Questions: A farmer has a rectangular garden plot surrounded by 220 feet of fence. Find the length and width of the garden if its area is 2,800 feet^2. width = 65, length = 45 width = 70, length = 40 width = 60, length = 50 width = 60, length = 30 none of these

A farmer has a rectangular garden plot surrounded by 220 feet of fence. Find the length and width of the garden if its area is 2,800 feet^2.
width = 65, length = 45
width = 70, length = 40
width = 60, length = 50
width = 60, length = 30
none of these

Transcript text: A farmer has a rectangular garden plot surrounded by 220 feet of fence. Find the length and width of the garden if its area is 2,800 feet $^{2}$. width $=65$, length $=45$ width $=70$, length $=40$ width $=60$, length $=50$ width $=60$, length $=30$ none of these

Solution

Solution Steps

To solve this problem, we need to use the given perimeter and area of the rectangular garden plot to find the length and width. The perimeter of a rectangle is given by $2 \times (\text{length} + \text{width})$ and the area is given by $\text{length} \times \text{width}$. We will check each provided pair of dimensions to see if they satisfy both the perimeter and area conditions.

Step 1: Define the Problem

We are given a rectangular garden plot with a perimeter of $ P = 220 $ feet and an area of $ A = 2800 $ square feet. We need to find the dimensions (length $ L $ and width $ W $) that satisfy these conditions.

Step 2: Set Up the Equations

The perimeter of a rectangle is given by the equation: \[ P = 2(L + W) \] Substituting the given perimeter: \[ 220 = 2(L + W) \implies L + W = 110 \]

The area of a rectangle is given by: \[ A = L \times W \] Substituting the given area: \[ 2800 = L \times W \]

Step 3: Solve the System of Equations

From the first equation, we can express $ L $ in terms of $ W $: \[ L = 110 - W \] Substituting this into the area equation: \[ 2800 = (110 - W) \times W \] Expanding and rearranging gives us a quadratic equation: \[ W^2 - 110W + 2800 = 0 \]

Step 4: Find the Roots of the Quadratic Equation

Using the quadratic formula $ W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $: \[ W = \frac{110 \pm \sqrt{110^2 - 4 \times 1 \times 2800}}{2 \times 1} \] Calculating the discriminant: \[ 110^2 - 4 \times 2800 = 12100 - 11200 = 900 \] Thus, \[ W = \frac{110 \pm 30}{2} \] Calculating the two possible values for $ W $: \[ W_1 = \frac{140}{2} = 70, \quad W_2 = \frac{80}{2} = 40 \]

Step 5: Determine Lengths

Using $ W = 70 $: \[ L = 110 - 70 = 40 \] Using $ W = 40 $: \[ L = 110 - 40 = 70 \]