The perimeter of a rectangle is 38m. If the area of the rectangle is 84 square meters, find its dimensions.

Solve this and show the solution, after that explain step by step in the most simplest way.

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.

The perimeter of a rectangle is 38m. If the area of the rectangle is 84 square meters, find its dimensions.

Solve for the dimensions of a rectangle with a perimeter of 38 meters and an area of 84 square meters. This solution explores key algebraic concepts, including the use of the quadratic equation to find the length and width of the rectangle. Suitable for students in Grades 7-9.

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