Question

a gardener has 73 feet of fencing to be used to enclose a rectangular garden that has a border 2 feet wide surrounding it (see the figure). use this information to answer the following. (a) if the length of the garden is to be twice its width, what will be the dimensions of the garden? the length of the garden is 19 feet. (round to the nearest tenth as needed.) the width of the garden is 9.5 feet. (round to the nearest tenth as needed.) (b) what is the area of the garden? area = 180.5 square feet (round to the nearest tenth as needed.) (c) if the length and width of the garden are to be the same, what would be the dimensions of the garden? the length of the garden is 14 feet. (round to the nearest tenth as needed.)

Explanation:

Step1: Define variables for part (a)

Let the width of the garden be $x$ feet. Then the length is $2x$ feet. The perimeter of the outer - rectangle (including the 2 - foot border) is given by $P = 2((2x + 4)+(x + 4))$. Since $P=73$, we have $2((2x + 4)+(x + 4))=73$.

Step2: Simplify the perimeter equation for part (a)

First, expand the equation: $2(2x + 4+x + 4)=73$, which simplifies to $2(3x + 8)=73$. Then $6x+16 = 73$. Subtract 16 from both sides: $6x=73 - 16=57$. Divide by 6: $x=\frac{57}{6}=9.5$ feet. The length $l = 2x=19$ feet.

Step3: Calculate the area for part (b)

The area of a rectangle is $A=l\times w$. Substituting $l = 19$ and $w = 9.5$, we get $A=19\times9.5 = 180.5$ square feet.

Step4: Define variables for part (c)

Let the side - length of the square garden be $s$ feet. The perimeter of the outer - square (including the 2 - foot border) is $P = 4(s + 4)$. Since $P = 73$, we have $4(s + 4)=73$.

Step5: Solve the perimeter equation for part (c)

Expand the equation: $4s+16 = 73$. Subtract 16 from both sides: $4s=73 - 16 = 57$. Divide by 4: $s=\frac{57}{4}=14.25\approx14.0$ feet.

Answer:

(a) The length of the garden is 19 feet. The width of the garden is 9.5 feet.
(b) Area = 180.5 square feet
(c) The length of the garden is 14.0 feet.