Grade 4 Area and Perimeter word problems | Grade1to6.com - Free Printable

Let's solve each problem step by step.

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Problem 1:

A square has sides that are 9 feet long. What is the area of the box?

#### Solution:
The formula for the area of a square is:
\[
\text{Area} = \text{side} \times \text{side}
\]
Given:
\[
\text{side} = 9 \text{ feet}
\]
\[
\text{Area} = 9 \times 9 = 81 \text{ square feet}
\]

#### Answer:
\[
\boxed{81 \text{ sq ft}}
\]

---

Problem 2:

A rectangular box is 7 centimeters long and 4 centimeters wide. What is its perimeter?

#### Solution:
The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width})
\]
Given:
\[
\text{length} = 7 \text{ cm}, \quad \text{width} = 4 \text{ cm}
\]
\[
\text{Perimeter} = 2 \times (7 + 4) = 2 \times 11 = 22 \text{ cm}
\]

#### Answer:
\[
\boxed{22 \text{ cm}}
\]

---

Problem 3:

Find the width of the book. The book's perimeter is 40 cm.

#### Solution:
Let the length of the book be \( l \) and the width be \( w \). The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (l + w)
\]
Given:
\[
\text{Perimeter} = 40 \text{ cm}
\]
Assume the length \( l \) is known or can be any value. For simplicity, let's assume the length is \( l \) and solve for \( w \):
\[
40 = 2 \times (l + w)
\]
Divide both sides by 2:
\[
20 = l + w
\]
Rearrange to solve for \( w \):
\[
w = 20 - l
\]

Without additional information about the length, the width \( w \) depends on the length \( l \). If more context is provided (e.g., the length is given), we can find a specific value for \( w \).

#### Answer:
\[
\boxed{w = 20 - l}
\]

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Problem 4:

The perimeter of a table is 70 cm. Find the width and length of the table.

#### Solution:
Let the length of the table be \( l \) and the width be \( w \). The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (l + w)
\]
Given:
\[
\text{Perimeter} = 70 \text{ cm}
\]
\[
70 = 2 \times (l + w)
\]
Divide both sides by 2:
\[
35 = l + w
\]
This equation shows that the sum of the length and width is 35 cm. Without additional information (e.g., the ratio of length to width or one of the dimensions), there are infinitely many solutions. For example:
- If \( l = 20 \text{ cm} \), then \( w = 15 \text{ cm} \).
- If \( l = 25 \text{ cm} \), then \( w = 10 \text{ cm} \).

#### Answer:
\[
\boxed{l + w = 35 \text{ cm}}
\]

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Problem 5:

A rectangular table is 24 centimeters long and 2 centimeters wide. What is its perimeter?

#### Solution:
The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width})
\]
Given:
\[
\text{length} = 24 \text{ cm}, \quad \text{width} = 2 \text{ cm}
\]
\[
\text{Perimeter} = 2 \times (24 + 2) = 2 \times 26 = 52 \text{ cm}
\]

#### Answer:
\[
\boxed{52 \text{ cm}}
\]

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Problem 6:

The lake around your house is 75 meters long and 35 meters wide. What is its perimeter?

#### Solution:
The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width})
\]
Given:
\[
\text{length} = 75 \text{ m}, \quad \text{width} = 35 \text{ m}
\]
\[
\text{Perimeter} = 2 \times (75 + 35) = 2 \times 110 = 220 \text{ m}
\]

#### Answer:
\[
\boxed{220 \text{ m}}
\]

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Final Answers:

1. \(\boxed{81 \text{ sq ft}}\)
2. \(\boxed{22 \text{ cm}}\)
3. \(\boxed{w = 20 - l}\)
4. \(\boxed{l + w = 35 \text{ cm}}\)
5. \(\boxed{52 \text{ cm}}\)
6. \(\boxed{220 \text{ m}}\)