Perimeter word problems worksheet for students to solve real-world scenarios involving rectangles, squares, and other shapes.

Problem: Finding Perimeters

The task involves solving perimeter problems for various shapes. Let's solve each problem step by step.

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#### Problem 1:
Jayson needed to figure how much construction tape was needed to close off the parking lot at the mall. He measured each side of the 4-sided lot. The sides measured 45 feet, 32.5 feet, 52 feet, and 37.5 feet. What is the perimeter of the parking lot?

- Solution:
The perimeter of a polygon is the sum of the lengths of all its sides.
\[
\text{Perimeter} = 45 + 32.5 + 52 + 37.5
\]
Adding these values:
\[
45 + 32.5 = 77.5
\]
\[
77.5 + 52 = 129.5
\]
\[
129.5 + 37.5 = 167
\]
Therefore, the perimeter of the parking lot is:
\[
\boxed{167 \text{ feet}}
\]

---

#### Problem 2:
Sheila wants to plant tulips along the perimeter of her school's playground. The playground is a square with one side measuring 30 yards. What is the perimeter of the playground?

- Solution:
The perimeter of a square is given by:
\[
\text{Perimeter} = 4 \times \text{side length}
\]
Here, the side length is 30 yards.
\[
\text{Perimeter} = 4 \times 30 = 120 \text{ yards}
\]
Therefore, the perimeter of the playground is:
\[
\boxed{120 \text{ yards}}
\]

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#### Problem 3:
Clyde wanted to know the perimeter of a stop sign. He measured one side and found it to be 7 inches long. What is the perimeter of the stop sign?

- Solution:
A stop sign is an octagon (8-sided polygon). The perimeter of a regular octagon is given by:
\[
\text{Perimeter} = 8 \times \text{side length}
\]
Here, the side length is 7 inches.
\[
\text{Perimeter} = 8 \times 7 = 56 \text{ inches}
\]
Therefore, the perimeter of the stop sign is:
\[
\boxed{56 \text{ inches}}
\]

---

#### Problem 4:
Bob needs to cover the edges of a piece of drywall with tape. The drywall section is a rectangle with a length measuring 8 feet and a width measuring 4 feet. What is the perimeter?

- Solution:
The perimeter of a rectangle is given by:
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width})
\]
Here, the length is 8 feet and the width is 4 feet.
\[
\text{Perimeter} = 2 \times (8 + 4) = 2 \times 12 = 24 \text{ feet}
\]
Therefore, the perimeter of the drywall section is:
\[
\boxed{24 \text{ feet}}
\]

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#### Problem 5:
Farmer Carlos has a new pig addition to his farm. He needs to add on to his old pig pen. The old square-shaped pen had a perimeter of 44 feet. Farmer Carlos added enough to make a rectangle with a perimeter of 54 feet. How many feet did the farmer add to his current pen?

- Solution:
First, find the side length of the original square pen. The perimeter of a square is given by:
\[
\text{Perimeter} = 4 \times \text{side length}
\]
For the original square pen:
\[
44 = 4 \times \text{side length}
\]
Solving for the side length:
\[
\text{side length} = \frac{44}{4} = 11 \text{ feet}
\]
So, the original square pen has sides of 11 feet each.

Next, consider the new rectangular pen. Let the dimensions of the rectangle be \( l \) (length) and \( w \) (width). The perimeter of a rectangle is given by:
\[
\text{Perimeter} = 2 \times (l + w)
\]
For the new rectangle:
\[
54 = 2 \times (l + w)
\]
Solving for \( l + w \):
\[
l + w = \frac{54}{2} = 27 \text{ feet}
\]

Since the farmer added to one dimension of the square, assume the width remains the same as the side of the square (11 feet). Then:
\[
l + 11 = 27
\]
Solving for \( l \):
\[
l = 27 - 11 = 16 \text{ feet}
\]

The farmer added:
\[
16 - 11 = 5 \text{ feet}
\]
Therefore, the farmer added:
\[
\boxed{5 \text{ feet}}
\]

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Final Answers:

1. \(\boxed{167 \text{ feet}}\)
2. \(\boxed{120 \text{ yards}}\)
3. \(\boxed{56 \text{ inches}}\)
4. \(\boxed{24 \text{ feet}}\)
5. \(\boxed{5 \text{ feet}}\)