Math Problem Statement

The area of the vegetable garden in the form shown in the figure was 80 m². A fence with a 1 m wide iron net was built around the green land. How many meters of wire are needed?

Solution

The problem presented involves finding how many meters of wire are needed to enclose a vegetable garden. The given conditions are as follows:

  • The area of the vegetable garden is 80 m².
  • The shape of the garden has one side marked as x+2x + 2 and another side marked as xx.
  • A 1-meter wide iron net fence is to be built around the perimeter of the garden.

Approach:

  1. Calculate dimensions:

    • The area AA of the garden is given by the formula for the area of a rectangle, which is: A=Length×WidthA = \text{Length} \times \text{Width}
    • The length and width are x+2x + 2 and xx, respectively. Hence: A=(x+2)×x=80m2A = (x + 2) \times x = 80 \, \text{m}^2 Expanding the equation: x2+2x=80x^2 + 2x = 80 Solving the quadratic equation: x2+2x80=0x^2 + 2x - 80 = 0
  2. Solve the quadratic equation: We can solve x2+2x80=0x^2 + 2x - 80 = 0 using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=2b = 2, and c=80c = -80.

  3. Find the perimeter: Once xx is determined, we can calculate the perimeter PP of the rectangle: P=2×(Length+Width)=2×((x+2)+x)P = 2 \times (\text{Length} + \text{Width}) = 2 \times ((x + 2) + x)

Let's proceed with solving the quadratic equation and find the perimeter of the garden.The value of xx is 8 meters, which means the dimensions of the garden are:

  • Width: x=8mx = 8 \, \text{m}
  • Length: x+2=10mx + 2 = 10 \, \text{m}

The total perimeter of the garden is: P=36metersP = 36 \, \text{meters}

Thus, 36 meters of wire are required to enclose the garden.

Would you like more details on any part of the solution or have any further questions?

Here are 5 related questions to expand your understanding:

  1. How do we derive the quadratic formula for solving quadratic equations?
  2. What would happen if the area of the garden was increased by 20%?
  3. How do you find the perimeter for other shapes, like a circle or triangle?
  4. Can we solve this problem graphically by plotting the quadratic equation?
  5. How do the properties of quadratic functions help in real-life situations like this?

Tip: Always check the discriminant in a quadratic equation to determine if real solutions exist before applying the quadratic formula.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Perimeter
Area

Formulas

Area of a rectangle: A = length × width
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
Perimeter of a rectangle: P = 2 × (length + width)

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10