Questions: The length of a rectangle is 7 ft more than double the width, and the area of the rectangle is 99 ft^2. Find the dimensions of the rectangle. Length: ft Width: ft

The length of a rectangle is 7 ft more than double the width, and the area of the rectangle is 99 ft^2. Find the dimensions of the rectangle.

Length: ft

Width: ft

Transcript text: The length of a rectangle is 7 ft more than double the width, and the area of the rectangle is $99 \mathrm{ft}^{2}$. Find the dimensions of the rectangle. Length : $\square$ ft Width : $\square$ ft

Solution

Solution Steps

Step 1: Define the Variables

Let the width of the rectangle be denoted as $ w $ (in feet). According to the problem, the length $ l $ can be expressed in terms of the width as: \[ l = 2w + 7 \]

Step 2: Set Up the Area Equation

The area $ A $ of the rectangle is given by the product of its length and width. Therefore, we can write the equation for the area as: \[ A = l \cdot w = 99 \] Substituting the expression for length, we have: \[ (2w + 7) \cdot w = 99 \]

Step 3: Solve the Equation

Expanding the equation gives: \[ 2w^2 + 7w - 99 = 0 \] Using the quadratic formula or factoring, we find the solutions for $ w $: \[ w = \frac{11}{2} \quad \text{or} \quad w = -9 \] Since width cannot be negative, we take the positive solution: \[ w = \frac{11}{2} \text{ ft} \]

Step 4: Calculate the Length

Now, substituting the value of $ w $ back into the expression for length: \[ l = 2 \left(\frac{11}{2}\right) + 7 = 18 \text{ ft} \]