Question

The rectangle shown has a perimeter of 80 cm and an area of 204cm squared. Its length is 4 more than five times its width. Write and solve a system of equations to find the dimensions of the rectangle.

Answer 1

To find the dimensions of the rectangle, let's define the variables:

- Let \( w \) be the width of the rectangle.
- Let \( l \) be the length of the rectangle.

We are given three pieces of information about the rectangle:

1. The perimeter \( P \) of the rectangle is 80 cm.
2. The area \( A \) of the rectangle is 204 cm².
3. The length is 4 more than five times its width.

From these statements, we can form the following equations:

1. The perimeter of a rectangle is given by:
\[
P = 2l + 2w = 80
\]
We can simplify this to:
\[
l + w = 40 \quad \text{(Equation 1)}
\]

2. The area of a rectangle is given by:
\[
A = l \cdot w = 204
\]
This gives us:
\[
lw = 204 \quad \text{(Equation 2)}
\]

3. The length is 4 more than five times the width:
\[
l = 5w + 4 \quad \text{(Equation 3)}
\]

Now, we will substitute Equation 3 into Equations 1 and 2.

**Substituting Equation 3 into Equation 1:**
\[
(5w + 4) + w = 40
\]
Combine like terms:
\[
6w + 4 = 40
\]
Subtract 4 from both sides:
\[
6w = 36
\]
Now, divide by 6:
\[
w = 6
\]

**Using the value of \( w \) to find \( l \):**
Substitute \( w = 6 \) into Equation 3:
\[
l = 5(6) + 4
\]
Calculating this gives:
\[
l = 30 + 4 = 34
\]

Now we have:
- Width \( w = 6 \) cm
- Length \( l = 34 \) cm

**Verification:**
1. **Perimeter:**
\[
P = 2l + 2w = 2(34) + 2(6) = 68 + 12 = 80 \, \text{cm} \quad \text{(correct)}
\]

2. **Area:**
\[
A = l \cdot w = 34 \cdot 6 = 204 \, \text{cm}^2 \quad \text{(correct)}
\]

Thus, the dimensions of the rectangle are:
- Width = 6 cm
- Length = 34 cm