Question

An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a livestock area of 25.0 perches by 4.00 perches. What was the total area in (a) the old unit of roods and (b) the more modern unit of square meters? Here, 1 acre is an area of 40 perches by 4 perches, 1 rood is an area of 40 perches by 1 perch, and 1 perch is the length 16.5 ft.

Short answer

1. Total area in old units of roods is${}^{\mathbf{14}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{roods}}$.

2. Total area in a modern unit of square meters is${}^{\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}{\mathbf{m}}^{\mathbf{2}}}$.

Step-by-step solution

Given data

$\mathbf{1}\mathbf{}\mathbf{acre}\mathbf{}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{perch}\mathbf{\times }\mathbf{}\mathbf{4}\mathbf{}\mathbf{perch}$

$\mathbf{1}\mathbf{}\mathbf{rood}\mathbf{}\mathbf{=}\mathbf{40}\mathbf{}{\mathbf{perch}}^{\mathbf{2}}$

$\mathbf{1}\mathbf{}\mathbf{perch}\mathbf{}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{ft}$

$\mathbf{1}\mathbf{}\mathbf{m}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{28}\mathbf{}\mathbf{ft}$

Understanding the unit conversion

Every unit of the area mentioned in the problem has a conversion factor between them and hence the area can be measured in units of rood and square meters.

(a) To find a total area in old units

The total area ${}^{\mathbf{A}}$can be calculated as:

$$\begin{gathered} \mathbf{A}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{acres}\mathbf{}\mathbf{+}\left(25\mathrm{perch}\times 4\mathrm{perch}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathbf{}\mathbf{acres}\mathbf{+}\mathbf{}\mathbf{100}\mathbf{}{\mathbf{perch}}^{\mathbf{2}} \end{gathered}$$

Now, you need to convert the acres into${}^{{\mathbf{perch}}^{\mathbf{2}}}$ .

role="math" localid="1654670503052" $$\begin{gathered} \mathbf{A}\mathbf{}\mathbf{=}\left(3\mathrm{acres}\times \frac{40\mathrm{perch}\times 4\mathrm{perch}}{1\mathrm{acres}}\right)\mathbf{}\mathbf{+}\mathbf{100}\mathbf{}{\mathbf{perch}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{480}\mathbf{}{\mathbf{perch}}^{\mathbf{2}\mathbf{}}\mathbf{+}\mathbf{}\mathbf{100}\mathbf{}{\mathbf{perch}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{580}\mathbf{}{\mathbf{perch}}^{\mathbf{2}} \end{gathered}$$

Now, you need to convert this area into roods. By using a conversion factor,

$\mathbf{1}\mathbf{}\mathbf{rood}\mathbf{}\mathbf{=}\mathbf{}\mathbf{40}\mathbf{}{\mathbf{perch}}^{\mathbf{2}}$

$\mathbf{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{580}\mathbf{}{\mathbf{perch}}^{\mathbf{2}}\mathbf{}\mathbf{\times }\frac{\mathbf{1}\mathbf{}\mathbf{rood}}{\mathbf{40}\mathbf{}{\mathbf{perch}}^{\mathbf{2}}}$

Thus, total area is ${}^{\mathbf{14}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{roods}}$.

(b) Determination of the total area in modern units

It is given that,${}^{\mathbf{1}\mathbf{}\mathbf{perch}\mathbf{}\mathbf{=}\mathbf{}\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{ft}}$

Therefore,

${}^{\mathbf{580}\mathbf{}\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{c}{\mathbf{h}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{58}\mathbf{,}\mathbf{177}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{f}{\mathbf{t}}^{\mathbf{2}}}$

You can say that total area is${}^{\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}{\mathbf{ft}}^{\mathbf{2}}}$.

Therefore, area in modern units is,

$$\begin{gathered} \mathbf{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}\mathbf{}{\mathbf{ft}}^{\mathbf{2}}}\mathbf{\times }{\left(\frac{1m}{3.28\mathrm{ft}}\right)}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}{\mathbf{m}}^{\mathbf{2}} \end{gathered}$$

Thus, the total area in modern units of square meters is${}^{\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}}$ .