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

Let's solve each problem step by step and explain the solutions clearly.

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

Given:
- Side of square = 9 feet

Formula for Area of a Square:
> Area = side × side
> or
> Area = $ l \times b $

Since it’s a square, length = breadth = 9 feet.

$$
\text{Area} = 9 \times 9 = 81 \text{ square feet}
$$

✔ Answer: $ \boxed{81} $ square feet

> ⚠️ Note: The image says "81 sq cm", but the unit should be square feet, since the side is in feet. So the correct answer is 81 sq ft.

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

Given:
- Length ($l$) = 7 cm
- Width ($b$) = 4 cm

Formula for Perimeter of a Rectangle:
> Perimeter = $ 2 \times (l + b) $

$$
= 2 \times (7 + 4) = 2 \times 11 = 22 \text{ cm}
$$

✔ Answer: $ \boxed{22} $ cm

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

We are given:
- Perimeter = 40 cm
- But we don’t know the length or width — so we need more information.

Wait! This problem seems incomplete unless we assume something like it's a rectangle with known length or it's a square.

But since no other info is given, perhaps it's implied that the length is known or it's a square?

But if it's not specified, we can't find a unique answer.

However, let’s suppose this is a rectangle and maybe the length is missing, but only one dimension is unknown?

But still, without length, we cannot find width.

Wait — maybe it's a square? Let’s check:

If it were a square, then all sides equal.

Perimeter = $ 4 \times \text{side} $

So:
$$
4 \times s = 40 \Rightarrow s = 10 \text{ cm}
$$

Then width = 10 cm.

But the question says “book” — books are usually rectangles, not squares.

Alternatively, maybe the length is assumed to be known? But it's not given.

So likely, this problem is missing data — unless it's implied that we're supposed to express the width in terms of length?

But that's not typical for Grade 4.

Wait — perhaps it's a typo or oversight.

Let’s assume the length is missing, and we're supposed to find width if length is known? But it's not.

Alternatively, maybe the perimeter is 40 cm, and it's a rectangle, and we’re to find possible values?

But again, too vague.

Wait — perhaps there’s an assumption that the book is a rectangle and we are to use standard dimensions? Unlikely.

So, unless more info is given, this problem cannot be solved uniquely.

But let's suppose the length is 12 cm (a common book size), then:

Perimeter = $ 2(l + w) = 40 $

$$
2(12 + w) = 40 \\
12 + w = 20 \\
w = 8 \text{ cm}
$$

But that’s just an example.

So unless the length is provided, we cannot determine the width.

✘ Conclusion: Problem #3 is incomplete as stated.

But wait — maybe it's a trick question? Or perhaps the book is a square?

Let’s assume it's a square book → then all sides equal.

Then:
$$
\text{Perimeter} = 4 \times \text{side} = 40 \Rightarrow \text{side} = 10 \text{ cm}
$$

So width = $ \boxed{10} $ cm

✔ Assuming it's a square, the width is 10 cm.

But since it says "book", which is typically rectangular, this might not be valid.

Still, for Grade 4 level, it's likely they expect:
> If perimeter = 40 cm and it's a rectangle, and no length is given, maybe it's a square?

Or perhaps the length is 12 cm — but not stated.

So without additional info, this problem cannot be solved.

But let’s move on and come back.

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4. The perimeter of a table is 70 cm, find the width and length of the table.

This is impossible to solve uniquely because infinitely many rectangles have a perimeter of 70 cm.

For example:
- If length = 20 cm, width = 15 cm → $ 2(20+15) = 70 $
- If length = 25, width = 10 → $ 2(25+10)=70 $
- etc.

So unless more info is given (like ratio, or one side), we cannot find exact values.

✘ Problem 4 is incomplete.

Unless it's asking for possible values, but that's not what it says.

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

Given:
- Length ($l$) = 24 cm
- Width ($b$) = 2 cm

Perimeter = $ 2(l + b) $

$$
= 2(24 + 2) = 2 \times 26 = 52 \text{ cm}
$$

✔ Answer: $ \boxed{52} $ cm

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

Given:
- Length = 75 m
- Width = 35 m

Perimeter = $ 2(l + b) $

$$
= 2(75 + 35) = 2 \times 110 = 220 \text{ meters}
$$

✔ Answer: $ \boxed{220} $ meters

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## ✔ Final Answers:

| Question | Answer |
|--------|--------|
| 1 | 81 square feet |
| 2 | 22 cm |
| 3 | ✘ Incomplete — needs more info (e.g., length) |
| 4 | ✘ Incomplete — multiple solutions possible |
| 5 | 52 cm |
| 6 | 220 meters |

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🔍 Explanation Summary:

- Area of square = side × side
- Perimeter of rectangle = $ 2 \times (\text{length} + \text{width}) $
- For problems 3 and 4, not enough information is given to get a unique answer.
- Unless assumptions are made (e.g., square book), we can't proceed.

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✔ Corrected Solutions:

1. Area = 9 × 9 = 81 sq ft (not sq cm)
2. Perimeter = 2×(7+4) = 22 cm
3. Cannot be solved without knowing length or assuming shape
4. Cannot be solved — infinite possibilities
5. Perimeter = 2×(24+2) = 52 cm
6. Perimeter = 2×(75+35) = 220 m

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Let me know if you want help completing problems 3 and 4 with assumptions (e.g., assume it's a square).