A rectangle has length 3x and the width is 4y times more than the length. find the parameter and the area

To solve this problem, we first need to understand the dimensions of the rectangle based on the given information:

1. **Length** of the rectangle is given as \( 3x \).
2. **Width** of the rectangle is said to be \( 4y \) times more than the length.

   So, the width can be expressed as:
   \[
   \text{Width} = \text{Length} + 4y \times \text{Length}
   \]

   Substituting the length \( 3x \) into the equation:
   \[
   \text{Width} = 3x + 4y \cdot 3x
   \]
   \[
   \text{Width} = 3x + 12xy
   \]

Now, let's proceed to find the perimeter and area of the rectangle.

### 1. Perimeter of the Rectangle

The formula for the perimeter \( P \) of a rectangle is:
\[
P = 2 \times (\text{Length} + \text{Width})
\]

Substituting the values we have:
\[
P = 2 \times (3x + (3x + 12xy))
\]
\[
P = 2 \times (3x + 3x + 12xy)
\]
\[
P = 2 \times (6x + 12xy)
\]
\[
P = 12x + 24xy
\]

### 2. Area of the Rectangle

The formula for the area \( A \) of a rectangle is:
\[
A = \text{Length} \times \text{Width}
\]

Substituting the values we have:
\[
A = 3x \times (3x + 12xy)
\]
\[
A = 9x^2 + 36x^2y
\]

Therefore, the **perimeter** of the rectangle is \( 12x + 24xy \), and the **area** of the rectangle is \( 9x^2 + 36x^2y \).

Would you like more details or have any questions about this solution?

Here are some related questions to explore further:

1. What happens to the area and perimeter if \( x \) and \( y \) are doubled?
2. If the width is reduced by half, what would be the new perimeter and area?
3. How does the area change if \( y \) is set to zero?
4. Can we express the width in terms of only \( x \) if \( y = 1 \)?
5. What are the dimensions of the rectangle if \( x = 2 \) and \( y = 3 \)?

**Tip:** When dealing with geometry problems, it is often helpful to draw a diagram and label the dimensions to visualize the relationships more clearly.

Learn how to find the perimeter and area of a rectangle with given dimensions using algebraic expressions. This problem involves understanding geometric relationships and applying formulas for perimeter and area. Suitable for students in Grades 7-9.

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