Question

The rectangle shown has a perimeter of 142 cm and the given area. Its length is 5 more than five times its width. Write and solve a system of equations to find the dimensions of the rectangle.

Answer 1

Let the width of the rectangle be \( w \) cm. According to the problem, the length \( l \) is given by the equation:

\[
l = 5 + 5w
\]

The perimeter \( P \) of a rectangle is calculated as:

\[
P = 2l + 2w
\]

We know the perimeter is 142 cm, so we can write:

\[
2l + 2w = 142
\]

We can simplify this equation:

\[
l + w = 71
\]

Now we have a system of two equations:

1. \( l = 5 + 5w \)
2. \( l + w = 71 \)

Now, we will substitute the expression for \( l \) from the first equation into the second equation:

\[
(5 + 5w) + w = 71
\]

Combine like terms:

\[
5 + 5w + w = 71
\]
\[
5 + 6w = 71
\]

Now, subtract 5 from both sides:

\[
6w = 66
\]

Now, divide by 6:

\[
w = 11
\]

Now that we have the width, we can find the length \( l \):

\[
l = 5 + 5w = 5 + 5(11) = 5 + 55 = 60
\]

Thus, the dimensions of the rectangle are:
- Width: \( 11 \) cm
- Length: \( 60 \) cm

To verify, we can check the perimeter:

\[
P = 2l + 2w = 2(60) + 2(11) = 120 + 22 = 142 \text{ cm}
\]

Therefore, the dimensions of the rectangle are correct:
- **Width:** 11 cm
- **Length:** 60 cm