Compound Shapes Area Worksheet

To solve the problem of finding the area of each compound shape, we will break down each figure into simpler geometric shapes (such as triangles, rectangles, 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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1)

The figure consists of a triangle with a circular hole in the middle.
- Triangle Area: The base is 29 ft, and the height is also 29 ft.
\[
\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 29 \times 29 = \frac{841}{2} = 420.5 \, \text{ft}^2
\]
- Circle Area: The radius of the circle is 9 ft.
\[
\text{Area of Circle} = \pi r^2 = \pi \times 9^2 = 81\pi \approx 254.47 \, \text{ft}^2
\]
- Total Area: Subtract the area of the circle from the area of the triangle.
\[
\text{Total Area} = 420.5 - 254.47 \approx 166.0 \, \text{ft}^2
\]

Answer for 1): $\boxed{166.0}$

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2)

The figure is a right triangle with a sector of a circle removed.
- Triangle Area: The base is 5 in, and the height is 11 in.
\[
\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 11 = 27.5 \, \text{in}^2
\]
- Sector Area: The sector is a quarter-circle (90° out of 360°) with a radius of 5 in.
\[
\text{Area of Sector} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \pi \times 5^2 = \frac{25\pi}{4} \approx 19.63 \, \text{in}^2
\]
- Total Area: Subtract the area of the sector from the area of the triangle.
\[
\text{Total Area} = 27.5 - 19.63 \approx 7.9 \, \text{in}^2
\]

Answer for 2): $\boxed{7.9}$

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3)

The figure is a square with a circular hole in the middle.
- Square Area: The side length is 16 in.
\[
\text{Area of Square} = \text{side}^2 = 16^2 = 256 \, \text{in}^2
\]
- Circle Area: The diameter is 8 in, so the radius is 4 in.
\[
\text{Area of Circle} = \pi r^2 = \pi \times 4^2 = 16\pi \approx 50.27 \, \text{in}^2
\]
- Total Area: Subtract the area of the circle from the area of the square.
\[
\text{Total Area} = 256 - 50.27 \approx 205.7 \, \text{in}^2
\]

Answer for 3): $\boxed{205.7}$

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4)

The figure is an L-shaped figure.
- Rectangle 1: Dimensions are 14 m × 7 m.
\[
\text{Area of Rectangle 1} = 14 \times 7 = 98 \, \text{m}^2
\]
- Rectangle 2: Dimensions are 7 m × 4 m (since the total width is 11 m and one part is already 7 m).
\[
\text{Area of Rectangle 2} = 7 \times 4 = 28 \, \text{m}^2
\]
- Total Area: Add the areas of the two rectangles.
\[
\text{Total Area} = 98 + 28 = 126 \, \text{m}^2
\]

Answer for 4): $\boxed{126}$

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5)

The figure is a rectangle with a semicircular cutout.
- Rectangle Area: The dimensions are 11 cm × 4 cm.
\[
\text{Area of Rectangle} = 11 \times 4 = 44 \, \text{cm}^2
\]
- Semicircle Area: The diameter is 4 cm, so the radius is 2 cm.
\[
\text{Area of Semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 2^2 = 2\pi \approx 6.28 \, \text{cm}^2
\]
- Total Area: Subtract the area of the semicircle from the area of the rectangle.
\[
\text{Total Area} = 44 - 6.28 \approx 37.7 \, \text{cm}^2
\]

Answer for 5): $\boxed{37.7}$

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6)

The figure is a combination of a rectangle and a trapezoid.
- Rectangle Area: The dimensions are 12 m × 12 m.
\[
\text{Area of Rectangle} = 12 \times 12 = 144 \, \text{m}^2
\]
- Trapezoid Area: The bases are 12 m and 6 m, and the height is 12 m.
\[
\text{Area of Trapezoid} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (12 + 6) \times 12 = \frac{1}{2} \times 18 \times 12 = 108 \, \text{m}^2
\]
- Total Area: Add the areas of the rectangle and the trapezoid.
\[
\text{Total Area} = 144 + 108 = 252 \, \text{m}^2
\]

Answer for 6): $\boxed{252}$

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7)

The figure is a triangle with a semicircular cutout.
- Triangle Area: The base is 5 yd, and the height is 5 yd.
\[
\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 5 = 12.5 \, \text{yd}^2
\]
- Semicircle Area: The diameter is 2 yd, so the radius is 1 yd.
\[
\text{Area of Semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 1^2 = \frac{\pi}{2} \approx 1.57 \, \text{yd}^2
\]
- Total Area: Subtract the area of the semicircle from the area of the triangle.
\[
\text{Total Area} = 12.5 - 1.57 \approx 10.9 \, \text{yd}^2
\]

Answer for 7): $\boxed{10.9}$

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8)

The figure is a larger square with a smaller square cutout.
- Larger Square Area: The side length is 6 ft.
\[
\text{Area of Larger Square} = 6^2 = 36 \, \text{ft}^2
\]
- Smaller Square Area: The side length is 3 ft.
\[
\text{Area of Smaller Square} = 3^2 = 9 \, \text{ft}^2
\]
- Total Area: Subtract the area of the smaller square from the area of the larger square.
\[
\text{Total Area} = 36 - 9 = 27 \, \text{ft}^2
\]

Answer for 8): $\boxed{27}$

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9)

The figure is a composite shape consisting of a triangle and a rectangle.
- Rectangle Area: The dimensions are 20 yd × 15 yd.
\[
\text{Area of Rectangle} = 20 \times 15 = 300 \, \text{yd}^2
\]
- Triangle Area: The base is 20 yd, and the height is 10 yd.
\[
\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 10 = 100 \, \text{yd}^2
\]
- Total Area: Add the areas of the rectangle and the triangle.
\[
\text{Total Area} = 300 + 100 = 400 \, \text{yd}^2
\]

Answer for 9): $\boxed{400}$

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Final Answers:

1. $\boxed{166.0}$
2. $\boxed{7.9}$
3. $\boxed{205.7}$
4. $\boxed{126}$
5. $\boxed{37.7}$
6. $\boxed{252}$
7. $\boxed{10.9}$
8. $\boxed{27}$
9. $\boxed{400}$