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 $x \mathrm{~m}$ and length $y \mathrm{~m}$.

Then each of these has area $x y \mathrm{~m}^{2}$.

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

Enclosure $A_{2}$ has width $x+x+x=3 x \mathrm{~m}$.

Since the area of enclosure $A_{2}$ is four times that of $A_{1}$, then its area is $4 x y \mathrm{~m}^{2}$.

Therefore, the length of the enclosure $A_{2}$ is its area divided by its width, or $\frac{4 x y}{3 x}=\frac{4}{3} y \mathrm{~m}$. (We use the notation (4/3) $y$ in the diagram.)

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

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

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-127.jpg?height=350&width=458&top_left_y=1934&top_left_x=888)

The total width of the field is $45 \mathrm{~m}$. 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

$$
\left(x+x+x+\frac{3}{2} x\right)+(x+x+x)+(x+x+x)+\left(x+x+x+\frac{3}{2} x\right)=15 x
$$

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

$$
\left(y+y+\frac{4}{3} y\right)+(y+y)+(y+y)+\left(y+y+\frac{4}{3} y\right)+\left(y+y+\frac{4}{3} y\right)=14 y
$$

which we calculate by going from top to bottom along each column from left to right.

Since the total length of fencing is $360 \mathrm{~m}$, then $15 x+14 y=360$.

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

Therefore, the area of enclosure $A_{1}$ is $x y=(10)(15)=150 \mathrm{~m}^{2}$.

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

ANswer: (B)
A farmer has a rectangular field with width 45 metres. He divides the field into smaller rectangular animal enclosures in three different sizes, as shown.

F - - 45 metres - - - -

| $A_{1}$ | $A_{1}$ | $A_{1}$ |  |
| :--- | :--- | :--- | :--- |
| $A_{1}$ | $A_{1}$ | $A_{1}$ | $A_{3}$ |
| $A_{2}$ |  |  |  |
|  |  |  |  |

Each enclosure labelled $A_{1}$ has the same dimensions. Also, the area of the enclosure labelled $A_{2}$ is 4 times the area of $A_{1}$, and the area of the enclosure labelled $A_{3}$ is 5 times the area of $A_{1}$. The lines in the diagram represent fences. The total length of all the fences is 360 metres. The area of $A_{1}$, in square metres, is closest to
(A) 143.4
(B) 150.0
(C) 175.2
(D) 162.7
(E) 405.0