Area and Perimeter of Rectangles Worksheets - Math Monks - Free Printable

Problem Set: Area and Perimeter of Rectangle - Word Problems

We will solve each problem step by step, using the formulas for the area and perimeter of a rectangle:
- Area of a rectangle: \( \text{Area} = \text{length} \times \text{width} \)
- Perimeter of a rectangle: \( \text{Perimeter} = 2 \times (\text{length} + \text{width}) \)

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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:
- Length (\( l \)) = 16 ft
- Width (\( w \)) = 12 ft

1. Calculate the Area:
\[
\text{Area} = l \times w = 16 \, \text{ft} \times 12 \, \text{ft} = 192 \, \text{ft}^2
\]

2. Calculate the Perimeter:
\[
\text{Perimeter} = 2 \times (l + w) = 2 \times (16 \, \text{ft} + 12 \, \text{ft}) = 2 \times 28 \, \text{ft} = 56 \, \text{ft}
\]

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:
- Length (\( l \)) = 40 cm
- Width (\( w \)) = 30 cm

1. Calculate the Area:
\[
\text{Area} = l \times w = 40 \, \text{cm} \times 30 \, \text{cm} = 1200 \, \text{cm}^2
\]

2. Calculate the Perimeter:
\[
\text{Perimeter} = 2 \times (l + w) = 2 \times (40 \, \text{cm} + 30 \, \text{cm}) = 2 \times 70 \, \text{cm} = 140 \, \text{cm}
\]

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

---

#### 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:
- Length (\( l \)) = 8 cm
- Width (\( w \)) = 6 cm

1. Calculate the Area:
\[
\text{Area} = l \times w = 8 \, \text{cm} \times 6 \, \text{cm} = 48 \, \text{cm}^2
\]

2. Calculate the Perimeter:
\[
\text{Perimeter} = 2 \times (l + w) = 2 \times (8 \, \text{cm} + 6 \, \text{cm}) = 2 \times 14 \, \text{cm} = 28 \, \text{cm}
\]

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:
- Length (\( l \)) = 500 m
- Width (\( w \)) = 250 m

1. Calculate the Perimeter:
\[
\text{Perimeter} = 2 \times (l + w) = 2 \times (500 \, \text{m} + 250 \, \text{m}) = 2 \times 750 \, \text{m} = 1500 \, \text{m}
\]

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

---

#### 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:
- Area = 150 m²
- Length (\( l \)) = 25 m
- Width (\( w \)) = ?

1. Find the Width:
\[
\text{Area} = l \times w \implies 150 \, \text{m}^2 = 25 \, \text{m} \times w \implies w = \frac{150 \, \text{m}^2}{25 \, \text{m}} = 6 \, \text{m}
\]

2. Calculate the Perimeter:
\[
\text{Perimeter} = 2 \times (l + w) = 2 \times (25 \, \text{m} + 6 \, \text{m}) = 2 \times 31 \, \text{m} = 62 \, \text{m}
\]

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 = 5200 ft²
- Width (\( w \)) = 75 ft
- Length (\( l \)) = ?

1. Find the Length:
\[
\text{Area} = l \times w \implies 5200 \, \text{ft}^2 = l \times 75 \, \text{ft} \implies l = \frac{5200 \, \text{ft}^2}{75 \, \text{ft}} = 69.33 \, \text{ft}
\]

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. \(\boxed{69.33 \, \text{ft}}\)