Quadrilateral from sides and diagonal

Give a carpenter four planks of fixed length and one diagonal brace, and there is only one four-sided frame they can build — no wobble, no second shape hiding in the same measurements. That rigidity is exactly what makes "quadrilateral from four sides and one diagonal" the first and most fundamental construction in the ICSE Class 8 Constructions chapter: it takes a shape you cannot build directly with a compass (a quadrilateral has no fixed angles to lock it in place) and splits it into two shapes you already know how to build perfectly — triangles. This article rebuilds that idea from the ground up: why a single diagonal turns an unstable four-sided figure into two rigid SSS triangles, how to carry out the construction accurately with only a ruler and compass, why the triangle inequality is the silent gatekeeper that decides whether the data is even usable, and how to avoid the handful of mistakes that turn a correct quadrilateral into a self-crossing mess.

The core insight

A diagonal splits any quadrilateral into two triangles that share that diagonal as a common side, so if all four sides and one diagonal are known, both triangles are fully determined by SSS and the quadrilateral can be constructed uniquely with ruler and compass alone.

The core idea

A quadrilateral by itself is not rigid. Four hinged rods of fixed length can flex into infinitely many different quadrilaterals — think of a rectangle you can push sideways into a slanted parallelogram without changing any side length. Knowing all four sides is therefore not enough to fix the shape; you also need to freeze one of the two diagonals to stop the flexing.

A diagonal turns flexibility into two rigid triangles. Draw quadrilateral $ABCD$ and add diagonal $AC$ . This single line splits the figure into $\triangle ABC$ and $\triangle ACD$ , and both triangles now share side $AC$ . A triangle, unlike a quadrilateral, cannot flex once its three sides are fixed — three given lengths determine exactly one triangle shape (up to mirror reflection), which is the SSS construction from Class 7.

Two SSS constructions sharing a side. If the sides $AB, BC, CD, DA$ and diagonal $AC$ are all given, then $\triangle ABC$ has all three of its sides known ( $AB$ , $BC$ , $AC$ ), and $\triangle ACD$ also has all three sides known ( $AC$ , $CD$ , $DA$ ). Building both triangles on opposite sides of the shared segment $AC$ reconstructs the whole quadrilateral, uniquely.

The triangle inequality is the built-in usability check. Before any arc is drawn, each of the two triangles must satisfy the rule that the sum of any two sides exceeds the third side:

AB + BC > AC, \qquad AC + CD > DA, \qquad AC + DA > CD

where: $AB, BC, CD, DA$ are the four given sides of the quadrilateral and $AC$ is the given diagonal. If any such inequality fails for either triangle, the two arcs used to locate the missing vertex will never meet, and no quadrilateral exists for that data — the construction is not "hard," it is impossible.

Figure

A quadrilateral ABCD with diagonal AC drawn splitting it into triangle ABC above the diagonal and triangle ACD below the diagonal

Why it works

The reasoning behind this construction is a chain of two simple geometric facts rather than a single algebraic formula, but it is exactly as load-bearing as one and deserves the same careful walk-through, since a student who understands why the method works never has to memorise the construction steps by rote — they can rebuild the procedure from scratch under exam pressure.

Step 1. A triangle with three known side lengths is unique. This is the SSS rigidity result from Class 7: given three lengths satisfying the triangle inequality, there is exactly one triangle shape they can form (mirror images count as the same shape, just flipped). No amount of pushing or pulling changes it once all three sides are fixed, unlike a quadrilateral, which can flex.

Step 2. A quadrilateral's diagonal creates two triangles that share that diagonal as a common side. Diagonal $AC$ of quadrilateral $ABCD$ divides it into $\triangle ABC$ and $\triangle ACD$ . Crucially, both triangles contain the segment $AC$ as one of their three sides, so once $AC$ is drawn and fixed in place, it acts as the shared "spine" that both triangles are built onto.

Step 3. If all four sides and the diagonal are known, both triangles individually satisfy SSS. $\triangle ABC$ has sides $AB$ , $BC$ , $AC$ — all given. $\triangle ACD$ has sides $AC$ , $CD$ , $DA$ — also all given. By Step 1, each triangle is therefore uniquely determined the moment its three side lengths are fixed.

Step 4. Building both unique triangles on opposite sides of the same fixed diagonal reconstructs one unique quadrilateral. Since $AC$ is shared and fixed, and each triangle is individually rigid, joining them along $AC$ (with $B$ and $D$ placed on opposite sides so the figure does not self-overlap) leaves no freedom left in the shape at all. This is why "4 sides + 1 diagonal" is enough data to pin down a quadrilateral completely, even though "4 sides" alone was not.

Worked example

Construct quadrilateral $ABCD$ in which $AB = 5.2$ cm, $BC = 4.8$ cm, $CD = 4.0$ cm, $DA = 4.5$ cm, and diagonal $AC = 6.0$ cm.

Step 1. Check both triangles satisfy the triangle inequality before touching a compass. For $\triangle ABC$ : sides $6.0, 5.2, 4.8$ , and $5.2 + 4.8 = 10.0 > 6.0$ — valid. For $\triangle ACD$ : sides $6.0, 4.5, 4.0$ , and $4.5 + 4.0 = 8.5 > 6.0$ — valid. The data is usable, so the construction will succeed.

Step 2. Draw a rough sketch labelling which side belongs to which triangle, then draw segment $AC = 6.0$ cm with a ruler as the fixed base for both triangles.

Step 3. Construct $\triangle ABC$ above $AC$ : with centre $A$ and radius $5.2$ cm draw an arc, and with centre $C$ and radius $4.8$ cm draw a second arc cutting the first at $B$ . Join $AB$ and $BC$ .

Step 4. Construct $\triangle ACD$ below $AC$ : with centre $A$ and radius $4.5$ cm draw an arc, and with centre $C$ and radius $4.0$ cm draw a second arc cutting the first at $D$ . Join $AD$ and $DC$ . Joining $AB, BC, CD, DA$ gives the required quadrilateral. Answer: quadrilateral $ABCD$ with $AB = 5.2$ cm, $BC = 4.8$ cm, $CD = 4.0$ cm, $DA = 4.5$ cm and diagonal $AC = 6.0$ cm, built from two SSS triangles on opposite sides of $AC$ .

Competitive stretch

If instead you were given $PQ = 3.5$ cm, $QR = 4.2$ cm, $RS = 5.0$ cm, $SP = 3.8$ cm, and diagonal $PR = 5.5$ cm, the same two checks apply before drawing: $\triangle PQR$ has sides $5.5, 3.5, 4.2$ with $3.5+4.2 = 7.7 > 5.5$ , and $\triangle PRS$ has sides $5.5, 3.8, 5.0$ with $3.8+5.0=8.8>5.5$ — both valid, so $Q$ and $S$ are built on opposite sides of $PR$ exactly as $B$ and $D$ were built on opposite sides of $AC$ above.

Common traps

Placing both triangles on the same side of the diagonal, which produces overlapping triangles instead of a simple, non-self-crossing quadrilateral.
Skipping the triangle inequality check on both triangles before construction — if the two shorter sides of either triangle do not sum to more than the third, the arcs never intersect.
Losing track of which of the four given sides belongs to which triangle, since the diagonal $AC$ is shared between both — always sketch a rough labelled figure first.
Erasing the construction arcs afterward; they are the evidence of an accurate SSS construction and should stay lightly visible.

Ask the tutor: Curious how this method changes when you are given two diagonals instead of one, or angles instead of a diagonal — ask the AI tutor to compare the related quadrilateral constructions side by side.

Why is one diagonal enough, but knowing all four angles instead is not part of this method?

One diagonal works because it converts the whole problem into two SSS triangle constructions, which Class 8 students can already do accurately with just a ruler and compass. Angles would require a protractor and a different construction strategy (SAS-style triangle building), and this particular topic in the Constructions chapter is specifically about the sides-and-diagonal method — the angle-based version is a separate construction case with its own steps.

What if I am given two diagonals instead of one, along with the four sides — does the same idea apply?

Not directly with a single diagonal split, since giving both diagonals over-determines four sides in a different way, and the construction sequence used to locate the vertices changes (you typically start from one diagonal and one adjacent triangle, then use the second diagonal to fix the last vertex). It is a related but distinct construction case in the same chapter, worth asking the tutor to walk through separately so you do not mix up the two methods.

Can any four side lengths and any diagonal length always be used to build a quadrilateral?

No. Each of the two triangles created by the diagonal must independently satisfy the triangle inequality — the sum of its two shorter sides must exceed its longest side. If either triangle fails this test, the two arcs used to locate that triangle's third vertex will never meet, no matter how carefully you draw, and the given data simply cannot form a real quadrilateral.