The length of a rectangle is four meters less than twice its width.

If the area of the rectangle is 96 m^2, what is the length and the width?

1 answer

Let the width of the rectangle be denoted as \( w \) meters. According to the problem, the length \( l \) can be expressed as:

\[ l = 2w - 4 \]

The area of the rectangle \( A \) is given by the formula:

\[ A = l \times w \]

Substituting the expression for \( l \) into the area formula, we get:

\[ 96 = (2w - 4) \times w \]

Expanding this equation, we have:

\[ 96 = 2w^2 - 4w \]

Rearranging the equation to set it to zero gives:

\[ 2w^2 - 4w - 96 = 0 \]

Dividing the entire equation by 2 simplifies it:

\[ w^2 - 2w - 48 = 0 \]

Next, we can factor the quadratic equation:

\[ (w - 8)(w + 6) = 0 \]

Setting each factor equal to zero gives us:

\[ w - 8 = 0 \quad \text{or} \quad w + 6 = 0 \]

Thus, we have:

\[ w = 8 \quad \text{or} \quad w = -6 \]

Since a width cannot be negative, we discard \( w = -6 \) and accept \( w = 8 \).

Now, substituting back to find the length:

\[ l = 2w - 4 = 2(8) - 4 = 16 - 4 = 12 \]

Therefore, the dimensions of the rectangle are:

  • Width: \( 8 \) meters
  • Length: \( 12 \) meters

In conclusion, the length and width of the rectangle are:

\[ \text{Length} = 12 \text{ m}, \quad \text{Width} = 8 \text{ m} \]