Math worksheet with word problems on calculating area and perimeter of rectangles.

Let's solve each problem step by step.

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Problem 1:

A piece of wood was cut such that its length becomes 16 ft and width 12 ft. Find the area and perimeter of the wood.

#### Solution:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
\[
\text{Area} = 16 \, \text{ft} \times 12 \, \text{ft} = 192 \, \text{ft}^2
\]

- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)
\[
\text{Perimeter} = 2 \times (16 \, \text{ft} + 12 \, \text{ft}) = 2 \times 28 \, \text{ft} = 56 \, \text{ft}
\]

#### Answer:
\[
\boxed{192 \, \text{ft}^2, 56 \, \text{ft}}
\]

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Problem 2:

A rectangular swimming pool is 30 cm wide and 40 cm in length. Find the total area and perimeter of the swimming pool.

#### Solution:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
\[
\text{Area} = 40 \, \text{cm} \times 30 \, \text{cm} = 1200 \, \text{cm}^2
\]

- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)
\[
\text{Perimeter} = 2 \times (40 \, \text{cm} + 30 \, \text{cm}) = 2 \times 70 \, \text{cm} = 140 \, \text{cm}
\]

#### Answer:
\[
\boxed{1200 \, \text{cm}^2, 140 \, \text{cm}}
\]

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Problem 3:

A rug had a length of 8 cm and a width of 6 cm. What is the area and perimeter of the rug?

#### Solution:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
\[
\text{Area} = 8 \, \text{cm} \times 6 \, \text{cm} = 48 \, \text{cm}^2
\]

- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)
\[
\text{Perimeter} = 2 \times (8 \, \text{cm} + 6 \, \text{cm}) = 2 \times 14 \, \text{cm} = 28 \, \text{cm}
\]

#### Answer:
\[
\boxed{48 \, \text{cm}^2, 28 \, \text{cm}}
\]

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Problem 4:

Smith is a farmer. His land is 250 m wide and 500 m long. He wants to install a fence all around his land. What is the perimeter of the fence that he needs to construct?

#### Solution:
- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)
\[
\text{Perimeter} = 2 \times (500 \, \text{m} + 250 \, \text{m}) = 2 \times 750 \, \text{m} = 1500 \, \text{m}
\]

#### Answer:
\[
\boxed{1500 \, \text{m}}
\]

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Problem 5:

The area of an aluminium sheet is 150 m². If the length of the sheet is 25 m. Find the width and perimeter of the sheet.

#### Solution:
1. Find the width:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
\[
150 \, \text{m}^2 = 25 \, \text{m} \times \text{width}
\]
\[
\text{width} = \frac{150 \, \text{m}^2}{25 \, \text{m}} = 6 \, \text{m}
\]

2. Find the perimeter:
- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)
\[
\text{Perimeter} = 2 \times (25 \, \text{m} + 6 \, \text{m}) = 2 \times 31 \, \text{m} = 62 \, \text{m}
\]

#### Answer:
\[
\boxed{6 \, \text{m}, 62 \, \text{m}}
\]

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Problem 6:

The area of an apple orchard is 5200 ft². Find the length of the rectangular orchard, if its width is 75 ft.

#### Solution:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
\[
5200 \, \text{ft}^2 = \text{length} \times 75 \, \text{ft}
\]
\[
\text{length} = \frac{5200 \, \text{ft}^2}{75 \, \text{ft}} = 69.33 \, \text{ft}
\]

#### Answer:
\[
\boxed{69.33 \, \text{ft}}
\]

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Final Answers:

1. \(\boxed{192 \, \text{ft}^2, 56 \, \text{ft}}\)
2. \(\boxed{1200 \, \text{cm}^2, 140 \, \text{cm}}\)
3. \(\boxed{48 \, \text{cm}^2, 28 \, \text{cm}}\)
4. \(\boxed{1500 \, \text{m}}\)
5. \(\boxed{6 \, \text{m}, 62 \, \text{m}}\)
6. \(\boxed{69.33 \, \text{ft}}\)