Quadrilaterals and Their Properties

Definitions and naming conventions

A parallelogram has two pairs of opposite parallel sides. A rectangle is a quadrilateral with four right angles. A rhombus has four equal sides. A square has both four equal sides and four right angles.

This page uses the same inclusive trapezoid convention as the existing CTS shapes page: a trapezoid has at least one pair of opposite parallel sides. Parallelograms therefore count as trapezoids here. Some classrooms instead require exactly one parallel pair. State that convention when it changes an answer.

A kite has two pairs of adjacent equal sides, with the pairs using all four sides. We include rhombi in this definition. The properties below concern ordinary convex kites; naming alone should not conceal a concave case.

A tilted rectangle remains a rectangle. Horizontal edges are not part of its definition. Similarly, a square drawn on a vertex remains a square, not a different family because its diagonals look horizontal and vertical.

Why opposite sides of a parallelogram are equal

Let $ABCD$ be a parallelogram, with vertices in boundary order, and draw diagonal $AC$. Since $AB\parallel CD$, alternate interior angles give $\angle BAC=\angle DCA$. Since $BC\parallel AD$, $\angle BCA=\angle DAC$. The triangles share $AC$.

Thus $\triangle BAC\cong\triangle DCA$ by ASA. Corresponding sides give $AB=CD$ and $BC=AD$. The same triangle comparison establishes equal opposite angles.

Adjacent angles are supplementary because a side acts as a transversal across parallel opposite sides. If one angle is $68^\circ$, the next is $112^\circ$, and the four angles alternate $68^\circ,112^\circ,68^\circ,112^\circ$.

Diagonals reveal different properties

The diagonals of a parallelogram bisect each other: each is divided into two equal segments at their intersection. This does not mean the diagonals themselves have equal lengths or meet at right angles.

A convex kite’s diagonal through the vertices where equal sides meet is a perpendicular bisector of the other diagonal. In a non-rhombus kite, the other diagonal need not bisect it. The distinction between “one bisects the other” and “they bisect each other” matters.

Recovering diagonal lengths

The diagonals of a parallelogram meet at $O$. If $AO=3x-1$ and $OC=x+7$, then

Both segments are $11$ units, so $AC=22$. It would be incorrect to report $11$ as the whole diagonal or to equate this value with half of the other diagonal without further information.

Conditions that prove a classification

Some useful tests are converses of familiar properties. If both pairs of opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram. If its diagonals bisect each other, it is also a parallelogram.

Once a parallelogram is established, equal diagonals prove it is a rectangle, and perpendicular diagonals prove it is a rhombus. If both conditions hold, it is a square. The starting hypothesis is essential. An arbitrary quadrilateral with perpendicular diagonals may be a kite rather than a rhombus.

Equal diagonals alone do not establish a rectangle. A non-rectangular isosceles trapezoid can also have equal diagonals. A useful question is not only “Does this family have the property?” but also “Can another family have it?”

A rhombus calculation

A rhombus has diagonals $10\text{ cm}$ and $24\text{ cm}$. They bisect at right angles, dividing the rhombus into four right triangles with legs $5\text{ cm}$ and $12\text{ cm}$. Every side therefore has length

Its perimeter is $52\text{ cm}$. Its area is the sum of the four triangles:

Equivalently, $A=d_1d_2/2$. The diagonal-area calculation works here because the diagonals are perpendicular. A product of diagonal lengths divided by two is not the general area formula for every quadrilateral.

A coordinate check without relying on appearance

Take $A(0,0)$, $B(4,0)$, $C(6,3)$ and $D(2,3)$. The opposite side vectors are $\overrightarrow{AB}=\overrightarrow{DC}=(4,0)$ and $\overrightarrow{AD}=\overrightarrow{BC}=(2,3)$, so the quadrilateral is a parallelogram.

Its adjacent side lengths are $4$ and $\sqrt{13}$, so it is not a rhombus. The adjacent direction vectors have dot product $4(2)+0(3)=8$, not zero, so they are not perpendicular and the figure is not a rectangle. A sketch helps locate the points, but the calculations establish the classification.

Related topics

Proof in Geometry explains property-and-converse reasoning. Coordinate Proofs establishes general quadrilateral results without selecting convenient special measurements.