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

Using the Quadratic Word Problems knowledge point
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 a rectangle is given by length multiplied by width:

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

Formulate the quadratic equation

Using the Solving Quadratic Equations knowledge point

$$
LATEXBLOCK0
$$

Solve the quadratic equation

Using the Solving Quadratic Equations knowledge point

$$
LATEXBLOCK1
$$

Since width must be positive:

$$
LATEXBLOCK2
$$

Answer:

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