Question

What are the dimensions of a rectangular room with a perimeter of 60 meters where the length is three times the width?

Answer 1

Let the width of the rectangular room be denoted as \( w \) meters. According to the problem, the length \( l \) is three times the width, so we can express the length as:

\[
l = 3w
\]

The formula for the perimeter \( P \) of a rectangle is given by:

\[
P = 2l + 2w
\]

We know from the problem that the perimeter is 60 meters, so we can set up the equation:

\[
2l + 2w = 60
\]

Substituting \( l = 3w \) into the perimeter equation, we get:

\[
2(3w) + 2w = 60
\]

This simplifies to:

\[
6w + 2w = 60
\]

Combining like terms gives:

\[
8w = 60
\]

To solve for \( w \), divide both sides by 8:

\[
w = \frac{60}{8} = 7.5 \text{ meters}
\]

Now that we have the width, we can find the length using \( l = 3w \):

\[
l = 3(7.5) = 22.5 \text{ meters}
\]

Thus, the dimensions of the room are:

\[
\text{Width} = 7.5 \text{ meters}, \quad \text{Length} = 22.5 \text{ meters}
\]