Question

The length of a rectangle is four times the width. The perimeter of the rectangle is 45 inches. Write a system of equations that represents this problem. What is the area of the rectangle?

Answer 1

Let's denote the width of the rectangle as \( w \) and the length as \( l \).

According to the problem, we have two key relationships:

1. The length is four times the width:
\[
l = 4w
\]

2. The perimeter of the rectangle is given as 45 inches. The formula for the perimeter \( P \) of a rectangle is:
\[
P = 2l + 2w
\]
Substituting the given perimeter:
\[
2l + 2w = 45
\]

Now we can rewrite the second equation using the first equation. Substitute \( l = 4w \) into the perimeter equation:
\[
2(4w) + 2w = 45
\]
This simplifies to:
\[
8w + 2w = 45
\]
\[
10w = 45
\]
Dividing both sides by 10, we find:
\[
w = 4.5
\]

Now, substituting \( w \) back into the equation for \( l \):
\[
l = 4w = 4(4.5) = 18
\]

Now we can find the area \( A \) of the rectangle, which is given by:
\[
A = l \times w
\]
Substituting the values we found:
\[
A = 18 \times 4.5
\]
Calculating this gives:
\[
A = 81
\]

Thus, the area of the rectangle is \( 81 \) square inches.

In summary, the system of equations is:
\[
\begin{align*}
1. & \quad l = 4w \\
2. & \quad 2l + 2w = 45
\end{align*}
\]

And the area of the rectangle is:
\[
\text{Area} = 81 \text{ square inches}
\]