same. The rhombus needs only two measurements because three of its sides are already determined.
Worked construction 3 — a square. Construct a square of side cm.
A square is a rhombus with a right angle, so construct at , cut cm, and complete with arcs of cm from and . Its diagonal must measure cm.
The feasibility limit for a rhombus. A diagonal must be shorter than twice the side, since the two half-sides have to reach across it: with side cm, a diagonal of cm would flatten the figure into a straight line and anything longer is impossible. Every construction in this chapter has such a limit, and it is always the triangle inequality in disguise.
What are the exact steps for a regular hexagon of a given side?
Draw a circle whose radius equals the required side, then step that same radius six times round the circumference.
Worked construction. Construct a regular hexagon of side cm.
- Open the compass to cm and draw a circle with centre
- Mark any point on the circle
- Without changing the compass, put the point on and cut the circle at ; from cut at ; and so on through , and
- The sixth arc should land exactly on . Join , , , , ,
Now the numbers that let you check it.
- Perimeter: cm
- Central angle of each of the six triangles:
- Interior angle of the hexagon:
- Long diagonal (across the centre): cm, because , and are in a straight line
- Short diagonal (skipping one vertex): cm
- Area: six equilateral triangles, so cm
Check the interior angle against the construction. Each interior angle of the hexagon is made of two base angles of neighbouring equilateral triangles, and , as required.
Why the same trick does not work for a pentagon. A regular pentagon has central angle , and its side is not equal to the radius — it is shorter. Stepping the radius round the circle would give you six marks, not five, so a pentagon needs a genuinely different and much harder construction.
The hexagon is the one regular polygon whose side happens to equal its radius, and that single coincidence is why it is the one the syllabus asks you to construct with compasses alone.
Worked construction. Construct a regular hexagon of side cm.
- Open the compass to cm and draw a circle with centre
- Mark any point on the circle
- Without changing the compass, put the point on and cut the circle at ; from cut at ; and so on through , and
- The sixth arc should land exactly on . Join , , , , ,
Now the numbers that let you check it.
- Perimeter: cm
- Central angle of each of the six triangles:
- Interior angle of the hexagon:
- Long diagonal (across the centre): cm, because , and are in a straight line
- Short diagonal (skipping one vertex): cm
- Area: six equilateral triangles, so cm
Check the interior angle against the construction. Each interior angle of the hexagon is made of two base angles of neighbouring equilateral triangles, and , as required.
Why the same trick does not work for a pentagon. A regular pentagon has central angle , and its side is not equal to the radius — it is shorter. Stepping the radius round the circle would give you six marks, not five, so a pentagon needs a genuinely different and much harder construction.
The hexagon is the one regular polygon whose side happens to equal its radius, and that single coincidence is why it is the one the syllabus asks you to construct with compasses alone.
Exam tip
How should a construction be presented in the ICSE paper?
Leave every arc on the page and write the given measurements beside the figure. Constructions are marked on the evidence of method, and an erased arc is an erased mark.
- Draw a rough sketch first, marked with the given lengths and angles. It shows the examiner you know the target and it tells you which line to draw first
- Never rub out construction arcs. They are the proof that you used a compass rather than a protractor or a guess
- Use a hard, sharp pencil and keep the compass setting fixed while a pair of arcs is being drawn. Two arcs at slightly different radii intersect in the wrong place
- Start from the longest given line — the diagonal in a quadrilateral, the longer diagonal in a parallelogram. Working outwards from a long line keeps the error small
- Check the triangle inequality before drawing. If two sides cannot reach across the diagonal, say so — the construction is not possible is sometimes the whole answer
- Verify by measuring one quantity you were not given: the second diagonal, or the fourth side. The calculation in each worked example above shows you what to expect
- Label every point as soon as you make it, and state the steps in order if the question asks for them
The habit that separates full marks from most marks. Write the one-line reason for the key step: the diagonals of a rhombus bisect each other at right angles, so I constructed the perpendicular bisector. A construction is a proof carried out with instruments, and the reason is what makes it one.
- Draw a rough sketch first, marked with the given lengths and angles. It shows the examiner you know the target and it tells you which line to draw first
- Never rub out construction arcs. They are the proof that you used a compass rather than a protractor or a guess
- Use a hard, sharp pencil and keep the compass setting fixed while a pair of arcs is being drawn. Two arcs at slightly different radii intersect in the wrong place
- Start from the longest given line — the diagonal in a quadrilateral, the longer diagonal in a parallelogram. Working outwards from a long line keeps the error small
- Check the triangle inequality before drawing. If two sides cannot reach across the diagonal, say so — the construction is not possible is sometimes the whole answer
- Verify by measuring one quantity you were not given: the second diagonal, or the fourth side. The calculation in each worked example above shows you what to expect
- Label every point as soon as you make it, and state the steps in order if the question asks for them
The habit that separates full marks from most marks. Write the one-line reason for the key step: the diagonals of a rhombus bisect each other at right angles, so I constructed the perpendicular bisector. A construction is a proof carried out with instruments, and the reason is what makes it one.
Did you know
Why do hexagons tile a honeycomb when pentagons cannot?
Look closely at a honeycomb, at the mesh of a wire-net fence, or at the hexagonal tiles on some footpaths. Hexagons fit together with no gaps. Pentagons never do, and the reason is the interior angle you calculated above.
For shapes to meet around a point with no gap and no overlap, the angles at that point must add to exactly . A regular hexagon's interior angle is , and
so three hexagons meet perfectly at every corner. An equilateral triangle has , and , so six triangles work. A square has , and four fit.
Now try a pentagon. Its interior angle is
and , which is not a whole number. Three pentagons leave a gap of , and four overlap. No arrangement of regular pentagons can tile a flat surface.
Running the same test on the regular heptagon, octagon and beyond gives no whole numbers either, so the triangle, the square and the hexagon are the only regular polygons that tile the plane — a complete answer from one division.
And that is why a honeycomb is hexagonal rather than square. Of those three tilings, the hexagon encloses the most area for a given total wall length, so the same quantity of wax holds more honey. The bees are not doing geometry, but the shape that survives is the one the geometry favours — and you can verify the area claim yourself by comparing a hexagon of perimeter cm, with area cm, against a square of the same perimeter, with area cm.
For shapes to meet around a point with no gap and no overlap, the angles at that point must add to exactly . A regular hexagon's interior angle is , and
so three hexagons meet perfectly at every corner. An equilateral triangle has , and , so six triangles work. A square has , and four fit.
Now try a pentagon. Its interior angle is
and , which is not a whole number. Three pentagons leave a gap of , and four overlap. No arrangement of regular pentagons can tile a flat surface.
Running the same test on the regular heptagon, octagon and beyond gives no whole numbers either, so the triangle, the square and the hexagon are the only regular polygons that tile the plane — a complete answer from one division.
And that is why a honeycomb is hexagonal rather than square. Of those three tilings, the hexagon encloses the most area for a given total wall length, so the same quantity of wax holds more honey. The bees are not doing geometry, but the shape that survives is the one the geometry favours — and you can verify the area claim yourself by comparing a hexagon of perimeter cm, with area cm, against a square of the same perimeter, with area cm.
Exam relevance
Do constructions matter for JEE, or only for the board paper?
Compasses do not appear in a competitive paper, but the reasoning behind these constructions does — and this is foundation work for two specific later topics.
Where it leads. Every construction here is a locus statement in disguise: an arc is the set of points at a fixed distance from a centre, and a perpendicular bisector is the set of points equidistant from two points. Those two sentences become the circle and the perpendicular bisector in Class 11 Coordinate Geometry, where JEE Main asks for the locus of a point satisfying a distance condition. A student who has drawn these figures reads locus as a procedure rather than a definition.
Where the hexagon leads. In Class 11 Complex Numbers, the sixth roots of unity are six points equally spaced round a unit circle at intervals — exactly the hexagon you stepped off with the compass, and the reason with gives a regular hexagon. JEE Advanced questions on roots of unity often reduce to a fact about that hexagon, such as the length of a side or a diagonal, which you can now compute: side , short diagonal , long diagonal .
Where the feasibility check leads. Deciding whether arcs can meet is the same as deciding whether a system has a solution — a condition on the data rather than a calculation with it. That habit is examined directly in JEE algebra questions asking for the range of a parameter for which a triangle or a circle exists.
Question types to expect. The ICSE paper asks for the construction, the steps and a measured answer. A competitive paper asks for a locus, a root of unity or a feasibility range — the same ideas with no drawing.
The single trap that costs marks. In the board paper it is erasing the arcs. In the competitive version it is forgetting that a distance condition can be unsatisfiable: a locus question whose answer is no such point exists is testing the same triangle-inequality thinking as an impossible construction.
Board versus competitive emphasis. This chapter is mostly worth marks now, but the locus reading of it is worth marks for years. When you draw an arc, say what set of points it represents — that sentence is the transferable part.
Where it leads. Every construction here is a locus statement in disguise: an arc is the set of points at a fixed distance from a centre, and a perpendicular bisector is the set of points equidistant from two points. Those two sentences become the circle and the perpendicular bisector in Class 11 Coordinate Geometry, where JEE Main asks for the locus of a point satisfying a distance condition. A student who has drawn these figures reads locus as a procedure rather than a definition.
Where the hexagon leads. In Class 11 Complex Numbers, the sixth roots of unity are six points equally spaced round a unit circle at intervals — exactly the hexagon you stepped off with the compass, and the reason with gives a regular hexagon. JEE Advanced questions on roots of unity often reduce to a fact about that hexagon, such as the length of a side or a diagonal, which you can now compute: side , short diagonal , long diagonal .
Where the feasibility check leads. Deciding whether arcs can meet is the same as deciding whether a system has a solution — a condition on the data rather than a calculation with it. That habit is examined directly in JEE algebra questions asking for the range of a parameter for which a triangle or a circle exists.
Question types to expect. The ICSE paper asks for the construction, the steps and a measured answer. A competitive paper asks for a locus, a root of unity or a feasibility range — the same ideas with no drawing.
The single trap that costs marks. In the board paper it is erasing the arcs. In the competitive version it is forgetting that a distance condition can be unsatisfiable: a locus question whose answer is no such point exists is testing the same triangle-inequality thinking as an impossible construction.
Board versus competitive emphasis. This chapter is mostly worth marks now, but the locus reading of it is worth marks for years. When you draw an arc, say what set of points it represents — that sentence is the transferable part.
Key takeaways
What should you be able to construct before moving on?
Every construction in this chapter reduces to triangles fixed by SSS or SAS, plus one property of the figure.
- A quadrilateral needs five measurements; draw the diagonal first and treat each half as an SSS triangle
- Four sides and one angle works the same way, with the first triangle fixed by SAS instead
- A parallelogram needs three measurements, because opposite sides are equal; from two diagonals and their angle, work outwards from the mid-point
- A rhombus needs two: from both diagonals, draw one and take the perpendicular bisector; from side and one diagonal, use equal arcs
- A regular hexagon needs one: its side equals the radius, so step the compass six times round the circle
- Hexagon facts: central angle , interior angle , long diagonal twice the side, short diagonal times the side
- Check the triangle inequality first — if two sides cannot reach across a diagonal, the construction does not exist
- Verify by measuring something you were not given, and leave every arc on the page
The quickest test of this chapter is the rhombus: given diagonals of cm and cm, predict the side before you draw, then construct it and measure. If the prediction and the measurement both say cm, your compass work and your reasoning agree.
- A quadrilateral needs five measurements; draw the diagonal first and treat each half as an SSS triangle
- Four sides and one angle works the same way, with the first triangle fixed by SAS instead
- A parallelogram needs three measurements, because opposite sides are equal; from two diagonals and their angle, work outwards from the mid-point
- A rhombus needs two: from both diagonals, draw one and take the perpendicular bisector; from side and one diagonal, use equal arcs
- A regular hexagon needs one: its side equals the radius, so step the compass six times round the circle
- Hexagon facts: central angle , interior angle , long diagonal twice the side, short diagonal times the side
- Check the triangle inequality first — if two sides cannot reach across a diagonal, the construction does not exist
- Verify by measuring something you were not given, and leave every arc on the page
The quickest test of this chapter is the rhombus: given diagonals of cm and cm, predict the side before you draw, then construct it and measure. If the prediction and the measurement both say cm, your compass work and your reasoning agree.