Propositiones ad Acuendos Juvenes/Problems/23 - De Campo Quadrangulo

Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $23$

A four-sided field measures $30$ perches down one side

and $32$ down the other;

it is $34$ perches across the top

and $32$ across the bottom.

How many acres are included in the field?

The sum of the lengths of the sides is $62$ perches, half of which is $31$.

The sum of the widths is $66$, half of which is $33$.

$31$ times this is $1023$.

Divide this into $12$ parts, as in $22$: De Campo Fastigioso, and there are $85$.

Again divide $85$ by $12$, and this gives $7$.

Therefore there are $7$ acres here.

$\blacksquare$

The formula being used here for a quadrilateral whose sides are $a$, $b$, $c$ and $d$ is the Egyptian Formula for Area of Quadrilateral:

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

which may be a fair approximation if the field is approximately rectangular.

But the area of a quadrilateral depends not only on the lengths of its sides but also its angles.

The maximum area is obtained when the quadrilateral is cyclic is the area, in which case Brahmagupta's Formula can be used:

$\AA^2 = \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}$

which in this case gives approximately $1022$ square units.