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.