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A rectangular parking lot has a length that is 7 meters greater than the width. The area of the parking lot is 120 square meters.

Find the length and the width.

Answer :

To solve this problem, we need to find the dimensions of a rectangular parking lot where the area is known, and the length is described in terms of the width.

Let's break down the problem:

1. Define Variables:
- Let [tex]\( w \)[/tex] be the width of the parking lot.
- Since the length is 7 meters greater than the width, the length can be expressed as [tex]\( w + 7 \)[/tex].

2. Set Up the Equation for the Area:
- The area [tex]\( A \)[/tex] of a rectangle is given by the formula:
[tex]\[
A = \text{length} \times \text{width}
\][/tex]
- We are given that the area is 120 square meters. Therefore, we write the equation:
[tex]\[
w \times (w + 7) = 120
\][/tex]

3. Solve the Quadratic Equation:
- Expand and rearrange the equation:
[tex]\[
w^2 + 7w - 120 = 0
\][/tex]
- This is a quadratic equation, and we can solve it using the quadratic formula:
[tex]\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 7 \)[/tex], and [tex]\( c = -120 \)[/tex].

4. Find the Solutions:
- Calculate the discriminant:
[tex]\[
b^2 - 4ac = 7^2 - 4 \times 1 \times (-120) = 49 + 480 = 529
\][/tex]
- Calculate the solutions for [tex]\( w \)[/tex]:
[tex]\[
w = \frac{-7 \pm \sqrt{529}}{2}
\][/tex]
[tex]\[
w = \frac{-7 \pm 23}{2}
\][/tex]

- This gives two potential solutions for [tex]\( w \)[/tex]:
[tex]\[
w = \frac{16}{2} = 8 \quad \text{or} \quad w = \frac{-30}{2} = -15
\][/tex]

5. Select the Appropriate Solution:
- Since a width cannot be negative, we take the positive solution [tex]\( w = 8 \)[/tex].

6. Calculate the Length:
- Substitute [tex]\( w = 8 \)[/tex] back into the expression for the length:
[tex]\[
\text{Length} = w + 7 = 8 + 7 = 15
\][/tex]

So, the width of the parking lot is 8 meters, and the length is 15 meters.

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