We are given the following information about the rectangle:

• The perimeter of the rectangle is 38 meters.
• The area of the rectangle is 84 square meters.

Let the length of the rectangle be $l$ meters and the width be $w$ meters. We can set up two equations based on the perimeter and area:

Step 1: Write the equations

1. Perimeter formula: The perimeter of a rectangle is given by $P = 2l + 2w$ Since the perimeter is 38 meters, we have: $2l + 2w = 38$ Simplifying, we get: $l + w = 19 \quad \text{(Equation 1)}$

2. Area formula: The area of a rectangle is given by $A = l \times w$ Since the area is 84 square meters, we have: $l \times w = 84 \quad \text{(Equation 2)}$

Step 2: Solve for one variable

From Equation 1, solve for $l$ in terms of $w$: $l = 19 - w$

Step 3: Substitute into the area equation

Substitute $l = 19 - w$ into Equation 2: $(19 - w) \times w = 84$ Expanding the equation: $19w - w^2 = 84$ Rearrange the equation: $w^2 - 19w + 84 = 0$

Step 4: Solve the quadratic equation

Now, solve the quadratic equation $w^2 - 19w + 84 = 0$ using the quadratic formula: $w = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(1)(84)}}{2(1)}$ $w = \frac{19 \pm \sqrt{361 - 336}}{2}$ $w = \frac{19 \pm \sqrt{25}}{2}$ $w = \frac{19 \pm 5}{2}$ So, we have two solutions for $w$: $w = \frac{19 + 5}{2} = 12 \quad \text{or} \quad w = \frac{19 - 5}{2} = 7$

Step 5: Find the corresponding lengths

If $w = 12$, then from Equation 1: $l = 19 - 12 = 7$

If $w = 7$, then: $l = 19 - 7 = 12$

Final Answer:

The dimensions of the rectangle are $12$ meters by $7$ meters.

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Step-by-Step Explanation:

1. Understand the problem: We have a rectangle with known perimeter and area. We need to find its dimensions.
2. Use perimeter formula: The perimeter is the sum of all sides, and for a rectangle, it’s $2l + 2w = 38$. Simplifying, we get $l + w = 19$.
3. Use area formula: The area is the product of length and width, so $l \times w = 84$.
4. Substitute and solve: Using the perimeter equation, we express $l$ in terms of $w$, substitute into the area equation, and solve the quadratic equation.
5. Check both solutions: Solving gives two possible widths, and the corresponding lengths. Both solutions are valid since $l$ and $w$ are interchangeable.

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Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How do you derive the quadratic formula?
2. Can the area of a rectangle ever be negative?
3. What happens if both length and width are equal in a rectangle?
4. How would you find the dimensions if given only the perimeter?
5. What is the difference between area and perimeter?

Tip: Always double-check your solutions by substituting them back into the original equations to ensure they work.