Third Grade Find the Perimeter Activity Sheet (teacher made) - Free Printable

To solve the problem of finding the perimeter of each shape, we need to calculate the total length of the boundary of each shape. The perimeter is the sum of the lengths of all the sides of a polygon.

Let's go through each shape step by step:

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Shape 1: Pentagon

- Sides: \(8 \, \text{in}, 8 \, \text{in}, 13 \, \text{in}, 13 \, \text{in}, 9 \, \text{in}\)
- Perimeter = \(8 + 8 + 13 + 13 + 9 = 51 \, \text{in}\)

Answer for Shape 1: \(51 \, \text{in}\)

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Shape 2: Rectangle

- Length = \(23 \, \text{in}\)
- Width = \(12 \, \text{in}\)
- Perimeter of a rectangle = \(2 \times (\text{Length} + \text{Width})\)
- Perimeter = \(2 \times (23 + 12) = 2 \times 35 = 70 \, \text{in}\)

Answer for Shape 2: \(70 \, \text{in}\)

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Shape 3: Irregular Pentagon

- Sides: \(14 \, \text{in}, 6 \, \text{in}, 9 \, \text{in}, 14 \, \text{in}, 6 \, \text{in}\)
- Perimeter = \(14 + 6 + 9 + 14 + 6 = 49 \, \text{in}\)

Answer for Shape 3: \(49 \, \text{in}\)

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Shape 4: Rectangle

- Length = \(26 \, \text{in}\)
- Width = \(10 \, \text{in}\)
- Perimeter of a rectangle = \(2 \times (\text{Length} + \text{Width})\)
- Perimeter = \(2 \times (26 + 10) = 2 \times 36 = 72 \, \text{in}\)

Answer for Shape 4: \(72 \, \text{in}\)

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Shape 5: Triangle

- Sides: \(15 \, \text{in}, 23 \, \text{in}, 23 \, \text{in}\)
- Perimeter = \(15 + 23 + 23 = 61 \, \text{in}\)

Answer for Shape 5: \(61 \, \text{in}\)

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Shape 6: Star (Irregular Polygon)

- All sides are equal.
- Each side = \(10 \, \text{in}\)
- A star typically has 10 sides.
- Perimeter = \(10 \times 10 = 100 \, \text{in}\)

Answer for Shape 6: \(100 \, \text{in}\)

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Final Answers:

1. \(51 \, \text{in}\)
2. \(70 \, \text{in}\)
3. \(49 \, \text{in}\)
4. \(72 \, \text{in}\)
5. \(61 \, \text{in}\)
6. \(100 \, \text{in}\)

Boxed Final Answer:
\[
\boxed{51 \, \text{in}, 70 \, \text{in}, 49 \, \text{in}, 72 \, \text{in}, 61 \, \text{in}, 100 \, \text{in}}
\]