Propositiones ad Acuendos Juvenes/Problems/28 - De Civitate Triangula

Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $28$

A triangular town measures $100$ feet along one side,

$100$ feet along another side

and $90$ feet along the front.

I want to build houses there,

each house being $20$ feet long

and $10$ feet wide.

How many houses can there be?

Adding the two [[Definition:Side of Polygon|sides] makes $200$ feet.

The half of $200$ is $100$.

The front measures $90$ and the half of $90$ is $45$.

Because each house is $20$ feet long and $20$ feet wide:

take one $20$th of $100$ which is $5$

and take one $10$th of $40$ which is $4$.

Then $5$ times $4$ is $20$.

That is the number of houses the town should contain.

According to the Roman formula the area of the triangle is $4500$ square feet.

This is $22 \frac 1 2$ house areas.

The actual area is approximately $4019$ square feet, which is just over $20$ house areas.

The marking down of $45$ to $40$ in the calculation may be an attempt to compensate for the inaccuracy of the formula.

However, fitting them into a triangle is not easy.

David Singmaster reports that he managed to fit $15$ in.

John Hadley, on the other hand, managed $18$, but bent the walls of some of the houses slightly.