Question

Solve the problem: 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.

Answers

GPT-4o mini
To solve the problem, we start by letting the length of the rectangle be \( l \) and the width be \( w \).

From the problem, we know two things:

1. The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2l + 2w
\]
In this case, the perimeter is 32 cm, so:
\[
2l + 2w = 32
\]
Dividing the entire equation by 2, we get:
\[
l + w = 16 \quad \text{(1)}
\]

2. The width \( w \) is described as being 2 cm longer than half the length:
\[
w = \frac{l}{2} + 2 \quad \text{(2)}
\]

Now, we can substitute equation (2) into equation (1):

Substituting \( w \) in (1):
\[
l + \left(\frac{l}{2} + 2\right) = 16
\]

Now simplify this:
\[
l + \frac{l}{2} + 2 = 16
\]
Combining the terms:
\[
\frac{2l}{2} + \frac{l}{2} + 2 = 16
\]
\[
\frac{3l}{2} + 2 = 16
\]
Subtracting 2 from both sides:
\[
\frac{3l}{2} = 14
\]
Multiplying both sides by \(\frac{2}{3}\):
\[
l = \frac{2 \times 14}{3} = \frac{28}{3} \approx 9.33 \text{ cm}
\]

Now, we can find the width \( w \) using equation (2):
\[
w = \frac{l}{2} + 2 = \frac{\frac{28}{3}}{2} + 2
\]
\[
w = \frac{28}{6} + 2 = \frac{28}{6} + \frac{12}{6} = \frac{40}{6} = \frac{20}{3} \approx 6.67 \text{ cm}
\]

Having determined the length and width, we can now find the area \( A \) of the rectangle:
\[
A = l \times w = \left(\frac{28}{3}\right) \times \left(\frac{20}{3}\right) = \frac{28 \times 20}{9} = \frac{560}{9} \text{ cm}^2 \approx 62.22 \text{ cm}^2
\]

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