We are tasked with finding the possible dimensions (length and width) of a rectangular plot of land that satisfies two conditions:

1. Khloe has 53 meters of fencing for three sides.
2. The area of the land is 306 square meters.

Step-by-step Approach:

1. Define Variables:

Let:

• $L$ = length of the side parallel to the river.
• $W$ = width of the side perpendicular to the river.

2. Form Equations:

• The total fencing covers three sides: two widths and one length. Thus, the perimeter constraint is: $L + 2W = 53$

• The area of the rectangle is given by: $L \times W = 306$

3. Solve the System of Equations:

From the first equation, solve for $L$: $L = 53 - 2W$

Substitute this into the area equation: $(53 - 2W) \times W = 306$ Expanding the equation: $53W - 2W^2 = 306$ Rearranging into standard quadratic form: $2W^2 - 53W + 306 = 0$

4. Solve the Quadratic Equation:

We can solve this quadratic using the quadratic formula: $W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 2$, $b = -53$, and $c = 306$.

First, calculate the discriminant: $\Delta = (-53)^2 - 4(2)(306) = 2809 - 2448 = 361$

Now, solve for $W$: $W = \frac{53 \pm \sqrt{361}}{4} = \frac{53 \pm 19}{4}$

Thus, the two possible values for $W$ are: $W = \frac{53 + 19}{4} = \frac{72}{4} = 18$ $W = \frac{53 - 19}{4} = \frac{34}{4} = 8.5$

5. Find Corresponding Lengths:

Using the equation $L = 53 - 2W$, we can now find the values of $L$ for each $W$:

• If $W = 18$: $L = 53 - 2(18) = 53 - 36 = 17$

• If $W = 8.5$: $L = 53 - 2(8.5) = 53 - 17 = 36$

Final Answer:

The two sets of possible dimensions for the field are:

1. Length = 17 meters, Width = 18 meters.
2. Length = 36 meters, Width = 8.5 meters.

Would you like a detailed explanation of any of these steps?

Questions:

1. How does the quadratic formula work when solving area problems like this?
2. Could the discriminant ever result in no possible dimensions for a problem like this?
3. What happens if more fencing were available? How would it affect the dimensions?
4. How would you interpret negative or unrealistic solutions for this problem?
5. Could this method be used for non-rectangular plots, and how would it change?

Tip:

When solving problems with two variables, always look for ways to express one variable in terms of the other—it simplifies the calculations and reduces complexity.