Let’s solve the problem step-by-step.

Problem:

A rectangle has:

• Perimeter = 23 cm
• Area = 33 cm²

We need to determine the dimensions of the rectangle.

Step 1: Write the formulas.

For a rectangle:

1. Perimeter: $P = 2(l + w)$
2. Area: $A = l \cdot w$

Here:

• $l$ is the length.
• $w$ is the width.

Step 2: Use the perimeter formula.

From $P = 2(l + w)$, we know: $2(l + w) = 23 \quad \Rightarrow \quad l + w = 11.5$ This gives: $w = 11.5 - l$

Step 3: Use the area formula.

From $A = l \cdot w$: $l \cdot w = 33$ Substitute $w = 11.5 - l$: $l \cdot (11.5 - l) = 33$ Simplify: $11.5l - l^2 = 33 \quad \Rightarrow \quad l^2 - 11.5l + 33 = 0$

Step 4: Solve the quadratic equation.

The equation is: $l^2 - 11.5l + 33 = 0$ Using the quadratic formula: $l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 1$, $b = -11.5$, $c = 33$

Substitute: $l = \frac{-(-11.5) \pm \sqrt{(-11.5)^2 - 4(1)(33)}}{2(1)}$ $l = \frac{11.5 \pm \sqrt{132.25 - 132}}{2}$ $l = \frac{11.5 \pm \sqrt{0.25}}{2}$ $l = \frac{11.5 \pm 0.5}{2}$

Solutions: $l = \frac{11.5 + 0.5}{2} = 6 \quad \text{or} \quad l = \frac{11.5 - 0.5}{2} = 5.5$

Step 5: Find $w$.

If $l = 6$: $w = 11.5 - 6 = 5.5$

If $l = 5.5$: $w = 11.5 - 5.5 = 6$

Thus, the dimensions are: $6 \, \text{cm} \, \text{by} \, 5.5 \, \text{cm}$

Would you like further clarification or details?
Here are 5 follow-up questions:

1. How does the quadratic formula help solve such problems?
2. Can you explain why the perimeter formula gives two possible dimensions?
3. How would this problem change if the area were different?
4. Why is it necessary to verify both solutions in the context of geometry?
5. Can you show how to solve this problem using trial and error instead of algebra?

Tip: Always double-check your quadratic equation for potential arithmetic errors!