Construction of quadrilaterals

A triangle is stubborn — give it three side lengths and there is only one shape it can possibly form, so a ruler and compass can draw it exactly. A quadrilateral is not stubborn at all. Four hinged rods of fixed length can flex into a rectangle, a parallelogram, a kite, or a squashed sliver, all with the same four sides. So how does one draw one particular quadrilateral accurately, without guessing a single angle by eye? The trick used throughout this chapter is to stop thinking of the quadrilateral as four sides and start thinking of it as two triangles glued along a diagonal. Once each triangle is pinned down by SSS, SAS, or ASA — the same rigid triangle constructions learned in Class 7 — the whole quadrilateral is pinned down too. This article works through the four standard data combinations the syllabus prescribes for a general quadrilateral, then shows how parallelograms, rectangles, rhombi, and squares need even less data because their own symmetry does part of the job for free.

The core insight

Split any quadrilateral into two triangles along one diagonal, fix each triangle with SSS/SAS/ASA using the given data, and the shared diagonal automatically stitches them into one uniquely determined quadrilateral.

The core idea

A quadrilateral has ten measurements but only five are independent. Count them: 4 sides, 2 diagonals, and 4 angles is ten numbers in total, but they are not all free to vary — once five suitably chosen ones are fixed, the rest are forced. That is why "construct a quadrilateral" problems always hand you exactly five pieces of data, never four and never six.

Why five, and not four like a triangle? A diagonal splits the quadrilateral into two triangles that share that diagonal as a common side. Each triangle needs three independent measurements to be rigid.

N = 3 + 3 - 1 = 5

where: the two $3$ 's are the measurements needed to fix each triangle by SSS/SAS/ASA, and the $-1$ removes the double-count of the shared diagonal, which belongs to both triangles at once. This is why four sides alone are never enough — with no diagonal or angle to pin the hinge, the shape can flex like a loose door frame.

The four standard data sets for a general quadrilateral are: four sides and one diagonal; three sides and two diagonals; three sides and two included angles; and two adjacent sides with three angles. Every one of them supplies exactly five independent numbers, and every one of them is handled the same way — locate a diagonal (given or constructible from the angle data), then build each of the two triangles on it using SSS, SAS, or ASA as the data allows.

Special quadrilaterals need less data. A parallelogram's opposite sides are equal and parallel, so fixing two adjacent sides and the included angle already forces the other two sides and both remaining angles — only 3 measurements, not 5. Rectangles add equal diagonals, rhombi add perpendicular diagonals that bisect each other, and squares combine both properties; each extra property removes one more degree of freedom, so squares can be constructed from as little as one side.

Derivation ALocating a vertex from two known sides

Show that once diagonal AC and the two sides meeting at an unknown vertex B are known, compass arcs from A and C alone are enough to pin down B exactly, because two circles of fixed radii intersect in at most two points and only one of those lies on the required side of AC.

Step 1. Draw diagonal $AC$ at its given length; this fixes the shared side of both triangles that make up the quadrilateral.

Step 2. Vertex $B$ must lie at distance $AB$ from $A$ and distance $CB$ from $C$ . The set of all points at distance $AB$ from $A$ is a circle of radius $AB$ centred at $A$ ; likewise for $C$ . So $B$ is a point common to both circles, meaning it lies on the arc:

\{P : |PA| = AB\} \cap \{P : |PC| = CB\}

Step 3. Two distinct circles intersect in at most two points, symmetric about line $AC$ — one above it and one below. Draw only the arcs, not full circles, on the side of $AC$ where $B$ is meant to lie, so the correct intersection point is the only one marked.

Step 4. The intersection point is $B$ ; join $AB$ and $CB$ with straight lines. Triangle $ABC$ is now fixed by SSS (all three sides known), so $B$ 's position is exact, not approximate — this is why the arcs must stay visible as proof of the construction, and the same two-arc method repeated below $AC$ locates the fourth vertex $D$ .

Worked example

Problem: Construct quadrilateral $PQRS$ with $PQ = 5\text{ cm}$ , $QR = 4.5\text{ cm}$ , $RS = 4\text{ cm}$ , $SP = 5.5\text{ cm}$ , and diagonal $PR = 6.5\text{ cm}$ .

Step 1. Draw diagonal $PR = 6.5\text{ cm}$ as a straight line segment — this is the shared side that will anchor both triangles.

Step 2. With centre $P$ and radius $5\text{ cm}$ , draw an arc above $PR$ ; with centre $R$ and radius $4.5\text{ cm}$ , draw another arc above $PR$ . Their intersection is $Q$ . Join $PQ$ and $RQ$ , fixing triangle $PQR$ by SSS since all three sides ( $PQ=5$ , $QR=4.5$ , $PR=6.5$ ) are now known.

Step 3. With centre $P$ and radius $5.5\text{ cm}$ , draw an arc below $PR$ ; with centre $R$ and radius $4\text{ cm}$ , draw another arc below $PR$ . Their intersection is $S$ . Join $PS$ and $RS$ , fixing triangle $PRS$ by SSS.

Step 4. Join up $PQRS$ . The required quadrilateral is $PQRS$ , with $PQ=5$ , $QR=4.5$ , $RS=4$ , $SP=5.5$ , and $PR=6.5$ (all cm), each measurement checkable directly against the given data.

Competitive stretch

For a parallelogram $ABCD$ with $AB = 6\text{ cm}$ , $AD = 4\text{ cm}$ , $\angle A = 60^\circ$ , only 3 measurements are needed instead of 5: draw $AB$ , construct $\angle A = 60^\circ$ to mark $D$ at $4\text{ cm}$ , then locate $C$ with arcs of radius $6\text{ cm}$ from $D$ and $4\text{ cm}$ from $B$ (since $DC \parallel AB$ and $BC \parallel AD$ in a parallelogram). The saving of 2 measurements is exactly the extra rigidity the parallelogram's equal-and-parallel-sides property supplies.

Common traps

Attempting a quadrilateral from four sides alone. Unlike a triangle, four sides with no diagonal or angle leave the shape free to flex — five independent measurements are the minimum, not four.
Mixing up which triangle to build first. For "3 sides + 2 diagonals" or "2 sides + 3 angles," the first rigid triangle to draw is different from the "4 sides + 1 diagonal" case — always identify the given data combination before starting.
Erasing the construction arcs. The compass arcs that locate each vertex are the geometric proof of accuracy; a clean-looking figure with the arcs rubbed out loses marks even if the shape is correct.
Assuming any five numbers automatically work. Each triangle used in the construction must itself be valid — no side can equal or exceed the sum of the other two sides of that triangle — so an arbitrary set of five measurements can fail to produce a real quadrilateral.

Ask the tutor: Want to work through the "three sides and two included angles" construction step by step, or try a rhombus given only its two diagonals?

Why can't I just draw all four sides in order and call it done

Four sides fixed in sequence still leave a hinge at each vertex, so the same four lengths can close up into many different quadrilaterals — a rectangle-ish shape or a squashed parallelogram-like sliver, depending on the angles you happen to draw by eye. Without a fifth measurement (a diagonal or an angle) to lock one of those hinges, the construction is not unique, and an examiner will mark it as underdetermined even if your specific drawing happens to look right.

What if the two arcs used to locate a vertex don't intersect at all

That means the five given measurements are geometrically inconsistent — most often because one side of the triangle being built is too long compared to the other two, violating the triangle inequality. Recheck the numbers: if, for instance, $AB$ is close to $AC + CB$ in the arc-intersection step, the arcs will barely touch or miss each other entirely, and no valid quadrilateral exists for that data.

Do rhombi and squares use the same two-triangle method as a general quadrilateral

Yes, but with far less data because their diagonals have special properties: a rhombus's diagonals bisect each other at right angles, so knowing both diagonals alone is enough to place all four vertices without any SSS or SAS step. A square is the same idea taken further, since its diagonals are also equal in length, so a single diagonal length can fix the entire figure.

Construction of quadrilaterals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads