2.Define the shape of the rectangular area by establishing a relationship between the length and width of the rectangle. For example, L = 2W + 5, or W = 3L – 4. Be sure to include the appropriate units (inches, feet, yards, miles, or meters).
3.Using the fact that A = LW, together with the relationship defined in step 2, eliminate one of the variables to set up a quadratic equation.
4.Solve the quadratic equation using any of the techniques learned in this unit. The solution(s) will be one of the dimensions; use step 2 to find the other.
5.Now determine the perimeter so that you will know how much fencing to buy.
To define the shape of the rectangular area, we establish a relationship between the length and width of the rectangle. Let's consider an example:
Let's say we have a rectangle with length L and width W. The relationship between the length and width can be expressed as L = 2W + 5.
In this example, the units could be in inches, feet, yards, miles, or meters, depending on the context.
Next, we can use the fact that the area (A) of a rectangle is given by A = LW. We will combine this with the relationship defined in step 2 to eliminate one of the variables and set up a quadratic equation.
Substituting L = 2W + 5 into the area equation, we get A = (2W + 5)W.
Simplifying the equation, we have A = 2W^2 + 5W.
Now we have a quadratic equation of the form ax^2 + bx + c = 0. In this case, a = 2, b = 5, and c = 0.
To solve the quadratic equation, we can use techniques like factoring, completing the square, or using the quadratic formula. Let's say we solve it using the quadratic formula.
The quadratic formula is given as x = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values from our equation, we have W = (-5 ± √(5^2 - 4(2)(0))) / (2(2)).
Simplifying further, we have W = (-5 ± √(25)) / 4.
This gives us two possible solutions for the width, W. Let's compute each:
W1 = (-5 + √25) / 4 = 0.5
W2 = (-5 - √25) / 4 = -1.25
Since the width cannot be negative in this case, we discard the negative solution.
With the width, W, found to be 0.5, we can substitute this back into the relationship defined in step 2 to find the length, L.
L = 2W + 5 = 2(0.5) + 5 = 6
So, the dimensions of the rectangular area are length = 6 units and width = 0.5 units.
To determine the perimeter, we use the formula for the perimeter of a rectangle, which is given by P = 2(L + W).
Plugging in the values, we have P = 2(6 + 0.5) = 13.
Therefore, you would need to buy 13 units of fencing to enclose the rectangular area.