A Rhombus Needs Only Two Measurements, Not Five

Because its own properties supply the rest.

A general quadrilateral needs five measurements, as the previous part showed. A rhombus needs only two — its two diagonals, say, or one side and one diagonal, or one side and one angle.

The reason is that all four sides are equal is already three pieces of information. Tell me one side and you have told me all four, so three of the five measurements arrive free with the word rhombus.

The same logic runs through this whole page. A rectangle needs two measurements, because all angles are right angles supplies the rest. A square needs just one. Every property a shape is defined by is a measurement you no longer have to be given — and knowing how many a figure needs is the quickest check that a question is complete. This page covers the second part of the ICSE Class 8 Mathematics chapter on constructions.

From two adjacent sides and the included angle, or from the two diagonals and the angle between them. Either way, three measurements suffice.

Worked example 1 — two sides and the included angle. Construct parallelogram $ABCD$ with $AB=5$ cm, $BC=4$ cm and $\angle B=60^{\circ }$.

- Draw $AB=5$ cm.
- At $B$, draw a ray at $60^{\circ }$ to $BA$ and cut $BC=4$ cm.
- With centre $A$ and radius $4$ cm, and centre $C$ and radius $5$ cm, draw arcs meeting at $D$.
- Join $CD$ and $DA$.

The two arcs use the opposite sides property: $AD=BC=4$ cm and $DC=AB=5$ cm. Those two lengths were never given to you — they came from the definition.

Predict before you measure. Since adjacent angles of a parallelogram are supplementary, $\angle A=180-60=120^{\circ }$, and $\angle C=60^{\circ }$, $\angle D=120^{\circ }$. The perimeter is $2(5+4)=18$ cm. Measuring $\angle A$ afterwards and finding $120^{\circ }$ verifies the construction.

Worked example 2 — from the diagonals. Construct a parallelogram whose diagonals are $8$ cm and $6$ cm, meeting at $60^{\circ }$.

The diagonals of a parallelogram bisect each other, so the halves are $4$ cm and $3$ cm.

- Draw $AC=8$ cm and mark its midpoint $O$, so $AO=OC=4$ cm.
- Through $O$, draw a line at $60^{\circ }$ to $AC$.
- Mark $B$ and $D$ on that line, each $3$ cm from $O$ and on opposite sides.
- Join $AB$, $BC$, $CD$ and $DA$.

Why the angle is needed. Two diagonals alone are not enough. Pin the two half-lengths at their midpoint and the frame swings freely, giving a different parallelogram at every angle. The diagonals fix the lengths; the angle between them fixes the shape. Any question giving only two diagonals and asking for a parallelogram is incomplete — and spotting that is worth as much as being able to draw the figure.

The one exception. If the angle between the diagonals happens to be $90^{\circ }$, the figure is forced to be a rhombus, and then the two diagonals really are enough on their own.

Two measurements, and the right-angle property of the diagonals does the work.

Worked example 1 — both diagonals given. Construct a rhombus with diagonals $8$ cm and $6$ cm.

The diagonals bisect each other at right angles, so:

- Draw $AC=8$ cm and mark its midpoint $O$.
- At $O$, construct a perpendicular to $AC$ with compass and ruler.
- Mark $B$ and $D$ on the perpendicular, each $3$ cm from $O$, on opposite sides.
- Join $AB$, $BC$, $CD$ and $DA$.

Predict the side before measuring it. The half-diagonals $4$ cm and $3$ cm form a right-angled triangle with the side as hypotenuse:

$$\text{side}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\text{ cm}$$

So every side should measure $5$ cm and the perimeter $20$ cm. The area is

$$\frac{1}{2}\times 8\times 6=24{\text{ cm}}^2$$

Measuring all four sides and finding $5$ cm each is a complete verification — and note that this construction needed no protractor at all, since the perpendicular is a compass construction.

Worked example 2 — one side and one diagonal. Construct a rhombus with side $5$ cm and one diagonal $8$ cm.

- Draw $AC=8$ cm.
- With centre $A$ and radius $5$ cm, and centre $C$ and radius $5$ cm, draw arcs above $AC$ meeting at $B$.
- Repeat below $AC$ to get $D$.
- Join the four sides.

This is the same rhombus as before. Half the given diagonal is $4$ cm, so the other half-diagonal is

$$\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3\text{ cm}$$

giving a second diagonal of $6$ cm — exactly the figure from worked example 1, reached from different information. Two constructions producing the same shape is the surest sign both were right.

Worked example 3 — one side and one angle. Construct a rhombus with side $4.5$ cm and $\angle A=75^{\circ }$.

- Draw $AB=4.5$ cm, then a ray at $75^{\circ }$ from $A$ and cut $AD=4.5$ cm.
- With centre $B$ and centre $D$, radius $4.5$ cm each, draw arcs meeting at $C$.
- Join $BC$ and $DC$.

The other angles follow from the properties: $\angle C=75^{\circ }$ and $\angle B=\angle D=105^{\circ }$. Check: $75+105+75+105=360^{\circ }$.

Never assume the diagonals of a rhombus are equal. They bisect each other at right angles, which is what makes the construction work, but $8$ cm and $6$ cm are plainly different. Equal diagonals would make it a square.

Use Pythagoras to find the missing side first, then construct with right angles.

Worked example 1 — a rectangle from a side and a diagonal. Construct a rectangle with one side $6$ cm and diagonal $10$ cm.

The side, the other side and the diagonal form a right-angled triangle:

$$\text{other side}=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ cm}$$

- Draw $AB=8$ cm.
- At $A$ and at $B$, construct perpendiculars to $AB$.
- Cut $AD=6$ cm and $BC=6$ cm along them, on the same side.
- Join $DC$.

The area is $8\times 6=48{\text{ cm}}^2$ and the perimeter $2(8+6)=28$ cm. Measuring $DC$ should give $8$ cm, and both diagonals should measure $10$ cm — two independent checks.

Worked example 2 — a square from its diagonal. Construct a square whose diagonal is $6$ cm.

The diagonals of a square are equal and bisect each other at right angles, so:

- Draw $AC=6$ cm and mark its midpoint $O$.
- Construct a perpendicular to $AC$ at $O$.
- Mark $B$ and $D$ on it, each $3$ cm from $O$, on opposite sides.
- Join the four sides.

Predict the side. Half of each diagonal is $3$ cm, so

$$\text{side}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}\approx 4.24\text{ cm}$$

The area can be checked two ways, and they must agree:

$${\text{side}}^2=(3\sqrt{2}{)}^2=18{\text{ cm}}^2,\frac{1}{2}{d}^2=\frac{1}{2}\times 36=18{\text{ cm}}^2$$

Worked example 3 — a square from its side. Construct a square of side $4.5$ cm.

Draw $AB=4.5$ cm, erect perpendiculars at $A$ and $B$, cut $4.5$ cm along each, and join. The diagonal should measure

$$4.5\sqrt{2}\approx 4.5\times 1.414=6.36\text{ cm}$$

Measuring both diagonals and getting about $6.4$ cm confirms the figure.

**Construct right angles with the compass, not the protractor, wherever a question says ruler and compass only.** A protractor reading is a measurement; a compass construction is a proof. The two look identical on paper, but only one earns the method marks when the instruction demanded it — and misreading that instruction is the commonest way to lose marks in an otherwise faultless construction.

Draw the longer parallel side first, set the angle at one end, then use the parallel property for the rest.

Worked example. Construct trapezium $ABCD$ with $AB\parallel DC$, $AB=7$ cm, $DC=4$ cm, $AD=5$ cm and $\angle A=60^{\circ }$.

- Draw $AB=7$ cm.
- At $A$, draw a ray at $60^{\circ }$ and cut $AD=5$ cm.
- Through $D$, draw a line parallel to $AB$. With ruler and compass, do this by copying the $60^{\circ }$ angle at $D$ using the corresponding-angles rule.
- Cut $DC=4$ cm along that parallel, in the same direction as $B$.
- Join $BC$.

Predict what you can, then measure it. Since $AB\parallel DC$, the angles $\angle A$ and $\angle D$ are co-interior and must be supplementary:

$$\angle D=180-60=120^{\circ }$$

So measuring $\angle D$ and finding $120^{\circ }$ verifies that your parallel line really is parallel — which is the one step in this construction most likely to drift.

**Cut $DC$ in the right direction.** Running the $4$ cm the wrong way along the parallel gives a figure whose sides cross, because $C$ then lands on the far side of $D$. Checking that $C$ sits roughly above the segment $AB$ before joining $BC$ takes a glance and prevents the whole drawing being wasted.

How many measurements a trapezium needs. Being a trapezium supplies only one condition — one pair of sides parallel — so it needs four measurements, and this question gave exactly four. Compare the counts across the whole chapter:

- General quadrilateral: five
- Trapezium: four
- Parallelogram: three
- Rhombus and rectangle: two
- Square: one

Read that list as a checklist. If a question gives a rhombus three measurements, one of them is redundant and should agree with the others; if it gives only one, the question is incomplete. Counting the measurements before you pick up the compass tells you whether the figure is even determined — and it is the same counting argument that made five the magic number for a general quadrilateral.