, 2013

We refer to distances in the horizontal direction as widths and distances in the vertical direction as lengths.

Suppose that each of the six enclosures labelled $A_1$ have width $x$ m and length $y$ m.

Then each of these has area $\text{xy m 2}$.

We start by determining the dimensions of the remaining enclosures in terms of these two variables.

Enclosure $A_2$ has width $x+x+x=3x$ m.

Since the area of enclosure $A_2$ is four times that of $A_1$, then its area is $\text{4xy m 2}$.

Therefore, the length of the enclosure $A_2$ is its area divided by its width, or $\dfrac{4xy}{3x}=\dfrac{4}{3}y$ m. (We use the notation $(4/3)y$ in the diagram.)

Thus, the length of enclosure $A_3$ is $y+y+\frac{4}{3}y=\frac{10}{3}y$ m.

Since the area of enclosure $A_3$ is $\text{5xy m 2}$, then its width is $\dfrac{5xy}{\frac{10}{3}y} = \dfrac{3}{2}x$ m. (We use the notation $(3/2)x$ in the diagram.)

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The total width of the field is 45 m. This can also be expressed (using the top fence) as $x+x+x+\frac{3}{2}x=\frac{9}{2}x$ m.

Since $\frac{9}{2}x = 45$, then $x = \frac{2}{9}(45)=10$.

In terms of $x$, the total length, in metres, of “horizontal" fencing is $$(x+x+x+\tfrac{3}{2}x)+(x+x+x)+(x+x+x)+(x+x+x+\tfrac{3}{2}x)=15x$$ which we calculate by going from left to right along each row from top to bottom.

In terms of $y$, the total length, in metres, of “vertical" fencing is $$(y+y+\tfrac{4}{3}y)+(y+y)+(y+y)+(y+y+\tfrac{4}{3}y)+(y+y+\tfrac{4}{3}y) = 14y$$ which we calculate by going from top to bottom along each column from left to right.

Since the total length of fencing is 360 m, then $15x+14y=360$.

Since $x=10$, then $150+14y=360$ or $14y=210$ and so $y=15$.

Therefore, the area of enclosure $A_1$ is $\text{xy = (10)(15)=150 m 2}$.

Of the given answers, this is closest to (in fact, equal to) 150.0.