Area Worksheets - Free Printable

To solve the problem of finding the area of each compound shape, we need to break down each figure into simpler geometric shapes (such as rectangles, triangles, circles, or semicircles) and then calculate their areas individually. 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 figure step by step:

---

Figure 1:

The shape consists of a rectangle with a triangle cut out from one corner.

- Rectangle dimensions: Length = 8 cm, Width = 5 cm
- Triangle dimensions: Base = 3 cm, Height = 2 cm

#### Step 1: Calculate the area of the rectangle.
\[
\text{Area of rectangle} = \text{Length} \times \text{Width} = 8 \times 5 = 40 \, \text{cm}^2
\]

#### Step 2: Calculate the area of the triangle.
\[
\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \times 2 = 3 \, \text{cm}^2
\]

#### Step 3: Subtract the area of the triangle from the area of the rectangle.
\[
\text{Total area} = \text{Area of rectangle} - \text{Area of triangle} = 40 - 3 = 37 \, \text{cm}^2
\]

Answer for Figure 1:
\[
\boxed{37}
\]

---

Figure 2:

The shape is a rectangle with a semicircle on top.

- Rectangle dimensions: Length = 10 cm, Width = 4 cm
- Semicircle diameter: Equal to the width of the rectangle, so \( \text{Diameter} = 4 \, \text{cm} \)
- Radius of semicircle: \( r = \frac{\text{Diameter}}{2} = \frac{4}{2} = 2 \, \text{cm} \)

#### Step 1: Calculate the area of the rectangle.
\[
\text{Area of rectangle} = \text{Length} \times \text{Width} = 10 \times 4 = 40 \, \text{cm}^2
\]

#### Step 2: Calculate the area of the semicircle.
\[
\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = \frac{1}{2} \pi \times 4 = 2\pi \, \text{cm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area of semicircle} \approx 2 \times 3.14 = 6.28 \, \text{cm}^2
\]

#### Step 3: Add the area of the semicircle to the area of the rectangle.
\[
\text{Total area} = \text{Area of rectangle} + \text{Area of semicircle} = 40 + 6.28 = 46.28 \, \text{cm}^2
\]

Answer for Figure 2:
\[
\boxed{46.3}
\]

---

Figure 3:

The shape is a rectangle with a quarter-circle cut out from one corner.

- Rectangle dimensions: Length = 6 cm, Width = 4 cm
- Quarter-circle radius: Equal to the width of the rectangle, so \( r = 4 \, \text{cm} \)

#### Step 1: Calculate the area of the rectangle.
\[
\text{Area of rectangle} = \text{Length} \times \text{Width} = 6 \times 4 = 24 \, \text{cm}^2
\]

#### Step 2: Calculate the area of the quarter-circle.
\[
\text{Area of quarter-circle} = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (4)^2 = \frac{1}{4} \pi \times 16 = 4\pi \, \text{cm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area of quarter-circle} \approx 4 \times 3.14 = 12.56 \, \text{cm}^2
\]

#### Step 3: Subtract the area of the quarter-circle from the area of the rectangle.
\[
\text{Total area} = \text{Area of rectangle} - \text{Area of quarter-circle} = 24 - 12.56 = 11.44 \, \text{cm}^2
\]

Answer for Figure 3:
\[
\boxed{11.4}
\]

---

Figure 4:

The shape is a rectangle with a semicircle on top.

- Rectangle dimensions: Length = 9 cm, Width = 3 cm
- Semicircle diameter: Equal to the width of the rectangle, so \( \text{Diameter} = 3 \, \text{cm} \)
- Radius of semicircle: \( r = \frac{\text{Diameter}}{2} = \frac{3}{2} = 1.5 \, \text{cm} \)

#### Step 1: Calculate the area of the rectangle.
\[
\text{Area of rectangle} = \text{Length} \times \text{Width} = 9 \times 3 = 27 \, \text{cm}^2
\]

#### Step 2: Calculate the area of the semicircle.
\[
\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1.5)^2 = \frac{1}{2} \pi \times 2.25 = 1.125\pi \, \text{cm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area of semicircle} \approx 1.125 \times 3.14 = 3.5325 \, \text{cm}^2
\]

#### Step 3: Add the area of the semicircle to the area of the rectangle.
\[
\text{Total area} = \text{Area of rectangle} + \text{Area of semicircle} = 27 + 3.5325 = 30.5325 \, \text{cm}^2
\]

Answer for Figure 4:
\[
\boxed{30.5}
\]

---

Figure 5:

The shape is a circle with a triangle cut out.

- Circle radius: \( r = 5 \, \text{cm} \)
- Triangle dimensions: Base = 8 cm, Height = 6 cm

#### Step 1: Calculate the area of the circle.
\[
\text{Area of circle} = \pi r^2 = \pi (5)^2 = 25\pi \, \text{cm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area of circle} \approx 25 \times 3.14 = 78.5 \, \text{cm}^2
\]

#### Step 2: Calculate the area of the triangle.
\[
\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 8 \times 6 = 24 \, \text{cm}^2
\]

#### Step 3: Subtract the area of the triangle from the area of the circle.
\[
\text{Total area} = \text{Area of circle} - \text{Area of triangle} = 78.5 - 24 = 54.5 \, \text{cm}^2
\]

Answer for Figure 5:
\[
\boxed{54.5}
\]

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Final Answers:

\[
\boxed{37, 46.3, 11.4, 30.5, 54.5}
\]