Area of compound shapes worksheet with six figures for calculating total area.

To solve the problem of finding the area of compound shapes, we need to break each shape into simpler geometric figures (such as rectangles, triangles, circles, etc.) and then calculate the area of each part. Finally, we combine these areas according to whether they are added or subtracted based on the structure of the compound shape.

Let's go through each shape step by step:

---

Shape A

#### Description:
- The shape consists of a rectangle with a semicircle removed from one side.
- Dimensions: Rectangle is \(10 \times 6\), and the diameter of the semicircle is \(6\) (same as the width of the rectangle).

#### Steps:
1. Area of the rectangle:
\[
\text{Area}_{\text{rectangle}} = \text{length} \times \text{width} = 10 \times 6 = 60
\]

2. Area of the semicircle:
- Radius of the semicircle: \( r = \frac{\text{diameter}}{2} = \frac{6}{2} = 3 \)
- Area of a full circle: \( \pi r^2 = \pi (3)^2 = 9\pi \)
- Area of the semicircle: \( \frac{1}{2} \times 9\pi = \frac{9\pi}{2} \)

3. Total area of the shape:
\[
\text{Area}_{\text{A}} = \text{Area}_{\text{rectangle}} - \text{Area}_{\text{semicircle}} = 60 - \frac{9\pi}{2}
\]

Using \( \pi \approx 3.14 \):
\[
\frac{9\pi}{2} \approx \frac{9 \times 3.14}{2} = \frac{28.26}{2} = 14.13
\]
\[
\text{Area}_{\text{A}} \approx 60 - 14.13 = 45.87
\]

Final Answer for Shape A:
\[
\boxed{45.87}
\]

---

Shape B

#### Description:
- The shape consists of a square with a semicircle added on top.
- Dimensions: Square side = \(8\), and the diameter of the semicircle is \(8\) (same as the side of the square).

#### Steps:
1. Area of the square:
\[
\text{Area}_{\text{square}} = \text{side}^2 = 8^2 = 64
\]

2. Area of the semicircle:
- Radius of the semicircle: \( r = \frac{\text{diameter}}{2} = \frac{8}{2} = 4 \)
- Area of a full circle: \( \pi r^2 = \pi (4)^2 = 16\pi \)
- Area of the semicircle: \( \frac{1}{2} \times 16\pi = 8\pi \)

3. Total area of the shape:
\[
\text{Area}_{\text{B}} = \text{Area}_{\text{square}} + \text{Area}_{\text{semicircle}} = 64 + 8\pi
\]

Using \( \pi \approx 3.14 \):
\[
8\pi \approx 8 \times 3.14 = 25.12
\]
\[
\text{Area}_{\text{B}} \approx 64 + 25.12 = 89.12
\]

Final Answer for Shape B:
\[
\boxed{89.12}
\]

---

Shape C

#### Description:
- The shape consists of a rectangle with a triangle added to one side.
- Dimensions: Rectangle is \(12 \times 5\), and the triangle has a base of \(5\) and a height of \(4\).

#### Steps:
1. Area of the rectangle:
\[
\text{Area}_{\text{rectangle}} = \text{length} \times \text{width} = 12 \times 5 = 60
\]

2. Area of the triangle:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4 = 10
\]

3. Total area of the shape:
\[
\text{Area}_{\text{C}} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{triangle}} = 60 + 10 = 70
\]

Final Answer for Shape C:
\[
\boxed{70}
\]

---

Shape D

#### Description:
- The shape consists of a trapezoid with a semicircle removed from the top.
- Dimensions: Trapezoid bases are \(10\) and \(6\), height is \(4\), and the diameter of the semicircle is \(6\) (same as the smaller base of the trapezoid).

#### Steps:
1. Area of the trapezoid:
\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (10 + 6) \times 4 = \frac{1}{2} \times 16 \times 4 = 32
\]

2. Area of the semicircle:
- Radius of the semicircle: \( r = \frac{\text{diameter}}{2} = \frac{6}{2} = 3 \)
- Area of a full circle: \( \pi r^2 = \pi (3)^2 = 9\pi \)
- Area of the semicircle: \( \frac{1}{2} \times 9\pi = \frac{9\pi}{2} \)

3. Total area of the shape:
\[
\text{Area}_{\text{D}} = \text{Area}_{\text{trapezoid}} - \text{Area}_{\text{semicircle}} = 32 - \frac{9\pi}{2}
\]

Using \( \pi \approx 3.14 \):
\[
\frac{9\pi}{2} \approx \frac{9 \times 3.14}{2} = \frac{28.26}{2} = 14.13
\]
\[
\text{Area}_{\text{D}} \approx 32 - 14.13 = 17.87
\]

Final Answer for Shape D:
\[
\boxed{17.87}
\]

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Final Answers

\[
\boxed{45.87, 89.12, 70, 17.87}
\]