Question

a gardener has 97 feet of fencing to be used to enclose a rectangular garden that has a border 2 - foot wide surrounding it (see the figure). (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 (round to the nearest tenth as needed.)

Explanation:

Step1: Let the width of the garden be $x$ feet.

Since the length of the garden is twice its width, the length is $2x$ feet. The border around the garden is 2 - foot - wide. So the outer - length of the fenced region is $2x + 2+2=2x + 4$ feet and the outer - width is $x + 2+2=x + 4$ feet. The total length of the fencing is 97 feet, which is the perimeter of the outer rectangle. The perimeter formula for a rectangle is $P = 2(l + w)$. So, $P=2((2x + 4)+(x + 4))$.

Step2: Simplify the perimeter equation.

$2((2x + 4)+(x + 4))=2(2x + 4+x + 4)=2(3x + 8)=6x+16$.
We know that $6x + 16=97$.

Step3: Solve for $x$.

Subtract 16 from both sides of the equation: $6x=97 - 16=81$.
Then divide both sides by 6: $x=\frac{81}{6}=13.5$ feet.

Step4: Find the length of the garden.

Since the length $l = 2x$, substituting $x = 13.5$ gives $l=2\times13.5 = 27$ feet.

Answer:

The width of the garden is 13.5 feet and the length of the garden is 27 feet.