Questions: A rectangular room is 2 times as long as it is wide, and its perimeter is 48 meters. Find the dimension of the room. The length is : meters and the width is meters.

Solution Steps

To find the dimensions of the room, we can use the relationship between the length and width given in the problem. Let the width be \( w \) meters. Then the length is \( 2w \) meters. The perimeter of a rectangle is given by the formula \( 2 \times (\text{length} + \text{width}) \). We can set up the equation \( 2 \times (2w + w) = 48 \) and solve for \( w \).

Let the width of the room be \( w \) meters. According to the problem, the length \( l \) is given by the relationship \( l = 2w \).

The perimeter \( P \) of a rectangle is calculated using the formula: \[ P = 2(l + w) \] Substituting the expression for length, we have: \[ P = 2(2w + w) = 2(3w) = 6w \] Given that the perimeter is 48 meters, we can set up the equation: \[ 6w = 48 \]

To find \( w \), we divide both sides of the equation by 6: \[ w = \frac{48}{6} = 8 \text{ meters} \]

Now, substituting \( w \) back into the equation for length: \[ l = 2w = 2 \times 8 = 16 \text{ meters} \]

Final Answer