Question

How should you begin your answer? The perimeter of a rectangle is 32 cm. If the width of the rectangle is 2 cm longer than half the length, find the area. ( Write Equation and Solve)

a. let A = area of the rectangle

b. let P = the perimeter of the rectangle

c. let W = width of the rectangle

d. let L = length of the rectangle

Answer 1

pick one

Answer 2

To solve the problem, we can start by defining the variables as follows:

- Let \( L \) = length of the rectangle.
- Let \( W \) = width of the rectangle.
- The perimeter \( P \) of the rectangle is given by the formula:
\[
P = 2L + 2W
\]

Given that the perimeter \( P = 32 \) cm, we can write:
\[
2L + 2W = 32
\]

We also know from the problem that the width \( W \) is 2 cm longer than half the length:
\[
W = \frac{1}{2}L + 2
\]

Now we have two equations:
1. \( 2L + 2W = 32 \)
2. \( W = \frac{1}{2}L + 2 \)

Next, we can substitute the expression for \( W \) from equation 2 into equation 1:

\[
2L + 2\left(\frac{1}{2}L + 2\right) = 32
\]

Now, simplify the equation:

\[
2L + L + 4 = 32
\]

Combine like terms:

\[
3L + 4 = 32
\]

Subtract 4 from both sides:

\[
3L = 28
\]

Divide by 3 to find \( L \):

\[
L = \frac{28}{3} \approx 9.33 \text{ cm}
\]

Now substituting \( L \) back to find \( W \):
\[
W = \frac{1}{2}\left(\frac{28}{3}\right) + 2 = \frac{14}{3} + 2 = \frac{14}{3} + \frac{6}{3} = \frac{20}{3} \approx 6.67 \text{ cm}
\]

Now that we have both dimensions, we can find the area \( A \) of the rectangle using the formula:
\[
A = L \times W
\]

Substituting in the values:

\[
A = \left(\frac{28}{3}\right) \times \left(\frac{20}{3}\right) = \frac{560}{9} \text{ cm}^2 \approx 62.22 \text{ cm}^2
\]

Thus, the area of the rectangle is:
\[
A = \frac{560}{9} \text{ cm}^2
\]

Answer 3

Let's go with option **c**.

**c. Let \( W \) = width of the rectangle.**

Now let’s solve the problem step by step:

1. **Set up the equations:**
- The perimeter of the rectangle is given by:
\[
P = 2L + 2W = 32
\]

- The width is defined as:
\[
W = \frac{1}{2}L + 2
\]

2. **Substitute for \( W \) in the perimeter equation:**
\[
2L + 2\left(\frac{1}{2}L + 2\right) = 32
\]

3. **Simplify:**
\[
2L + L + 4 = 32
\]
\[
3L + 4 = 32
\]

4. **Solve for \( L \):**
\[
3L = 32 - 4
\]
\[
3L = 28
\]
\[
L = \frac{28}{3} \approx 9.33 \, \text{cm}
\]

5. **Find \( W \):**
\[
W = \frac{1}{2}\left(\frac{28}{3}\right) + 2 = \frac{14}{3} + \frac{6}{3} = \frac{20}{3} \approx 6.67 \, \text{cm}
\]

6. **Calculate area \( A \):**
\[
A = L \times W = \left(\frac{28}{3}\right) \times \left(\frac{20}{3}\right) = \frac{560}{9} \, \text{cm}^2 \approx 62.22 \, \text{cm}^2
\]

So, the area of the rectangle is:
\[
A = \frac{560}{9} \, \text{cm}^2
\]