Propositiones ad Acuendos Juvenes/Problems/24 - De Campo Triangulo

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Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $24$

De Campo Triangulo
A Triangular Field
A triangular field is $30$ perches along $2$ sides
and $18$ perches along the bottom.
How many acres must be enclosed?


Solution

Join the $2$ long sides and make $60$.

Half of $60$ is $30$.

Because the bottom is $18$, take half of this which is $9$.

$9$ times $30$ is $270$.

Divide $270$ by $12$ which makes $22 \tfrac 1 2$.

Again divide by $12$, giving $1$ acre and $12$ perches.

$\blacksquare$


Historical Note

When the answer is given in perches, it is clear that square perches is really meant.

Note that the acre as given is defined as $144$ square perches rather than the modern definition as $160$ of them.


As David Singmaster points out, the answer is in fact numerically incorrect: $22 \frac 1 2$ acre is in fact $126$ square perches.


However, using Heron's Formula:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where:

\(\ds s\) \(=\) \(\ds \dfrac {30 + 30 + 18} 2\) \(\ds = 39\)
\(\ds s - a\) \(=\) \(\ds 39 - 30\) \(\ds = 9\)
\(\ds s - b\) \(=\) \(\ds 39 - 30\) \(\ds = 9\)
\(\ds s - b\) \(=\) \(\ds 39 - 18\) \(\ds = 21\)

we obtain:

\(\ds \AA\) \(=\) \(\ds \sqrt {39 \times 9 \times 9 \times 21}\)
\(\ds \) \(=\) \(\ds \sqrt {66 \, 339}\)
\(\ds \) \(\approx\) \(\ds 257.56\)

At $144$ square perches to the acre, this equals $1$ acre and $113.56$ square perches.


Sources

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