This geometry worksheet offers practice problems for calculating the area of regular polygons like pentagons and hexagons, as well as composite shapes.

To solve the problems in the worksheet, we need to calculate the areas of various regular polygons and composite figures. Let's go through each problem step by step.

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Problem 1: Rectangle

The given figure is a rectangle with dimensions:
- Length = 10.9 m
- Width = 7.5 m

The area of a rectangle is given by:
\[
\text{Area} = \text{Length} \times \text{Width}
\]
\[
\text{Area} = 10.9 \, \text{m} \times 7.5 \, \text{m} = 81.75 \, \text{m}^2
\]

Answer for Problem 1:
\[
\boxed{81.75 \, \text{m}^2}
\]

---

Problem 2: Regular Pentagon

The given figure is a regular pentagon with:
- Side length = 13.2 yd
- Apothem = 3.5 yd

The area of a regular polygon is given by:
\[
\text{Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}
\]
For a pentagon, the perimeter is:
\[
\text{Perimeter} = 5 \times \text{Side length} = 5 \times 13.2 \, \text{yd} = 66 \, \text{yd}
\]
Thus, the area is:
\[
\text{Area} = \frac{1}{2} \times 66 \, \text{yd} \times 3.5 \, \text{yd} = 33 \, \text{yd} \times 3.5 \, \text{yd} = 115.5 \, \text{yd}^2
\]

Answer for Problem 2:
\[
\boxed{115.5 \, \text{yd}^2}
\]

---

Problem 3: Regular Hexagon

The given figure is a regular hexagon with:
- Side length = 12 cm
- Apothem = \(5\sqrt{3} \, \text{cm}\)

The area of a regular polygon is given by:
\[
\text{Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}
\]
For a hexagon, the perimeter is:
\[
\text{Perimeter} = 6 \times \text{Side length} = 6 \times 12 \, \text{cm} = 72 \, \text{cm}
\]
Thus, the area is:
\[
\text{Area} = \frac{1}{2} \times 72 \, \text{cm} \times 5\sqrt{3} \, \text{cm} = 36 \, \text{cm} \times 5\sqrt{3} \, \text{cm} = 180\sqrt{3} \, \text{cm}^2
\]

Answer for Problem 3:
\[
\boxed{180\sqrt{3} \, \text{cm}^2}
\]

---

Problem 4: Triangle

The given figure is a triangle with:
- Base = 22 in
- Height = 16 in

The area of a triangle is given by:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
\[
\text{Area} = \frac{1}{2} \times 22 \, \text{in} \times 16 \, \text{in} = 11 \, \text{in} \times 16 \, \text{in} = 176 \, \text{in}^2
\]

Answer for Problem 4:
\[
\boxed{176 \, \text{in}^2}
\]

---

Problem 5: Composite Figure (Rectangle + Semi-circle)

The given figure consists of a rectangle and a semi-circle:
- Rectangle dimensions: Length = 32 ft, Width = 18 ft
- Semi-circle diameter = 18 ft (same as the width of the rectangle)

#### Step 1: Area of the rectangle
\[
\text{Area}_{\text{rectangle}} = \text{Length} \times \text{Width} = 32 \, \text{ft} \times 18 \, \text{ft} = 576 \, \text{ft}^2
\]

#### Step 2: Area of the semi-circle
The radius of the semi-circle is:
\[
r = \frac{\text{Diameter}}{2} = \frac{18 \, \text{ft}}{2} = 9 \, \text{ft}
\]
The area of a full circle is:
\[
\text{Area}_{\text{circle}} = \pi r^2 = \pi (9 \, \text{ft})^2 = 81\pi \, \text{ft}^2
\]
The area of the semi-circle is half of this:
\[
\text{Area}_{\text{semi-circle}} = \frac{1}{2} \times 81\pi \, \text{ft}^2 = 40.5\pi \, \text{ft}^2
\]

#### Step 3: Total area
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semi-circle}} = 576 \, \text{ft}^2 + 40.5\pi \, \text{ft}^2
\]

Answer for Problem 5:
\[
\boxed{576 + 40.5\pi \, \text{ft}^2}
\]

---

Problem 6: Trapezoid

The given figure is a trapezoid with:
- Bases: \(b_1 = 9 \, \text{mm}\), \(b_2 = 9 \, \text{mm}\) (both bases are equal, so it's actually a parallelogram)
- Height = 12 mm

The area of a trapezoid (or parallelogram) is given by:
\[
\text{Area} = \text{Base} \times \text{Height}
\]
\[
\text{Area} = 9 \, \text{mm} \times 12 \, \text{mm} = 108 \, \text{mm}^2
\]

Answer for Problem 6:
\[
\boxed{108 \, \text{mm}^2}
\]

---

Problem 7: Composite Figure (Rectangle + Triangle)

The given figure consists of a rectangle and a triangle:
- Rectangle dimensions: Length = 10 km, Width = 6 km
- Triangle base = 4.8 km, Height = 6 km (same as the width of the rectangle)

#### Step 1: Area of the rectangle
\[
\text{Area}_{\text{rectangle}} = \text{Length} \times \text{Width} = 10 \, \text{km} \times 6 \, \text{km} = 60 \, \text{km}^2
\]

#### Step 2: Area of the triangle
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4.8 \, \text{km} \times 6 \, \text{km} = 2.4 \, \text{km} \times 6 \, \text{km} = 14.4 \, \text{km}^2
\]

#### Step 3: Total area
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{triangle}} = 60 \, \text{km}^2 + 14.4 \, \text{km}^2 = 74.4 \, \text{km}^2
\]

Answer for Problem 7:
\[
\boxed{74.4 \, \text{km}^2}
\]

---

Problem 8: Composite Figure (Square - Triangle)

The given figure consists of a square with a triangle removed:
- Square side length = 16 cm
- Triangle base = 11.5 cm, Height = 8.5 cm

#### Step 1: Area of the square
\[
\text{Area}_{\text{square}} = \text{Side}^2 = 16 \, \text{cm} \times 16 \, \text{cm} = 256 \, \text{cm}^2
\]

#### Step 2: Area of the triangle
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 11.5 \, \text{cm} \times 8.5 \, \text{cm} = 5.75 \, \text{cm} \times 8.5 \, \text{cm} = 48.875 \, \text{cm}^2
\]

#### Step 3: Total area
\[
\text{Total Area} = \text{Area}_{\text{square}} - \text{Area}_{\text{triangle}} = 256 \, \text{cm}^2 - 48.875 \, \text{cm}^2 = 207.125 \, \text{cm}^2
\]

Answer for Problem 8:
\[
\boxed{207.125 \, \text{cm}^2}
\]

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Final Answers:

1. \(\boxed{81.75 \, \text{m}^2}\)
2. \(\boxed{115.5 \, \text{yd}^2}\)
3. \(\boxed{180\sqrt{3} \, \text{cm}^2}\)
4. \(\boxed{176 \, \text{in}^2}\)
5. \(\boxed{576 + 40.5\pi \, \text{ft}^2}\)
6. \(\boxed{108 \, \text{mm}^2}\)
7. \(\boxed{74.4 \, \text{km}^2}\)
8. \(\boxed{207.125 \, \text{cm}^2}\)