Math Core

Rectangles, rhombuses and squares are parallelograms with something extra: right angles, equal sides, or both. That extra condition gives their diagonals special properties, and those properties are what let a carpenter check that a frame is square or let you classify a shape from its coordinates.

Three special parallelograms

Definition

Rectangle, rhombus, square

A rectangle is a parallelogram with four right angles.
A rhombus is a parallelogram with four congruent sides.
A square is a parallelogram with four right angles and four congruent sides.

You don't have to show a quadrilateral is a parallelogram first. A quadrilateral with four right angles has both pairs of opposite angles congruent, so it is automatically a parallelogram, and therefore a rectangle. A quadrilateral with four congruent sides has both pairs of opposite sides congruent, so it is automatically a rhombus.

The definitions nest inside each other:

Every rectangle, rhombus and square is a parallelogram, so each one has all the parallelogram properties.
A square is both a rectangle and a rhombus, so it has all of their properties.
A rectangle is a square only if its sides are all equal, and a rhombus is a square only if its angles are all right angles.

Diagonals of a rectangle

In rectangle $ABCD$ , compare $\triangle ABC$ and $\triangle DCB$ . They share $\overline{BC}$ , they have $\overline{AB} \cong \overline{DC}$ (opposite sides), and $\angle ABC \cong \angle DCB$ (both right angles). By SAS they are congruent, so by CPCTC, $\overline{AC} \cong \overline{DB}$ .

Rectangle ABCD. Its diagonals are congruent and bisect each other, so AE = BE = CE = DE.

Since the diagonals are congruent and bisect each other, all four half-diagonals are equal: $AE = BE = CE = DE$ . That makes $\triangle AEB$ and the other three small triangles isosceles.

Diagonals of a rhombus

In rhombus $ABCD$ with diagonals meeting at $E$ , look at $\triangle ABE$ and $\triangle ADE$ . $AB = AD$ (all sides of a rhombus are equal), $BE = DE$ (diagonals bisect each other), and $\overline{AE}$ is shared. By SSS the triangles are congruent. So $\angle AEB \cong \angle AED$ . These two angles form a linear pair, so each is $90^\circ$ . By CPCTC, $\angle BAE \cong \angle DAE$ as well, so diagonal $\overline{AC}$ bisects $\angle A$ .

Rhombus ABCD with side 5. Its diagonals are perpendicular and cut it into four congruent right triangles.

Diagonal properties of special parallelograms

A parallelogram is a rectangle if and only if its diagonals are congruent.
A parallelogram is a rhombus if and only if its diagonals are perpendicular.
A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles.
A square has all of these properties.

Property	Parallelogram	Rectangle	Rhombus	Square
Diagonals bisect each other	yes	yes	yes	yes
Diagonals congruent		yes		yes
Diagonals perpendicular			yes	yes
Diagonals bisect the angles			yes	yes

Common mistake

The "if" direction only works for parallelograms. A quadrilateral with congruent diagonals is not necessarily a rectangle; an isosceles trapezoid has congruent diagonals too. First show the shape is a parallelogram, then use its diagonals to decide what kind. Also keep the two special facts straight: rectangles have congruent diagonals, rhombuses have perpendicular diagonals.

Worked example: Side of a rhombus from its diagonals

The diagonals of a rhombus measure $24$ cm and $10$ cm. Find its perimeter.

The diagonals bisect each other at right angles, so they form four congruent right triangles with legs $12$ cm and $5$ cm. Each side of the rhombus is a hypotenuse:

s = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}.

The perimeter is $4 \cdot 13 = 52$ cm.

Worked example: Angles in a rhombus

In rhombus $ABCD$ , $m\angle ABC = 110^\circ$ and the diagonals meet at $E$ . Find $m\angle ABD$ , $m\angle BAC$ and $m\angle AEB$ .

Diagonal $\overline{BD}$ bisects $\angle ABC$ , so $m\angle ABD = 110^\circ \div 2 = 55^\circ$ .

Consecutive angles are supplementary, so $m\angle BAD = 70^\circ$ , and diagonal $\overline{AC}$ bisects it: $m\angle BAC = 35^\circ$ .

The diagonals are perpendicular, so $m\angle AEB = 90^\circ$ . Check with $\triangle ABE$ : $55 + 35 + 90 = 180$ .

Worked example: Diagonals of a rectangle

In rectangle $ABCD$ , $AC = 5x - 7$ and $BD = 2x + 11$ . The diagonals meet at $E$ . Find $AE$ .

The diagonals of a rectangle are congruent: $5x - 7 = 2x + 11$ , so $3x = 18$ and $x = 6$ . Then $AC = 23$ , and since the diagonals bisect each other, $AE = 23 \div 2 = 11.5$ .

Classifying on the coordinate plane

To classify a quadrilateral from its vertices, work from general to specific: first show it is a parallelogram, then test the diagonals.

Worked example: What kind of quadrilateral?

Classify $ABCD$ with $A(1, 1)$ , $B(5, 4)$ , $C(2, 8)$ and $D(-2, 5)$ as precisely as possible.

Parallelogram? The midpoint of $\overline{AC}$ is $(1.5, 4.5)$ and the midpoint of $\overline{BD}$ is $\left(\dfrac{5 - 2}{2}, \dfrac{4 + 5}{2}\right) = (1.5, 4.5)$ . The diagonals bisect each other, so yes.

Rectangle? $AC = \sqrt{1^2 + 7^2} = \sqrt{50}$ and $BD = \sqrt{7^2 + 1^2} = \sqrt{50}$ . The diagonals are congruent, so yes.

Rhombus? The slope of $\overline{AC}$ is $\dfrac{7}{1} = 7$ and the slope of $\overline{BD}$ is $\dfrac{5 - 4}{-2 - 5} = -\dfrac{1}{7}$ . The product is $-1$ , so the diagonals are perpendicular. Yes.

It is a rectangle and a rhombus, so $ABCD$ is a square. (Check: $AB = \sqrt{4^2 + 3^2} = 5$ and $BC = \sqrt{3^2 + 4^2} = 5$ .)

Square ABCD. Its diagonals are congruent and perpendicular.Open in grapher →

Tip

Carpenters check that a rectangular frame is truly rectangular by measuring both diagonals. If the frame's opposite sides are equal (so it is a parallelogram) and the diagonals match, the corners are right angles.

Practice

Practice 1

Which statement is always true?

Practice 2

In rectangle $JKLM$ , $JL = 3y + 4$ and $KM = 5y - 8$ . What is $KM$ ?