Polygon construction

Handed a ruler and a pair of compasses — no protractor allowed — how do you draw a rhombus with an exact $8\text{ cm}$ diagonal, or a hexagon whose six sides all measure precisely $4\text{ cm}$ ? This is the puzzle behind polygon construction: building a shape from a short list of given lengths and angles, using only the classical tools, so that the final figure is provably exact rather than eyeballed. The trick is always the same — split the polygon into triangles, because a triangle is rigid the moment you fix three pieces of data (SSS, SAS or ASA), and a triangle built from arcs and straight edges is automatically exact. A quadrilateral is two triangles glued along a diagonal; a regular hexagon is six equilateral triangles fanned out from a centre. This article works through how much data a quadrilateral actually needs, why the parallelogram and rhombus need less than that, and why marking off the side length itself around a circle is enough to close a perfect hexagon.

The core insight

Every ruler-and-compass polygon construction reduces to building rigid triangles from given side/angle data — a general quadrilateral needs 5 independent measurements (two SSS/SAS triangles sharing a diagonal), while a regular hexagon needs only its side length, because that side subtends a central angle of exactly $360^\circ/6 = 60^\circ$ .

The core idea

Why a triangle is the building block. A triangle is rigid: once three independent pieces of data are fixed (three sides, or two sides and the included angle, or two angles and the included side), exactly one triangle shape exists, and a compass can locate its third vertex as the intersection of two arcs. Every polygon construction is really a sequence of triangle constructions stitched together along shared sides or diagonals.

How much data a quadrilateral needs. Draw one diagonal of a quadrilateral $ABCD$ — say $AC$ — and it splits into $\triangle ABC$ and $\triangle ACD$ . Triangle one needs 3 independent measurements; triangle two shares the side $AC$ with triangle one, so it needs only 2 more (since $AC$ is already fixed). That gives a total of

N_{\text{quad}} = 3 + 2 = 5. \qquad (1)

where: $N_{\text{quad}}$ is the count of independent measurements (sides, diagonals, or angles) that fully determine a general quadrilateral up to congruence — for example, 4 sides plus 1 diagonal, or 3 sides plus 2 included angles. Fewer than 5 leaves the shape flexible (think of a rhombus with fixed sides that can still be squashed, like a hinge); this is the same "degrees of freedom" idea a $2$ -triangle split always gives, since each extra triangle glued on a shared side after the first only ever adds 2 new numbers, not 3.

Special cases need less. A parallelogram's opposite sides are automatically equal, so two adjacent sides plus one diagonal (or plus the included angle) is enough — 3 numbers, not 5. A rhombus has all four sides equal, so one side plus one diagonal fixes it — just 2 numbers. A regular hexagon needs only its side length, for the geometric reason unpacked in the derivation below.

Figure

A regular hexagon inscribed in a circle showing the six equilateral triangles formed with the centre and the sixty degree central angle at one vertex

Derivation AWhy six equal arcs close a regular hexagon

The goal is to show, using only the sum-of-angles-round-a-point fact and the equilateral-triangle test, that stepping a compass six times around a circle — always at the same opening as the circle's own radius — must land exactly back on the starting point and produce six sides of that same length, with no protractor and no trial and error anywhere in the process.

Step 1. Draw a circle of radius $r$ centred at $O$ , and mark a point $A$ on it. Open the compass to the same length $r$ and, starting at $A$ , step around the circle: with centre $A$ draw an arc of radius $r$ cutting the circle at $B$ ; with centre $B$ draw the next arc at radius $r$ cutting the circle at $C$ ; continue to $D$ , $E$ , $F$ , and back toward $A$ .

Step 2. Look at one wedge, $\triangle OAB$ . Since $A$ and $B$ both lie on the circle, $OA = OB = r$ . Since the arc from $A$ to $B$ was drawn with the compass fixed at radius $r$ , the chord $AB = r$ too. So $\triangle OAB$ has all three sides equal to $r$ — it is equilateral, and every equilateral triangle has all three angles equal to $60^\circ$ . In particular the angle at the centre, $\angle AOB = 60^\circ$ .

Step 3. The same argument applies to every wedge $\triangle OBC$ , $\triangle OCD$ , and so on — each is equilateral with a central angle of exactly

\angle AOB = \angle BOC = \cdots = \frac{360^\circ}{6} = 60^\circ. \qquad (2)

where: the $360^\circ$ is the full angle around the centre $O$ , and $6$ is the number of equal wedges a hexagon splits into, since $n$ equal arcs around a full circle always produce a central angle of $360^\circ/n$ for a regular $n$ -gon.

Step 4. Six wedges of $60^\circ$ each sum to exactly $6 \times 60^\circ = 360^\circ$ — a full turn — so the sixth arc, stepped from $F$ , must land back on $A$ with no gap and no overlap. Joining the six points in order, $AB, BC, CD, DE, EF, FA$ , gives six chords all equal to $r$ : a regular hexagon, closed automatically, with no angle ever measured. Verified: $6 \times (360^\circ/6) = 360^\circ$ exactly, so the construction always closes for any radius $r$ ; each wedge is equilateral because $OA=OB=AB=r$ by the fixed compass opening.

Worked example

Construct a rhombus $ABCD$ with side $AB = 5\text{ cm}$ and diagonal $AC = 8\text{ cm}$ , then extend it to find the second diagonal $BD$ .

Step 1. Draw $AC = 8\text{ cm}$ with a ruler — this is the shared diagonal that will split the rhombus into two triangles, $\triangle ABC$ and $\triangle ACD$ .

Step 2. With centre $A$ and radius $5\text{ cm}$ , draw an arc above $AC$ ; with centre $C$ and the same radius $5\text{ cm}$ (since all four sides of a rhombus are equal, $BC = 5\text{ cm}$ too), draw a second arc above $AC$ . The arcs meet at $B$ . Join $AB$ and $CB$ — this fixes $\triangle ABC$ by SSS, using $AC = 8$ , $AB = 5$ , $BC = 5$ .

Step 3. Repeat the same pair of arcs (radius $5\text{ cm}$ from $A$ , radius $5\text{ cm}$ from $C$ ) below line $AC$ to locate $D$ , then join $AD$ and $CD$ . This fixes $\triangle ACD$ by SSS in the same way, sharing side $AC$ with the first triangle — confirming the count from equation (1): $3$ measurements for the first triangle, $2$ more ( $AB$ -style radius reused, no new number needed since the rhombus rule already fixes $AD=CD=5$ ) for the second.

Step 4. $ABCD$ is the required rhombus, with $AB = BC = CD = DA = 5\text{ cm}$ by construction. Measuring the completed figure confirms the rhombus property: the second diagonal $BD$ crosses $AC$ at right angles and bisects it, cutting $AC$ into two $4\text{ cm}$ halves — a check you can perform with the compass alone, without ever having drawn $BD$ as a construction step. The rhombus $ABCD$ has $AB=BC=CD=DA=5\text{ cm}$ , diagonal $AC=8\text{ cm}$ , and diagonal $BD$ bisecting $AC$ perpendicularly.

Common traps

Reaching for a protractor on the hexagon. The whole point of the six-equal-arcs method is that the $60^\circ$ central angle in equation (2) is produced automatically by matching the compass opening to the radius — measuring an angle defeats the method and invites error.
Assuming any four measurements fix a quadrilateral. Equation (1) shows a general quadrilateral needs 5 independent numbers, not 4; four sides alone (with no diagonal or angle) leave the shape free to flex, like a hinged frame.
Using different radii for a rhombus's four sides. All four sides are equal by definition, so every arc drawn to locate a rhombus vertex must use the same compass opening — a common slip is redrawing the compass width partway through.
Skipping a rough sketch. Attempting the arcs directly, without first sketching which diagonal to draw and where each triangle should sit, often produces two arcs that do not intersect where expected.

Ask the tutor: Want to walk through constructing a specific quadrilateral from your own given data, or double-check a hexagon construction step by step? Ask the AI tutor for a guided example.

Why does splitting a quadrilateral with a diagonal always work, no matter which diagonal I pick?

Either diagonal splits the quadrilateral into two triangles that share that diagonal as a side, and a triangle is rigid once three of its measurements are known — so the method itself does not depend on which diagonal you choose. What can differ is how convenient the given data is: if you are given diagonal $AC$ and the two sides touching it, drawing $AC$ first is the natural choice, because both triangles can then be built by SSS in one pass. Choosing the "wrong" diagonal for the given data does not make the construction impossible, just clumsier, since you would need to work out extra lengths first.

Does the 5-measurements rule mean I can pick any 5 numbers and get a valid quadrilateral?

Not quite — the 5 numbers must be geometrically consistent, meaning each of the two triangles formed by the diagonal split must individually satisfy the triangle inequality (each side less than the sum of the other two), and the two triangles must actually meet correctly along the shared diagonal. In practice, for the well-behaved data given in ICSE problems (sides, one diagonal, or two included angles from parallelogram/rhombus properties), this is never an issue, but it is why "any 5 lengths" is not automatically constructible in general.

Why does a regular hexagon need only one measurement, its side, while a general hexagon would need far more

A regular hexagon has extra built-in symmetry: all six sides are equal and all six central angles are forced to be $60^\circ$ each by equation (2), so fixing the single side length $r$ pins down the entire shape via the six equilateral triangles. A general, irregular hexagon has no such symmetry — it would need many more independent measurements (sides and angles) to fix uniquely, the same way an irregular quadrilateral needs 5 while a rhombus needs only 2, because each extra symmetry removes one piece of data you would otherwise have to specify.

Polygon construction

The core idea

Worked example

Common traps

Before this

Where this leads