Five Measurements Fix a Quadrilateral, and Four Never Can

Because a four-sided frame can be pushed out of shape without any side changing length.

Build a triangle from three rods joined at the corners and it is rigid — you cannot flex it at all. Build a quadrilateral from four rods and it wobbles: a square flattens into a rhombus, and every side is still exactly as long as it was.

So four sides describe a family of quadrilaterals, not one. To pin down a single figure you must also control the flexing, which takes one more measurement — a diagonal, or an angle.

That is why every construction in this chapter comes with five measurements. Not four, not six. Five is the exact number a quadrilateral needs, and understanding why makes the constructions themselves straightforward. This page covers the first part of the ICSE Class 8 Mathematics chapter on constructions.

Draw the diagonal first. It splits the quadrilateral into two triangles, and each triangle then has three known sides.

Worked example. Construct quadrilateral $ABCD$ with $AB=4.5$ cm, $BC=5.5$ cm, $CD=5$ cm, $DA=6$ cm and diagonal $AC=7$ cm.

Check it is possible before you begin. The diagonal $AC$ belongs to both triangles, so both must satisfy the triangle inequality — any two sides must together exceed the third.

Triangle $ABC$ has sides $4.5$, $5.5$ and $7$:

$$4.5+5.5=10>7,4.5+7=11.5>5.5,5.5+7=12.5>4.5$$

Triangle $ACD$ has sides $5$, $6$ and $7$:

$$5+6=11>7,5+7=12>6,6+7=13>5$$

Both pass, so the figure exists.

Steps of construction.

- Draw $AC=7$ cm with a ruler.
- With centre $A$ and radius $4.5$ cm, draw an arc above $AC$.
- With centre $C$ and radius $5.5$ cm, draw an arc cutting the first. Label the crossing point $B$.
- With centre $A$ and radius $6$ cm, draw an arc below $AC$.
- With centre $C$ and radius $5$ cm, draw an arc cutting it. Label the crossing point $D$.
- Join $AB$, $BC$, $CD$ and $DA$.

Why the second pair of arcs goes below the line. $B$ and $D$ must sit on opposite sides of the diagonal, or the figure folds over on itself and stops being a simple quadrilateral. Drawing both above the line is the commonest mistake in this construction, and it produces a shape where two sides cross.

Worked example 2 — a construction that cannot be done. Try $AB=3$ cm, $BC=4$ cm with diagonal $AC=8$ cm.

$$3+4=7,\text{and }7<8$$

The two arcs from $A$ and $C$ never meet — they fall short of each other by $1$ cm. No such triangle exists, so no such quadrilateral does either. Checking the inequality first takes ten seconds and saves you from a page of careful drawing that was always going to fail.

Start with the diagonal that sits in a triangle you can complete, then locate the fourth vertex from two known distances.

Worked example. Construct quadrilateral $ABCD$ with $AB=4$ cm, $BC=5$ cm, $CD=4.5$ cm, and diagonals $AC=6.5$ cm and $BD=7$ cm.

Here $AB$, $BC$ and $AC$ are all known, so triangle $ABC$ can be drawn straight away. Checking it first:

$$4+5=9>6.5,4+6.5=10.5>5,5+6.5=11.5>4$$

The triangle exists. Then $D$ is fixed by being $4.5$ cm from $C$ and $7$ cm from $B$.

Steps of construction.

- Draw $AC=6.5$ cm.
- With centre $A$ and radius $4$ cm, and centre $C$ and radius $5$ cm, draw arcs meeting at $B$.
- With centre $C$ and radius $4.5$ cm, draw an arc on the other side of $AC$ from $B$.
- With centre $B$ and radius $7$ cm, draw an arc cutting it at $D$.
- Join $AB$, $BC$, $CD$, $DA$ and, if asked, the diagonals.

How to spot which triangle to draw first. Look for a set of three given lengths that form a closed triangle. Here $AB$, $BC$ and $AC$ do. The remaining two measurements — $CD$ and $BD$ — are exactly the two distances needed to place $D$, which is why five measurements is always the right number and never one too few.

Notice what the second diagonal did. $BD$ was not part of the figure's outline at all, yet without it the position of $D$ would have been free to swing along an arc. A diagonal is a measurement like any other, and it controls the flexing that the sides alone cannot.

Draw the side that lies between two known angles, then build outwards using a protractor and compass.

Worked example 1 — four sides and one angle. Construct $ABCD$ with $AB=5$ cm, $BC=4.5$ cm, $CD=5.5$ cm, $DA=6$ cm and $\angle B=80^{\circ }$.

- Draw $AB=5$ cm.
- At $B$, use a protractor to draw a ray at $80^{\circ }$ to $BA$.
- Cut $BC=4.5$ cm along that ray.
- With centre $A$ and radius $6$ cm, and centre $C$ and radius $5.5$ cm, draw arcs meeting at $D$.
- Join $CD$ and $DA$.

The angle did the job a diagonal would otherwise have done: it fixed the position of $C$ relative to $AB$, after which two arcs located $D$.

Worked example 2 — three sides and two included angles. Construct $ABCD$ with $AB=4.5$ cm, $AD=5$ cm, $BC=5.5$ cm, $\angle A=110^{\circ }$ and $\angle B=70^{\circ }$.

- Draw $AB=4.5$ cm.
- At $A$, draw a ray at $110^{\circ }$ and cut $AD=5$ cm.
- At $B$, draw a ray at $70^{\circ }$ on the same side and cut $BC=5.5$ cm.
- Join $DC$.

Something worth noticing here. $\angle A+\angle B=110+70=180^{\circ }$. Two angles on the same side of a transversal summing to $180^{\circ }$ means $AD$ and $BC$ are parallel — so this construction produces a trapezium, and you can predict that before drawing a single line. Checking whether the two given angles are supplementary tells you what shape to expect.

Worked example 3 — three angles and two included sides. Construct $ABCD$ with $AB=5$ cm, $BC=4.5$ cm, $\angle A=100^{\circ }$, $\angle B=80^{\circ }$ and $\angle C=110^{\circ }$.

First find the fourth angle, since the four must total $360^{\circ }$:

$$\angle D=360-(100+80+110)=360-290=70^{\circ }$$

- Draw $AB=5$ cm, then a ray at $80^{\circ }$ from $B$ and cut $BC=4.5$ cm.
- At $A$, draw a ray at $100^{\circ }$; at $C$, draw a ray at $110^{\circ }$.
- Those two rays meet at $D$. Measuring $\angle D$ should give $70^{\circ }$, which verifies the construction.

A question that describes nothing. Suppose three angles are given as $120^{\circ }$, $130^{\circ }$ and $140^{\circ }$. Their sum is $390^{\circ }$, so the fourth angle would be

$$360-390=-30^{\circ }$$

A negative angle is impossible, and no such quadrilateral can be drawn. The angle sum is as strong a feasibility test as the triangle inequality, and it costs one addition.

One numbered step per action, in the order your hands perform it, each naming the instrument, the centre, the radius or the angle.

The steps carry marks of their own, separately from the drawing. A perfect figure with no written steps loses those marks, and so does a list too vague to follow.

A good step reads: *With centre $A$ and radius $6$ cm, draw an arc below $AC$.* It states the centre, the radius, and the side of the line. Someone else could reproduce your figure exactly from it.

A poor step reads: *Draw an arc from $A$.* How long? In which direction? The reader cannot follow it, and neither could you tomorrow.

What every set of steps should contain:

- The first line drawn, with its length — usually a diagonal or the side between two known angles
- Each arc, with its centre and radius
- Each angle, with its vertex and size
- The labelling of every intersection point as it appears
- The joining of the final sides

Leave all your construction arcs visible. They are the evidence that you used a compass rather than measuring with a ruler, and rubbing them out can cost the method marks even when the figure is right.

Then verify by measurement. In the last worked example, $\angle D$ was predicted to be $70^{\circ }$ before being drawn, so measuring it afterwards is a genuine test. A reading of $69^{\circ }$ or $71^{\circ }$ is fine — pencil thickness alone accounts for that. A reading of $60^{\circ }$ means a real error, and it is far better to find it yourself than to have it found for you.

Rough sketch first, always. Draw the quadrilateral freehand, mark on the five given measurements, and the order of construction usually becomes obvious: you are looking for the triangle you can complete immediately. That sketch takes half a minute and prevents the most expensive mistake of all, which is starting with the wrong line.