Math Problem Statement

Khloe has 53 m of fencing to build a three-sided fence around a rectangular plot of land that sits on a riverbank. (The fourth side of the enclosure would be the river.) The area of the land is 306 square meters. List each set of possible dimensions (length and width) of the field. list the possible dementions

Solution

To solve this problem, we can use the given information to set up a system of equations. Here's the breakdown:

Step 1: Define Variables

  • Let LL be the length of the rectangular field (the side parallel to the river).
  • Let WW be the width of the rectangular field (the two sides perpendicular to the river).

Step 2: Set up Equations

  • The total length of fencing used is 53 meters for three sides (two widths and one length), so: 2W+L=532W + L = 53
  • The area of the land is 306 square meters, so: L×W=306L \times W = 306

Step 3: Solve for One Variable

From the first equation, solve for LL: L=532WL = 53 - 2W

Step 4: Substitute into the Area Equation

Substitute L=532WL = 53 - 2W into the area equation: (532W)×W=306(53 - 2W) \times W = 306

Step 5: Expand and Solve the Quadratic Equation

53W2W2=30653W - 2W^2 = 306 2W253W+306=02W^2 - 53W + 306 = 0

Now, use the quadratic formula to solve for WW: W=(53)±(53)24×2×3062×2W = \frac{-(-53) \pm \sqrt{(-53)^2 - 4 \times 2 \times 306}}{2 \times 2} W=53±280924484W = \frac{53 \pm \sqrt{2809 - 2448}}{4} W=53±3614W = \frac{53 \pm \sqrt{361}}{4} W=53±194W = \frac{53 \pm 19}{4}

Step 6: Calculate the Possible Values for WW

W=53+194=724=18W = \frac{53 + 19}{4} = \frac{72}{4} = 18 W=53194=344=8.5W = \frac{53 - 19}{4} = \frac{34}{4} = 8.5

Step 7: Find the Corresponding LL Values

Now, substitute the possible values of WW back into L=532WL = 53 - 2W:

  • For W=18W = 18: L=532(18)=5336=17L = 53 - 2(18) = 53 - 36 = 17
  • For W=8.5W = 8.5: L=532(8.5)=5317=36L = 53 - 2(8.5) = 53 - 17 = 36

Final Answer:

The two possible sets of dimensions for the field are:

  • L=17L = 17 meters, W=18W = 18 meters
  • L=36L = 36 meters, W=8.5W = 8.5 meters

Would you like more details on the calculations or any clarification?

Here are 5 related questions:

  1. How would the dimensions change if the fencing length were 60 meters?
  2. What if the area was larger, say 400 square meters?
  3. How can we express the area in terms of WW and the fencing constraints?
  4. What happens if the shape of the field were different, like a triangle?
  5. Can this type of problem be solved graphically using a parabola?

Tip: Always verify quadratic roots by substituting them back into the original equation to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry

Formulas

Perimeter of a rectangle (2W + L)
Area of a rectangle (L × W)
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-10