To determine the dimensions of the rectangular parking lot, we need to find the values of [tex]\( x \)[/tex] that satisfy the given conditions.
Given:
- Length of the parking lot, [tex]\( L = x + 7 \)[/tex]
- Width of the parking lot, [tex]\( W = x + 10 \)[/tex]
- Area of the parking lot, [tex]\( A = 180 \, \text{m}^2 \)[/tex]
The area of a rectangle is calculated using the formula:
[tex]\[ A = \text{Length} \times \text{Width} \][/tex]
Substituting the given expressions for the length and width into the area formula, we get:
[tex]\[ 180 = (x + 7)(x + 10) \][/tex]
Now, we need to solve the equation [tex]\( (x + 7)(x + 10) = 180 \)[/tex].
We start by expanding the left-hand side:
[tex]\[ (x + 7)(x + 10) = x^2 + 10x + 7x + 70 \][/tex]
[tex]\[ x^2 + 17x + 70 = 180 \][/tex]
Next, we rearrange the equation to standard quadratic form:
[tex]\[ x^2 + 17x + 70 - 180 = 0 \][/tex]
[tex]\[ x^2 + 17x - 110 = 0 \][/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 17 \)[/tex], and [tex]\( c = -110 \)[/tex].
Substituting these values into the quadratic formula:
[tex]\[ x = \frac{-17 \pm \sqrt{17^2 - 4(1)(-110)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-17 \pm \sqrt{289 + 440}}{2} \][/tex]
[tex]\[ x = \frac{-17 \pm \sqrt{729}}{2} \][/tex]
[tex]\[ x = \frac{-17 \pm 27}{2} \][/tex]
Solving for the two values of [tex]\( x \)[/tex]:
1. [tex]\( x = \frac{-17 + 27}{2} = \frac{10}{2} = 5 \)[/tex]
2. [tex]\( x = \frac{-17 - 27}{2} = \frac{-44}{2} = -22 \)[/tex]
So, the solutions for [tex]\( x \)[/tex] are [tex]\( x = 5 \)[/tex] and [tex]\( x = -22 \)[/tex].
Now, we find the dimensions for each value of [tex]\( x \)[/tex]:
For [tex]\( x = 5 \)[/tex]:
- Length: [tex]\( L = 5 + 7 = 12 \, \text{m} \)[/tex]
- Width: [tex]\( W = 5 + 10 = 15 \, \text{m} \)[/tex]
For [tex]\( x = -22 \)[/tex]:
- Length: [tex]\( L = -22 + 7 = -15 \, \text{m} \)[/tex]
- Width: [tex]\( W = -22 + 10 = -12 \, \text{m} \)[/tex]
Thus, the possible dimensions of the parking lot are:
1. [tex]\( 12 \, \text{m} \times 15 \, \text{m} \)[/tex]
2. [tex]\( -15 \, \text{m} \times -12 \, \text{m} \)[/tex]
In a practical context, negative dimensions do not make sense. Hence, the valid dimensions of the parking lot are:
[tex]\[ 12 \, \text{m} \text{ in length and } 15 \, \text{m} \text{ in width.} \][/tex]
