Question

a gardener has 85 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. the length of the garden is 23.0 feet (round to the nearest tenth as needed.) the width of the garden is 11.5 feet (round to the nearest tenth as needed.) (b) what is the area of the garden? area = 264.5 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 17.3 feet (round to the nearest tenth as needed.) the width of the garden is

Explanation:

Step1: Recall the area formula for a rectangle

The area formula for a rectangle is $A = l\times w$, where $A$ is the area, $l$ is the length and $w$ is the width. Given $l = 23.0$ feet and $w=11.5$ feet.

Step2: Calculate the area

$A=l\times w=23.0\times11.5 = 264.5$ square - feet. This matches the given area in the problem, confirming our understanding.

Step3: Solve part (c)

Let the new length and width be $x$. The perimeter of the outer - rectangle (including the 2 - foot border) is related to the available fencing. The perimeter of the outer - rectangle $P = 2(x + 4)+2(x + 4)$ (since there is a 2 - foot border on both sides for length and width). The available fencing is 85 feet. So $P=85$. But $P = 4(x + 4)$. Then $4(x + 4)=85$, $4x+16 = 85$, $4x=85 - 16=69$, $x=\frac{69}{4}=17.25\approx17.3$ feet.

Answer:

The width of the garden in part (a) is 11.5 feet.
The length of the garden in part (a) is 23.0 feet.
The area of the garden in part (b) is 264.5 square feet.
The length and width of the garden in part (c) are 17.3 feet.