Area and perimeter of composite figures worksheet - Free Printable

To solve the problem, we need to calculate the area and perimeter of each composite figure. Let's go through each figure step by step.

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Figure 1:

#### Dimensions:
- Top rectangle: \(10 \, \text{cm} \times 7 \, \text{cm}\)
- Bottom rectangle: \(5 \, \text{cm} \times 4 \, \text{cm}\)

#### Area:
The area of a rectangle is given by:
\[
\text{Area} = \text{length} \times \text{width}
\]

- Area of the top rectangle:
\[
10 \, \text{cm} \times 7 \, \text{cm} = 70 \, \text{cm}^2
\]

- Area of the bottom rectangle:
\[
5 \, \text{cm} \times 4 \, \text{cm} = 20 \, \text{cm}^2
\]

- Total area:
\[
70 \, \text{cm}^2 + 20 \, \text{cm}^2 = 90 \, \text{cm}^2
\]

#### Perimeter:
To find the perimeter, we need to sum the lengths of all the outer edges. The figure can be visualized as a single shape with the following sides:
- Top side: \(10 \, \text{cm}\)
- Right side: \(7 \, \text{cm} + 4 \, \text{cm} = 11 \, \text{cm}\)
- Bottom side: \(5 \, \text{cm}\)
- Left side: \(3 \, \text{cm} + 7 \, \text{cm} = 10 \, \text{cm}\)

- Total perimeter:
\[
10 \, \text{cm} + 11 \, \text{cm} + 5 \, \text{cm} + 10 \, \text{cm} = 36 \, \text{cm}
\]

#### Answers for Figure 1:
\[
\text{Area} = 90 \, \text{cm}^2, \quad \text{Perimeter} = 36 \, \text{cm}
\]

---

Figure 2:

#### Dimensions:
- Main rectangle: \(10 \, \text{mm} \times 8 \, \text{mm}\)
- Cut-out rectangle: \(6 \, \text{mm} \times 6 \, \text{mm}\)

#### Area:
- Area of the main rectangle:
\[
10 \, \text{mm} \times 8 \, \text{mm} = 80 \, \text{mm}^2
\]

- Area of the cut-out rectangle:
\[
6 \, \text{mm} \times 6 \, \text{mm} = 36 \, \text{mm}^2
\]

- Total area (subtract the cut-out area):
\[
80 \, \text{mm}^2 - 36 \, \text{mm}^2 = 44 \, \text{mm}^2
\]

#### Perimeter:
The perimeter of the composite figure is the same as the perimeter of the main rectangle because the cut-out does not affect the outer boundary.
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width}) = 2 \times (10 \, \text{mm} + 8 \, \text{mm}) = 2 \times 18 \, \text{mm} = 36 \, \text{mm}
\]

#### Answers for Figure 2:
\[
\text{Area} = 44 \, \text{mm}^2, \quad \text{Perimeter} = 36 \, \text{mm}
\]

---

Figure 3:

#### Dimensions:
- Top rectangle: \(5 \, \text{m} \times 6 \, \text{m}\)
- Bottom rectangle: \(8 \, \text{m} \times 3 \, \text{m}\)

#### Area:
- Area of the top rectangle:
\[
5 \, \text{m} \times 6 \, \text{m} = 30 \, \text{m}^2
\]

- Area of the bottom rectangle:
\[
8 \, \text{m} \times 3 \, \text{m} = 24 \, \text{m}^2
\]

- Total area:
\[
30 \, \text{m}^2 + 24 \, \text{m}^2 = 54 \, \text{m}^2
\]

#### Perimeter:
To find the perimeter, we sum the lengths of all the outer edges. The figure can be visualized as a single shape with the following sides:
- Top side: \(5 \, \text{m}\)
- Right side: \(6 \, \text{m} + 3 \, \text{m} = 9 \, \text{m}\)
- Bottom side: \(8 \, \text{m}\)
- Left side: \(5 \, \text{m} + 8 \, \text{m} = 13 \, \text{m}\)

- Total perimeter:
\[
5 \, \text{m} + 9 \, \text{m} + 8 \, \text{m} + 13 \, \text{m} = 35 \, \text{m}
\]

#### Answers for Figure 3:
\[
\text{Area} = 54 \, \text{m}^2, \quad \text{Perimeter} = 35 \, \text{m}
\]

---

Figure 4:

#### Dimensions:
- Main rectangle: \(8 \, \text{cm} \times 6 \, \text{cm}\)
- Cut-out rectangle: \(4 \, \text{cm} \times 5 \, \text{cm}\)

#### Area:
- Area of the main rectangle:
\[
8 \, \text{cm} \times 6 \, \text{cm} = 48 \, \text{cm}^2
\]

- Area of the cut-out rectangle:
\[
4 \, \text{cm} \times 5 \, \text{cm} = 20 \, \text{cm}^2
\]

- Total area (subtract the cut-out area):
\[
48 \, \text{cm}^2 - 20 \, \text{cm}^2 = 28 \, \text{cm}^2
\]

#### Perimeter:
The perimeter of the composite figure is the same as the perimeter of the main rectangle because the cut-out does not affect the outer boundary.
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width}) = 2 \times (8 \, \text{cm} + 6 \, \text{cm}) = 2 \times 14 \, \text{cm} = 28 \, \text{cm}
\]

#### Answers for Figure 4:
\[
\text{Area} = 28 \, \text{cm}^2, \quad \text{Perimeter} = 28 \, \text{cm}
\]

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Final Answers:

1. \(\text{Area} = 90 \, \text{cm}^2\), \(\text{Perimeter} = 36 \, \text{cm}\)
2. \(\text{Area} = 44 \, \text{mm}^2\), \(\text{Perimeter} = 36 \, \text{mm}\)
3. \(\text{Area} = 54 \, \text{m}^2\), \(\text{Perimeter} = 35 \, \text{m}\)
4. \(\text{Area} = 28 \, \text{cm}^2\), \(\text{Perimeter} = 28 \, \text{cm}\)

\[
\boxed{9036443654352828}
\]