Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda/Solution 2

Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $29$

There is a round town, $8000$ feet in circumference.

I want to build houses there,

each house being $20$ feet long

and $10$ feet wide.

How many houses must it contain,

each house being $30$ feet long

and $20$ feet wide?

The boundary of the town is $8000$ feet.

Take a quarter of $8000$, getting $2000$.

Again take a third of $8000$, getting $2666$.

Take the half of $2000$, which is $1000$, and the half of $2666$, which is $1333$.

Not the thirtieth part of $1333$ is $44$.

Likewise the twentieth part of $1000$ is $50$.

$50$ times $44$ is $2200$.

Then form $2200$ times $4$ whch is $8800$.

This is the total number of houses.

The actual area of the town is approximately $8488$ house areas.

Using $\AA = \dfrac {C^2} {16}$ as in $25$: De Campo Rotundo, the area is approximately $6667$ house areas.

Alcuin assumes the circle contains a $1600 \times 2400$ rectangle, but such a circle could hold over $10 \, 000$ house areas.

David Singmaster reports that he managed to fit $8307$ in, but suspects it may be possible to do better.