area of a rectangle is \( 66 \mathrm{~m}^{2} \), and the length of the rectangle is 1 m more than double the width. Find the dimensions of the rectangle. Length : \( \square \mathrm{m} \) Width : \( \square \mathrm{m} \)

To find the dimensions of the rectangle, we can set up the equation based on the information provided. Let the width be \( w \) meters. According to the problem, the length \( l \) can be expressed as \( l = 2w + 1 \). The area \( A \) of a rectangle is given by the formula \( A = l \times w \). Given that the area is \( 66 \mathrm{~m}^{2} \), we can write: \[ 66 = (2w + 1) \times w \] Expanding this gives us: \[ 66 = 2w^2 + w \] Rearranging the equation, we get: \[ 2w^2 + w - 66 = 0 \] This is a quadratic equation that we can solve using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2, b = 1, c = -66 \). Calculating the discriminant: \[ b^2 - 4ac = 1^2 - 4 \times 2 \times (-66) = 1 + 528 = 529 \] Now, applying the quadratic formula: \[ w = \frac{-1 \pm \sqrt{529}}{4} \] Since \( \sqrt{529} = 23 \), we have: \[ w = \frac{-1 \pm 23}{4} \] Taking the positive root: \[ w = \frac{22}{4} = 5.5 \text{ m} \] Now substituting back to find the length: \[ l = 2(5.5) + 1 = 11 + 1 = 12 \text{ m} \] So the dimensions of the rectangle are: Length: \( 12 \mathrm{~m} \) Width: \( 5.5 \mathrm{~m} \)