Geometry Worksheets | Area Worksheets - Free Printable

To solve the problem of finding the area of each compound shape, we will break down each figure into simpler geometric shapes (such as rectangles, triangles, circles, etc.) and calculate their areas individually. Then, we will combine these areas to find the total area of the compound shape. Let's go through each figure step by step.

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Figure 1:

The shape is a trapezoid.
- Formula for the area of a trapezoid:
\[
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
- Here:
- Base₁ = 7 in
- Base₂ = 14 in
- Height = 14 in

\[
\text{Area} = \frac{1}{2} \times (7 + 14) \times 14 = \frac{1}{2} \times 21 \times 14 = 147 \text{ square inches}
\]

Answer for Figure 1: \( \boxed{147} \)

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Figure 2:

The shape is a rectangle with a semicircle on top.
- Rectangle area:
\[
\text{Area}_{\text{rectangle}} = \text{Length} \times \text{Width} = 18 \times 18 = 324 \text{ square cm}
\]
- Semicircle area:
\[
\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi r^2
\]
- Radius \( r = \frac{9}{2} = 4.5 \) cm
\[
\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi (4.5)^2 = \frac{1}{2} \pi (20.25) \approx \frac{1}{2} \times 3.14 \times 20.25 \approx 31.8 \text{ square cm}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semicircle}} = 324 + 31.8 \approx 355.8 \text{ square cm}
\]

Answer for Figure 2: \( \boxed{355.8} \)

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Figure 3:

The shape is a rectangle with a quarter-circle cut out.
- Rectangle area:
\[
\text{Area}_{\text{rectangle}} = \text{Length} \times \text{Width} = 20 \times 15 = 300 \text{ square yd}
\]
- Quarter-circle area:
\[
\text{Area}_{\text{quarter-circle}} = \frac{1}{4} \pi r^2
\]
- Radius \( r = 6 \) yd
\[
\text{Area}_{\text{quarter-circle}} = \frac{1}{4} \pi (6)^2 = \frac{1}{4} \pi (36) \approx \frac{1}{4} \times 3.14 \times 36 \approx 28.3 \text{ square yd}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} - \text{Area}_{\text{quarter-circle}} = 300 - 28.3 \approx 271.7 \text{ square yd}
\]

Answer for Figure 3: \( \boxed{271.7} \)

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Figure 4:

The shape is a circle with a triangle cut out.
- Circle area:
\[
\text{Area}_{\text{circle}} = \pi r^2
\]
- Radius \( r = \frac{8}{2} = 4 \) m
\[
\text{Area}_{\text{circle}} = \pi (4)^2 = 16\pi \approx 16 \times 3.14 \approx 50.2 \text{ square m}
\]
- Triangle area:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 8 \times 12 = 48 \text{ square m}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{\text{circle}} - \text{Area}_{\text{triangle}} = 50.2 - 48 \approx 2.2 \text{ square m}
\]

Answer for Figure 4: \( \boxed{2.2} \)

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Figure 5:

The shape is composed of three rectangles.
- Rectangle 1 (top):
\[
\text{Area}_{1} = 12 \times 12 = 144 \text{ square in}
\]
- Rectangle 2 (middle):
\[
\text{Area}_{2} = 20 \times 12 = 240 \text{ square in}
\]
- Rectangle 3 (bottom):
\[
\text{Area}_{3} = 12 \times 12 = 144 \text{ square in}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{1} + \text{Area}_{2} + \text{Area}_{3} = 144 + 240 + 144 = 528 \text{ square in}
\]

Answer for Figure 5: \( \boxed{528} \)

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Figure 6:

The shape is a combination of a rectangle and a trapezoid.
- Rectangle area:
\[
\text{Area}_{\text{rectangle}} = 20 \times 20 = 400 \text{ square cm}
\]
- Trapezoid area:
\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
- Base₁ = 20 cm
- Base₂ = 10 cm
- Height = 20 cm
\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times (20 + 10) \times 20 = \frac{1}{2} \times 30 \times 20 = 300 \text{ square cm}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{trapezoid}} = 400 + 300 = 700 \text{ square cm}
\]

Answer for Figure 6: \( \boxed{700} \)

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Figure 7:

The shape is a right triangle.
- Formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
- Here:
- Base = 6 ft
- Height = 13 ft
\[
\text{Area} = \frac{1}{2} \times 6 \times 13 = 39 \text{ square ft}
\]

Answer for Figure 7: \( \boxed{39} \)

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Figure 8:

The shape is a combination of a triangle and a rectangle.
- Triangle area:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
- Base = 6 yd
- Height = 9 yd
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times 6 \times 9 = 27 \text{ square yd}
\]
- Rectangle area:
\[
\text{Area}_{\text{rectangle}} = \text{Length} \times \text{Width} = 6 \times 3 = 18 \text{ square yd}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{rectangle}} = 27 + 18 = 45 \text{ square yd}
\]

Answer for Figure 8: \( \boxed{45} \)

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Figure 9:

The shape is a large rectangle with a smaller rectangle cut out.
- Large rectangle area:
\[
\text{Area}_{\text{large}} = 11 \times 8 = 88 \text{ square ft}
\]
- Small rectangle area:
\[
\text{Area}_{\text{small}} = 3 \times 8 = 24 \text{ square ft}
\]
- Total area:
\[
\text{Total Area} = \text{Area}_{\text{large}} - \text{Area}_{\text{small}} = 88 - 24 = 64 \text{ square ft}
\]

Answer for Figure 9: \( \boxed{64} \)

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Final Answers:

1. \( \boxed{147} \)
2. \( \boxed{355.8} \)
3. \( \boxed{271.7} \)
4. \( \boxed{2.2} \)
5. \( \boxed{528} \)
6. \( \boxed{700} \)
7. \( \boxed{39} \)
8. \( \boxed{45} \)
9. \( \boxed{64} \)