A rectangular room is 2 times as long as it is wide, and its perimeter is 48 meters. Find the dimensions of the room.

To find the dimensions of the room, let's start by defining the variables for the width and length of the room.

Let:
- \( w \) be the width of the room (in meters).
- \( l \) be the length of the room (in meters).

We are told that the room is 2 times as long as it is wide, so we can express the length in terms of the width:
\[ l = 2w \]

The perimeter of a rectangle is given by the formula:
\[ P = 2l + 2w \]

We are given that the perimeter is 48 meters, so:
\[ 2l + 2w = 48 \]

Substituting the expression for \( l \) into the perimeter equation:
\[ 2(2w) + 2w = 48 \]

Simplify and solve for \( w \):
\[ 4w + 2w = 48 \]
\[ 6w = 48 \]
\[ w = \frac{48}{6} = 8 \text{ meters} \]

Now that we have the width, we can find the length:
\[ l = 2w = 2(8) = 16 \text{ meters} \]

Thus, the dimensions of the room are:
- Width: \( 8 \) meters
- Length: \( 16 \) meters

Would you like further details on this solution or have any other questions?

Here are some related questions to consider:
1. What would the dimensions be if the perimeter were increased to 60 meters?
2. How would the problem change if the room were 3 times as long as it is wide?
3. What is the area of the room with the given dimensions?
4. If a border of 1 meter wide was added inside the room, what would be the new dimensions?
5. How would the solution change if the room were square with the same perimeter?

**Tip:** When solving problems involving the perimeter of a rectangle, it's often helpful to express one dimension in terms of the other to simplify the equation.

Learn how to find the dimensions of a rectangle when given its perimeter as 48 meters. Step-by-step solution provided, covering geometry concepts and linear equations. Suitable for students in Grades 7-9.

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