Rhombus

A rhombus is a quadrilateral whose four sides all have the same length. That one condition forces everything else about the shape.

Every rhombus is a parallelogram. Four equal sides guarantee that both pairs of opposite sides are parallel, so a rhombus inherits all the parallelogram properties:

• Two pairs of parallel sides — opposite sides are parallel and congruent.
• Opposite angles are equal.
• Consecutive angles are supplementary: they sum to $180°$.
• The diagonals bisect each other.

The converse fails: a parallelogram becomes a rhombus only when all four sides are equal, so a $3 \times 5$ rectangle is a parallelogram but not a rhombus.

The diagonals. Two extra facts separate a rhombus from a general parallelogram:

• The diagonals are perpendicular — they cross at $90°$.
• Each diagonal bisects the two vertex angles it passes through.

The diagonals are not equal in general; equal diagonals occur only in the special case where the rhombus is a square. Because they bisect each other at right angles, the two half-diagonals and one side form a right triangle, so for side $a$ and diagonals $d_1$ and $d_2$:

$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2.$

Rhombus vs square. Every square is a rhombus, since a square has four equal sides — but not every rhombus is a square, because a square additionally requires four right angles. In fact a rhombus with even one right angle must be a square: consecutive angles are supplementary, so a single $90°$ angle forces all four to be $90°$. "A rhombus with four right angles" is therefore just another name for a square, and any non-square rhombus has two acute and two obtuse angles.

Area. Two formulas, depending on which measurements you have:

$A = \frac{1}{2} d_1 d_2 \qquad \text{(from the two diagonals)}$

$A = b h = a^2 \sin\theta \qquad \text{(base and height, or side and any interior angle)}$

The perimeter is always $P = 4a$.

Worked example. A rhombus has diagonals $d_1 = 6$ and $d_2 = 8$. Its area is $\tfrac{1}{2}(6)(8) = 24$. Half-diagonals of $3$ and $4$ give the side $a = \sqrt{3^2 + 4^2} = 5$, so the perimeter is $20$ and the height is $h = A / a = 24 / 5 = 4.8$.