Problem 244

Solution 1:

The length of the expanded garden is $(5+2\times2)\text{ m}=9\text{ m}$, and the
width is $4 \text{ m}$.

Thus, the total area of the expanded garden is $\$9\text{ m}\times4\text{ m}=36\text{}$
m}^2$.

Solution 2:

The area of the original $5 $\text{}$
m}$by$4 $\text{}$ m}$ garden
is $\$5\text{ m}\times4\text{ m}=20\text{}$
m}^2$.

Each additional $2 \text{ m}$ by
$4 \text{ m}$ plot has area $2\text{ m}\times4\text{ m}=8\text{ m}^2$,
and so the total area of the expanded garden is $\$(20+2\times8)\text{ m}^2=36\text{}$
m}^2$.
Solution 1:

The combined garden and path has length $9\text{ m}+1\text{ m}=10\text{ m}$, and
width $\$(4+2\times1)\text{ m}=6\text{}$
m}$.

Thus, the area of the garden and the path is $\$10\text{ m}\times6\text{ m}=60\text{}$
m}^2$.

Solution 2:

Consider splitting the path into three rectangles, as shown.

[[IMAGE0]]

Each of the rectangles above and below the garden has dimensions
$9\text{ m}$ by $1\text{ m}$, and thus each has area $\$9\text{ m}\times1\text{ m}=9\text{}$
m}^2$.

The remaining section of the path has height$(4+2\times1)\text{ m}=6\text{ m}$ and
width $1\text{ m}$, and thus has area
$\$6\text{ m}\times1\text{ m}=6\text{}$
m}^2. The area of the expanded garden is 36 m^2$, and so the total combined area of the
garden and the path is $\$(36+2\times9+6)\text{ m}^2=60\text{}$ m}^2$.
Solution 1:

Each of the new plots has length $2 $\text{}$
m}$, and so$n$ plots
increase the $9\text{ m}$ length of
the garden by $2n\text{ m}$.

Thus, the combined length of the garden and the path is $\$(9+2n+2\times1)\text{ m}=(2n+11)\text{}$
m}. The combined width of the garden and the path is $(4+2\times1)\text{ m}=6\text{}$ m}$.

Thus in $\text{m}^2$, the total
combined area of the garden and the path is $6\times(2n+11)$.

Solving $6\times(2n+11)=150$, we get
$2n+11=\frac{150}{6}=25$ or $2n=14$, and so $n=7$.

Solution 2:

Consider splitting the combined area of the garden and path into
three rectangles, as shown.

[[IMAGE1]]

Each of the rectangles to the left and right of the garden has height
$(4+2\times1)\text{ m}=6\text{ m}$,
width $1\text{ m}$, and thus each has
area $\$6\text{ m}\times1\text{ m}=6\text{}$
m}^2$.

The remaining rectangle, which combines the garden and the remaining
sections of the path, also has height $6\text{ m}$.

Each of the new plots has length $2 $\text{}$
m}$, and so$n$ plots
increase the $9\text{ m}$ length of
the garden by $2n\text{ m}$.

Thus, the length of this remaining rectangle is $(2n+9)\text{ m}$.

Measured in $\text{m}^2$, the total
combined area of the garden and the path is $2\times6+6\times(2n+9)$ or $12+6\times(2n+9)$.

Solving $12+6\times(2n+9)=150$, we
get $6\times(2n+9)=138$ or $2n+9=\frac{138}{6}=23$ or $2n=14$, and so $n=7$.