Greatest Area of Quadrilateral with Sides in Arithmetic Sequence

Theorem

Let $Q$ be a quadrilateral whose sides $a$, $b$, $c$ and $d$ are in arithmetic sequence.

Let $\AA$ be the area of $Q$.

Let $Q$ be such that $\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.

Then:

$\AA = \sqrt {a b c d}$

Proof

We are given that $\AA$ is the greatest possible for a quadrilateral whose sides are $a$, $b$, $c$ and $d$.

Hence $\AA$ can be found using Brahmagupta's Formula.

Let $s$ denote the semiperimeter of $Q$:

$s = \dfrac {a + b + c + d} 2$

We are given that $a$, $b$, $c$ and $d$ are in arithmetic sequence.

Without loss of generality, that means there exists $k$ such that:

 $\ds b$ $=$ $\ds a + k$ $\ds c$ $=$ $\ds a + 2 k$ $\ds d$ $=$ $\ds a + 3 k$

where $k$ is the common difference.

Then:

 $\ds s$ $=$ $\ds \dfrac {a + b + c + d} 2$ Definition of Semiperimeter $\ds$ $=$ $\ds \dfrac {a + \paren {a + k} + \paren {a + 2 k} + \paren {a + 3 k} } 2$ $\ds$ $=$ $\ds \dfrac {4 a + 6 k} 2$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds 2 a + 3 k$

and so:

 $\ds \AA$ $=$ $\ds \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$ Brahmagupta's Formula $\ds$ $=$ $\ds \sqrt {\paren {a + 3 k} \times \paren {a + 2 k} \times \paren {a + k} \times a}$ substituting $s = 2 a + 3 k$ from $(1)$ and simplifying $\ds$ $=$ $\ds \sqrt {a b c d}$ from above

$\blacksquare$