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
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)