Perimeter of a Rectangle Sheet 4A - find the perimeters of six rectangles with given dimensions.

To solve the problem of finding the perimeters of the rectangles, we will use the formula for the perimeter of a rectangle:

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]

Let's solve each rectangle step by step.

---

1)

- Length: \(6 \, \text{cm}\)
- Width: \(2 \frac{1}{2} \, \text{cm} = 2.5 \, \text{cm}\)

\[
\text{Perimeter} = 2 \times (6 + 2.5) = 2 \times 8.5 = 17 \, \text{cm}
\]

Answer: \(17 \, \text{cm}\)

---

2)

- Length: \(3 \frac{1}{2} \, \text{m} = 3.5 \, \text{m}\)
- Width: \(3 \frac{1}{2} \, \text{m} = 3.5 \, \text{m}\)

\[
\text{Perimeter} = 2 \times (3.5 + 3.5) = 2 \times 7 = 14 \, \text{m}
\]

Answer: \(14 \, \text{m}\)

---

3)

- Length: \(80 \, \text{cm}\)
- Width: \(50 \, \text{cm}\)

\[
\text{Perimeter} = 2 \times (80 + 50) = 2 \times 130 = 260 \, \text{cm}
\]

Answer: \(260 \, \text{cm}\)

---

4)

- Length: \(4 \, \text{m}\)
- Width: \(2 \frac{1}{4} \, \text{m} = 2.25 \, \text{m}\)

\[
\text{Perimeter} = 2 \times (4 + 2.25) = 2 \times 6.25 = 12.5 \, \text{m}
\]

Answer: \(12.5 \, \text{m}\)

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5)

- Length: \(6 \frac{1}{2} \, \text{km} = 6.5 \, \text{km}\)
- Width: \(\frac{1}{2} \, \text{km} = 0.5 \, \text{km}\)

\[
\text{Perimeter} = 2 \times (6.5 + 0.5) = 2 \times 7 = 14 \, \text{km}
\]

Answer: \(14 \, \text{km}\)

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6)

- Length: \(6 \frac{1}{4} \, \text{m} = 6.25 \, \text{m}\)
- Width: \(4 \frac{1}{4} \, \text{m} = 4.25 \, \text{m}\)

\[
\text{Perimeter} = 2 \times (6.25 + 4.25) = 2 \times 10.5 = 21 \, \text{m}
\]

Answer: \(21 \, \text{m}\)

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Final Answers:

1. \(17 \, \text{cm}\)
2. \(14 \, \text{m}\)
3. \(260 \, \text{cm}\)
4. \(12.5 \, \text{m}\)
5. \(14 \, \text{km}\)
6. \(21 \, \text{m}\)

\[
\boxed{17 \, \text{cm}, 14 \, \text{m}, 260 \, \text{cm}, 12.5 \, \text{m}, 14 \, \text{km}, 21 \, \text{m}}
\]