Math Problem Statement

A rectangular parking lot has a length that is 9 yards greater than the width. The area of the parking lot is 360 square yards. Find the length and the width. Question content area bottom Part 1 The parking lot has a width of    enter your response here yards.

Solution

Let's denote the width of the parking lot as ww yards.

The length of the parking lot is 9 yards greater than the width, so the length is w+9w + 9 yards.

The area of the parking lot is given as 360 square yards, and the area of a rectangle is found using the formula:

Area=Length×Width\text{Area} = \text{Length} \times \text{Width}

Substituting the values we know:

w×(w+9)=360w \times (w + 9) = 360

Now, expand the equation:

w2+9w=360w^2 + 9w = 360

Rearrange this into a standard quadratic equation:

w2+9w360=0w^2 + 9w - 360 = 0

We can solve this quadratic equation using the quadratic formula:

w=b±b24ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation w2+9w360=0w^2 + 9w - 360 = 0, the coefficients are:

  • a=1a = 1
  • b=9b = 9
  • c=360c = -360

Now substitute these values into the quadratic formula:

w=9±924(1)(360)2(1)w = \frac{-9 \pm \sqrt{9^2 - 4(1)(-360)}}{2(1)} w=9±81+14402w = \frac{-9 \pm \sqrt{81 + 1440}}{2} w=9±15212w = \frac{-9 \pm \sqrt{1521}}{2} w=9±392w = \frac{-9 \pm 39}{2}

This gives two possible solutions:

w=9+392=302=15w = \frac{-9 + 39}{2} = \frac{30}{2} = 15 or w=9392=482=24w = \frac{-9 - 39}{2} = \frac{-48}{2} = -24

Since the width cannot be negative, we take the positive solution:

w=15yardsw = 15 \, \text{yards}

Thus, the width of the parking lot is 15 yards.

Finding the length:

The length is 9 yards more than the width, so:

Length=15+9=24yards\text{Length} = 15 + 9 = 24 \, \text{yards}

Final answer:

  • Width: 15 yards
  • Length: 24 yards

Would you like more details or have any questions?

Related Questions:

  1. How do you verify the area given the dimensions?
  2. What is the perimeter of the parking lot based on the dimensions?
  3. Can the quadratic formula be used in all such area-related problems?
  4. How would the solution change if the area was given in square meters instead of yards?
  5. What would happen if the width was greater than the length in this scenario?

Tip:

Always check the discriminant (the part inside the square root of the quadratic formula) to ensure it gives a real number solution before solving.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry

Formulas

Area of a rectangle = length × width
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic Formula
Area Formula for Rectangles

Suitable Grade Level

Grades 8-10