The width of a rectangle is 9 less than twice its length. If the area of the rectangle is \( 174 \mathrm{~cm}^{2} \), what is the length of the diagonal? The length of the diagonal is Give your answer to 2 decimal places.

Let's start by defining the length of the rectangle as \( l \). According to the problem, the width \( w \) can be expressed as \( w = 2l - 9 \). The area of the rectangle is given by the equation: \[ \text{Area} = l \times w = 174 \text{ cm}^2. \] Substituting the width into the area equation gives us: \[ l(2l - 9) = 174. \] This simplifies to: \[ 2l^2 - 9l - 174 = 0. \] Now, we can use the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -9 \), and \( c = -174 \): \[ l = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(-174)}}{2(2)} = \frac{9 \pm \sqrt{81 + 1392}}{4} = \frac{9 \pm \sqrt{1473}}{4}. \] Calculating \( \sqrt{1473} \) gives approximately \( 38.39 \). So: \[ l = \frac{9 \pm 38.39}{4}. \] Choosing the positive root for length: \[ l = \frac{47.39}{4} \approx 11.85 \text{ cm}. \] Now calculating \( w \): \[ w = 2(11.85) - 9 \approx 14.7 - 9 = 5.7 \text{ cm}. \] To find the diagonal \( d \) of the rectangle, we use the Pythagorean theorem: \[ d = \sqrt{l^2 + w^2} = \sqrt{(11.85)^2 + (5.7)^2} \approx \sqrt{140.6225 + 32.49} \approx \sqrt{173.1125} \approx 13.15 \text{ cm}. \] Thus, the length of the diagonal is approximately \( \boxed{13.15} \) cm.