Area, Perimeter, & Volume Worksheets For Grade 5 - www ... - Free Printable

To solve the problem of calculating the perimeter \( S \) for each rectilinear object, we need to sum up the lengths of all the sides of each shape. Let's go through each figure step by step.

---

1. Diamond (Rhombus)

The given side length is \( 6 \, \text{cm} \).

- A rhombus has four equal sides.
- Perimeter \( S = 4 \times \text{side length} \).

\[
S = 4 \times 6 = 24 \, \text{cm}
\]

Answer for 1:
\[
\boxed{24}
\]

---

2. Rectangle

The given dimensions are:
- Length \( l = 12 \, \text{cm} \)
- Width \( w = 8 \, \text{cm} \)

- The formula for the perimeter of a rectangle is:
\[
S = 2 \times (\text{length} + \text{width})
\]

\[
S = 2 \times (12 + 8) = 2 \times 20 = 40 \, \text{cm}
\]

Answer for 2:
\[
\boxed{40}
\]

---

3. Irregular Polygon

The given side lengths are:
- \( 12 \, \text{cm}, 12 \, \text{cm}, 12 \, \text{cm}, 12 \, \text{cm}, 12 \, \text{cm}, 12 \, \text{cm} \)

- All sides are equal, so this is a regular hexagon.
- Perimeter \( S = 6 \times \text{side length} \).

\[
S = 6 \times 12 = 72 \, \text{cm}
\]

Answer for 3:
\[
\boxed{72}
\]

---

4. L-shaped Figure

The given dimensions are:
- Top horizontal segment: \( 10 \, \text{cm} \)
- Vertical segments: \( 8 \, \text{cm} \) and \( 13 \, \text{cm} \)
- Bottom horizontal segment: \( 35 \, \text{cm} \)

- To find the perimeter, add up all the outer edges:
\[
S = 10 + 8 + 13 + 35 + 8 + 10
\]

\[
S = 84 \, \text{cm}
\]

Answer for 4:
\[
\boxed{84}
\]

---

5. Compound Shape (Rectangle with Cut-out)

The given dimensions are:
- Outer rectangle: \( 45 \, \text{cm} \times 35 \, \text{cm} \)
- Inner rectangle (cut-out): \( 32 \, \text{cm} \times 7 \, \text{cm} \)

- The perimeter of the compound shape includes:
1. The outer perimeter of the large rectangle.
2. The two additional vertical sides of the cut-out rectangle (since they are exposed).

- Perimeter of the outer rectangle:
\[
2 \times (45 + 35) = 2 \times 80 = 160 \, \text{cm}
\]

- Additional vertical sides from the cut-out:
\[
2 \times 7 = 14 \, \text{cm}
\]

- Total perimeter:
\[
S = 160 + 14 = 174 \, \text{cm}
\]

Answer for 5:
\[
\boxed{174}
\]

---

6. Arrow-shaped Figure

The given dimensions are:
- Horizontal segments: \( 17 \, \text{cm}, 17 \, \text{cm}, 17 \, \text{cm} \)
- Vertical segments: \( 12 \, \text{cm}, 12 \, \text{cm}, 9 \, \text{cm}, 9 \, \text{cm} \)

- Sum up all the outer edges:
\[
S = 17 + 12 + 17 + 9 + 17 + 12 + 9 + 17
\]

\[
S = 112 \, \text{cm}
\]

Answer for 6:
\[
\boxed{112}
\]

---

Final Answers:

1. \( \boxed{24} \)
2. \( \boxed{40} \)
3. \( \boxed{72} \)
4. \( \boxed{84} \)
5. \( \boxed{174} \)
6. \( \boxed{112} \)