Area of a Rhombus

Consider a rhombus $A B C D$ and draw its two diagonals. Recall that the diagonals bisect each other at a right angle. Let $E$ be the point of intersection between the diagonals.

The area of the rhombus can be written as the sum of the areas of

△ A C D

and

△ A C B .

These two triangles share the base

\overline{A C}

and their heights are

\overline{D E}

and

\overline{B E},

respectively.

A A = A_{A C D} + A_{A C B} ⇓ = \frac{1}{2} A C \cdot D E + \frac{1}{2} A C \cdot B E

Finally, use the fact that the heights of the triangles add up to

d_{1}

and that the base is equal to

d_{2}

to derive a formula for the area of a rhombus in terms of its diagonals.

A = \frac{1}{2} A C \cdot D E + \frac{1}{2} A C \cdot B E

▼

Simplify

Substitute

$A C = d_{2}$

A = \frac{1}{2} d_{2} \cdot D E + \frac{1}{2} d_{2} \cdot B E

FactorOut

Factor out $\frac{1}{2} d_{2}$

A = \frac{1}{2} d_{2} (D E + B E)

Substitute

$D E + B E = d_{1}$

A = \frac{1}{2} d_{2} d_{1}

CommutativePropMult

Commutative Property of Multiplication

A = \frac{1}{2} d_{1} d_{2} ✓

Area of a Rhombus

Proof

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