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)

Question

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

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 2

pick one

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: \[ A = \frac{560}{9} , \text{cm}^2 \]