Question

The length of a rectangle is four meters less than twice its width. If the area of the rectangle is 92 meters squared, what is the length and the width?

Answer 1

Let the width of the rectangle be \( w \) meters. According to the problem, the length \( l \) is given by:

\[
l = 2w - 4
\]

The area \( A \) of the rectangle can be expressed as:

\[
A = l \times w
\]

We know the area is 92 square meters:

\[
l \times w = 92
\]

Substituting the expression for \( l \):

\[
(2w - 4) \times w = 92
\]

Expanding this equation:

\[
2w^2 - 4w = 92
\]

Rearranging the equation gives:

\[
2w^2 - 4w - 92 = 0
\]

To simplify, we can divide the entire equation by 2:

\[
w^2 - 2w - 46 = 0
\]

Now, we will use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -2 \), and \( c = -46 \):

Calculating the discriminant:

\[
b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (-46) = 4 + 184 = 188
\]

Now applying the quadratic formula:

\[
w = \frac{-(-2) \pm \sqrt{188}}{2 \cdot 1} = \frac{2 \pm \sqrt{188}}{2}
\]

Simplifying \( \sqrt{188} \):

\[
\sqrt{188} = \sqrt{4 \cdot 47} = 2\sqrt{47}
\]

Substituting back:

\[
w = \frac{2 \pm 2\sqrt{47}}{2} = 1 \pm \sqrt{47}
\]

Since width cannot be negative, we take the positive root:

\[
w = 1 + \sqrt{47}
\]

Now we calculate the approximate value of \( \sqrt{47} \):

\[
\sqrt{47} \approx 6.86
\]
Thus,

\[
w \approx 1 + 6.86 \approx 7.86 \text{ meters}
\]

Next, we find the length \( l \):

\[
l = 2w - 4 = 2(1 + \sqrt{47}) - 4 = 2 + 2\sqrt{47} - 4 = -2 + 2\sqrt{47}
\]

Calculating the numeric value:

\[
l \approx -2 + 2 \cdot 6.86 \approx -2 + 13.72 \approx 11.72 \text{ meters}
\]

Thus, the dimensions of the rectangle are approximately:

\[
\text{Width } w \approx 7.86 \text{ meters}, \quad \text{Length } l \approx 11.72 \text{ meters}
\]

Final values (rounded reasonably):

\[
\boxed{l = 11.72 \text{ meters}, \, w = 7.86 \text{ meters}}
\]