Question

question 18 (3 points)

the length of a rectangle is 1 cm longer than twice its width. what are the dimensions of the rectangle if the area is 351 \text{ cm}^2?

Explanation:

Define variables and set up the equation

Let \(w\) represent the width of the rectangle in centimeters.
The length \(l\) is \(1\text{ cm}\) longer than twice its width:

$$l = 2w + 1$$

The area of the rectangle is \(351\text{ cm}^2\):

$$w(2w + 1) = 351$$

$$2w^2 + w - 351 = 0$$

Solve the quadratic equation for width

Using the quadratic formula where \(a = 2\), \(b = 1\), and \(c = -351\):

$$w = \frac{-1 \pm \sqrt{1^2 - 4(2)(-351)}}{2(2)}$$

$$w = \frac{-1 \pm \sqrt{1 + 2808}}{4}$$

$$w = \frac{-1 \pm \sqrt{2809}}{4}$$

$$w = \frac{-1 \pm 53}{4}$$

Since width must be positive:

$$w = \frac{52}{4} = 13$$

Calculate the length

Substitute \(w = 13\) back into the expression for length:

$$l = 2(13) + 1 = 27$$

Answer:

Width is \(13\text{ cm}\) and length is \(27\text{ cm}\).