Let's break down the given problem step by step to find the length of the rectangle.

We are told:

• The length of the rectangle is 5 cm greater than twice the width.
• The area of the rectangle is 33 cm².

Let:

• $w$ be the width of the rectangle.
• $l$ be the length of the rectangle.

From the first condition, we can write the length in terms of the width: $l = 2w + 5$ The area of a rectangle is given by: $\text{Area} = l \times w$ Substitute the value of $l$ from the first equation: $33 = (2w + 5) \times w$ This simplifies to: $33 = 2w^2 + 5w$ Rearrange this into a standard quadratic form: $2w^2 + 5w - 33 = 0$

Now, we solve this quadratic equation using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = 5$, and $c = -33$.

Substitute these values into the formula: $w = \frac{-5 \pm \sqrt{5^2 - 4(2)(-33)}}{2(2)}$ $w = \frac{-5 \pm \sqrt{25 + 264}}{4}$ $w = \frac{-5 \pm \sqrt{289}}{4}$ $w = \frac{-5 \pm 17}{4}$

So, we get two solutions: $w = \frac{-5 + 17}{4} = \frac{12}{4} = 3 \quad \text{or} \quad w = \frac{-5 - 17}{4} = \frac{-22}{4} = -5.5$

Since the width cannot be negative, we have $w = 3$ cm.

Now, substitute $w = 3$ back into the equation for the length: $l = 2(3) + 5 = 6 + 5 = 11 \, \text{cm}$

Thus, the length of the rectangle is 11 cm.

The correct answer is B) 11 cm.

Would you like more details, or do you have any questions?

Here are 5 related questions to consider:

1. How would the solution change if the area were different?
2. What happens if the relationship between length and width changes?
3. Can you solve for the width if only the length and area are given?
4. What are the implications of a quadratic equation having two solutions in this context?
5. How can the quadratic formula help solve other geometry problems?

Tip: When solving quadratic equations, always check both potential solutions to see if they make sense within the context of the problem.