To solve the problem of finding the perimeter of the given irregular shapes, we need to calculate the sum of the lengths of all the sides for each shape. Let's go through each shape step by step.
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1. Square (Top Left)
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Shape: Square
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Side lengths: All sides are equal, and each side is 5 ft.
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Formula for perimeter of a square: \( P = 4 \times \text{side length} \)
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Calculation:
\[
P = 4 \times 5 \, \text{ft} = 20 \, \text{ft}
\]
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Answer: Perimeter =
20 ft
---
2. Triangle (Top Right)
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Shape: Triangle
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Side lengths: 8 in, 6 in, 10 in
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Formula for perimeter of a triangle: Sum of all side lengths
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Calculation:
\[
P = 8 \, \text{in} + 6 \, \text{in} + 10 \, \text{in} = 24 \, \text{in}
\]
-
Answer: Perimeter =
24 in
---
3. Triangle (Middle Left)
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Shape: Triangle
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Side lengths: 5 cm, 7 cm, 6 cm
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Formula for perimeter of a triangle: Sum of all side lengths
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Calculation:
\[
P = 5 \, \text{cm} + 7 \, \text{cm} + 6 \, \text{cm} = 18 \, \text{cm}
\]
-
Answer: Perimeter =
18 cm
---
4. Rhombus (Middle Right)
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Shape: Rhombus
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Side lengths: All sides are equal, and each side is 3 m.
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Formula for perimeter of a rhombus: \( P = 4 \times \text{side length} \)
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Calculation:
\[
P = 4 \times 3 \, \text{m} = 12 \, \text{m}
\]
-
Answer: Perimeter =
12 m
---
5. Rectangle (Bottom Left)
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Shape: Rectangle
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Side lengths: Length = 5 ft, Width = 4 ft
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Formula for perimeter of a rectangle: \( P = 2 \times (\text{length} + \text{width}) \)
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Calculation:
\[
P = 2 \times (5 \, \text{ft} + 4 \, \text{ft}) = 2 \times 9 \, \text{ft} = 18 \, \text{ft}
\]
-
Answer: Perimeter =
18 ft
---
6. Hexagon (Bottom Right)
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Shape: Regular hexagon
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Side lengths: All sides are equal, and each side is 2 in.
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Formula for perimeter of a regular hexagon: \( P = 6 \times \text{side length} \)
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Calculation:
\[
P = 6 \times 2 \, \text{in} = 12 \, \text{in}
\]
-
Answer: Perimeter =
12 in
---
Final Answers
1. Perimeter of the square:
20 ft
2. Perimeter of the triangle (top right):
24 in
3. Perimeter of the triangle (middle left):
18 cm
4. Perimeter of the rhombus:
12 m
5. Perimeter of the rectangle:
18 ft
6. Perimeter of the hexagon:
12 in
\[
\boxed{20 \, \text{ft}, 24 \, \text{in}, 18 \, \text{cm}, 12 \, \text{m}, 18 \, \text{ft}, 12 \, \text{in}}
\]
Parent Tip: Review the logic above to help your child master the concept of perimeter and area of irregular shapes worksheet.