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

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

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

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

Substituting the values we know:

$w \times (w + 9) = 360$

Now, expand the equation:

$w^2 + 9w = 360$

Rearrange this into a standard quadratic equation:

$w^2 + 9w - 360 = 0$

We can solve this quadratic equation using the quadratic formula:

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

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

• $a = 1$
• $b = 9$
• $c = -360$

Now substitute these values into the quadratic formula:

$w = \frac{-9 \pm \sqrt{9^2 - 4(1)(-360)}}{2(1)}$ $w = \frac{-9 \pm \sqrt{81 + 1440}}{2}$ $w = \frac{-9 \pm \sqrt{1521}}{2}$ $w = \frac{-9 \pm 39}{2}$

This gives two possible solutions:

$w = \frac{-9 + 39}{2} = \frac{30}{2} = 15$ or $w = \frac{-9 - 39}{2} = \frac{-48}{2} = -24$

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

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

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

Finding the length:

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

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

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.