Practice worksheet for calculating the area of irregular shapes using given dimensions.

The task in the image involves calculating the area of various irregular shapes. Each shape is labeled with dimensions, and you are required to compute the area for each one. Below, I will solve each problem step by step.

---

Problem 1:

#### Shape:
A rectangle with a semicircle on top.
- Rectangle dimensions: \( L = 120 \) mm, \( W = 60 \) mm
- Semicircle diameter: \( D = 60 \) mm (same as the width of the rectangle)

#### Solution:
1. Area of the rectangle:
\[
\text{Area}_{\text{rectangle}} = L \times W = 120 \times 60 = 7200 \, \text{mm}^2
\]

2. Area of the semicircle:
- Radius of the semicircle: \( r = \frac{D}{2} = \frac{60}{2} = 30 \, \text{mm} \)
- Area of a full circle: \( \pi r^2 \)
- Area of the semicircle: \( \frac{1}{2} \pi r^2 \)
\[
\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi (30)^2 = \frac{1}{2} \pi (900) = 450\pi \, \text{mm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area}_{\text{semicircle}} \approx 450 \times 3.14 = 1413 \, \text{mm}^2
\]

3. Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semicircle}} = 7200 + 1413 = 8613 \, \text{mm}^2
\]

Answer for Problem 1:
\[
\boxed{8613}
\]

---

Problem 2:

#### Shape:
A rectangle with a triangular cutout.
- Rectangle dimensions: \( L = 100 \) mm, \( W = 50 \) mm
- Triangle base: \( b = 50 \) mm, Triangle height: \( h = 25 \) mm

#### Solution:
1. Area of the rectangle:
\[
\text{Area}_{\text{rectangle}} = L \times W = 100 \times 50 = 5000 \, \text{mm}^2
\]

2. Area of the triangle:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 50 \times 25 = 625 \, \text{mm}^2
\]

3. Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} - \text{Area}_{\text{triangle}} = 5000 - 625 = 4375 \, \text{mm}^2
\]

Answer for Problem 2:
\[
\boxed{4375}
\]

---

Problem 3:

#### Shape:
A trapezoid.
- Top base: \( b_1 = 80 \) mm
- Bottom base: \( b_2 = 120 \) mm
- Height: \( h = 50 \) mm

#### Solution:
1. Area of the trapezoid:
\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (80 + 120) \times 50
\]
\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times 200 \times 50 = 5000 \, \text{mm}^2
\]

Answer for Problem 3:
\[
\boxed{5000}
\]

---

Problem 4:

#### Shape:
A rectangle with a circular hole.
- Rectangle dimensions: \( L = 150 \) mm, \( W = 80 \) mm
- Circle diameter: \( D = 40 \) mm

#### Solution:
1. Area of the rectangle:
\[
\text{Area}_{\text{rectangle}} = L \times W = 150 \times 80 = 12000 \, \text{mm}^2
\]

2. Area of the circle:
- Radius of the circle: \( r = \frac{D}{2} = \frac{40}{2} = 20 \, \text{mm} \)
\[
\text{Area}_{\text{circle}} = \pi r^2 = \pi (20)^2 = 400\pi \, \text{mm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area}_{\text{circle}} \approx 400 \times 3.14 = 1256 \, \text{mm}^2
\]

3. Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} - \text{Area}_{\text{circle}} = 12000 - 1256 = 10744 \, \text{mm}^2
\]

Answer for Problem 4:
\[
\boxed{10744}
\]

---

Problem 5:

#### Shape:
A parallelogram.
- Base: \( b = 100 \) mm
- Height: \( h = 60 \) mm

#### Solution:
1. Area of the parallelogram:
\[
\text{Area}_{\text{parallelogram}} = b \times h = 100 \times 60 = 6000 \, \text{mm}^2
\]

Answer for Problem 5:
\[
\boxed{6000}
\]

---

Problem 6:

#### Shape:
A sector of a circle.
- Radius: \( r = 50 \) mm
- Central angle: \( \theta = 90^\circ \)

#### Solution:
1. Area of the sector:
- The formula for the area of a sector is:
\[
\text{Area}_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2
\]
- Here, \( \theta = 90^\circ \):
\[
\text{Area}_{\text{sector}} = \frac{90^\circ}{360^\circ} \times \pi (50)^2 = \frac{1}{4} \times \pi (2500) = 625\pi \, \text{mm}^2
\]
- Using \( \pi \approx 3.14 \):
\[
\text{Area}_{\text{sector}} \approx 625 \times 3.14 = 1962.5 \, \text{mm}^2
\]

Answer for Problem 6:
\[
\boxed{1962.5}
\]

---

Problem 7:

#### Shape:
A triangle.
- Base: \( b = 120 \) mm
- Height: \( h = 80 \) mm

#### Solution:
1. Area of the triangle:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 120 \times 80 = 4800 \, \text{mm}^2
\]

Answer for Problem 7:
\[
\boxed{4800}
\]

---

Problem 8:

#### Shape:
A composite shape consisting of a rectangle and a semicircle.
- Rectangle dimensions: \( L = 100 \) mm, \( W = 50 \) mm
- Semicircle diameter: \( D = 50 \) mm (same as the width of the rectangle)

#### Solution:
1. Area of the rectangle:
\[
\text{Area}_{\text{rectangle}} = L \times W = 100 \times 50 = 5000 \, \text{mm}^2
\]

2. Area of the semicircle:
- Radius of the semicircle: \( r = \frac{D}{2} = \frac{50}{2} = 25 \, \text{mm} \)
- Area of a full circle: \( \pi r^2 \)
- Area of the semicircle: \( \frac{1}{2} \pi r^2 \)
\[
\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi (25)^2 = \frac{1}{2} \pi (625) = 312.5\pi \, \text{mm}^2
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area}_{\text{semicircle}} \approx 312.5 \times 3.14 = 981.25 \, \text{mm}^2
\]

3. Total area:
\[
\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semicircle}} = 5000 + 981.25 = 5981.25 \, \text{mm}^2
\]

Answer for Problem 8:
\[
\boxed{5981.25}
\]

---

Final Answers:

\[
\boxed{8613, 4375, 5000, 10744, 6000, 1962.5, 4800, 5981.25}
\]