Questions: The area of a rectangle is 99 ft², and the length of the rectangle is 7 ft more than twice the width. Find the dimensions of the rectangle.

Solution Steps

To solve this problem, we need to set up a quadratic equation based on the given conditions. Let's denote the width of the rectangle as \( w \). According to the problem, the length \( l \) is \( 7 \) feet more than twice the width, so \( l = 2w + 7 \). The area of the rectangle is given as \( 99 \) square feet, so we can set up the equation \( w \times (2w + 7) = 99 \). This simplifies to a quadratic equation \( 2w^2 + 7w - 99 = 0 \). We can solve this quadratic equation to find the width \( w \), and then use it to find the length \( l \).

Let the width of the rectangle be \( w \). The length \( l \) is given by the expression \( l = 2w + 7 \). The area of the rectangle is given as \( 99 \, \text{ft}^2 \). Therefore, we can set up the equation: \[ w \cdot (2w + 7) = 99 \] This simplifies to the quadratic equation: \[ 2w^2 + 7w - 99 = 0 \]

Using the quadratic formula, we find the roots of the equation \( 2w^2 + 7w - 99 = 0 \). The solutions are: \[ w = -9 \quad \text{and} \quad w = \frac{11}{2} \] Since width cannot be negative, we take the positive solution: \[ w = 5.5 \, \text{ft} \]

Now, we can find the length using the width: \[ l = 2w + 7 = 2 \cdot 5.5 + 7 = 18 \, \text{ft} \]

Final Answer