Answer key showing step-by-step solutions for perimeter and area of geometric shapes.

The image you provided contains a worksheet with problems related to the perimeter and area of parallelograms, triangles, trapezoids, and other geometric shapes. Below, I will solve each problem step by step and explain the solution.

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Problem 1: Perimeter of a Parallelogram

Given:
- One side = 7 cm
- Adjacent side = 14 cm

Formula for Perimeter of a Parallelogram:
\[ \text{Perimeter} = 2 \times (\text{Length} + \text{Width}) \]

Solution:
\[ \text{Perimeter} = 2 \times (7 + 14) \]
\[ \text{Perimeter} = 2 \times 21 \]
\[ \text{Perimeter} = 42 \, \text{cm} \]

Answer:
\[ \boxed{42 \, \text{cm}} \]

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Problem 2: Area of a Triangle

Given:
- Base = 2.5 m
- Height = 3.8 m

Formula for Area of a Triangle:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]

Solution:
\[ \text{Area} = \frac{1}{2} \times 2.5 \times 3.8 \]
\[ \text{Area} = 1.25 \times 3.8 \]
\[ \text{Area} = 4.75 \, \text{m}^2 \]

Answer:
\[ \boxed{4.75 \, \text{m}^2} \]

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Problem 3: Perimeter of a Trapezoid

Given:
- Sides: 2.2 m, 1.6 m, 2.0 m, 4.8 m

Formula for Perimeter of a Trapezoid:
\[ \text{Perimeter} = \text{Sum of all sides} \]

Solution:
\[ \text{Perimeter} = 2.2 + 1.6 + 2.0 + 4.8 \]
\[ \text{Perimeter} = 10.6 \, \text{m} \]

Answer:
\[ \boxed{10.6 \, \text{m}} \]

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Problem 4: Perimeter of a Rectangle

Given:
- Length = 24 mm
- Width = 11 mm

Formula for Perimeter of a Rectangle:
\[ \text{Perimeter} = 2 \times (\text{Length} + \text{Width}) \]

Solution:
\[ \text{Perimeter} = 2 \times (24 + 11) \]
\[ \text{Perimeter} = 2 \times 35 \]
\[ \text{Perimeter} = 70 \, \text{mm} \]

Answer:
\[ \boxed{70 \, \text{mm}} \]

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Problem 5: Area of a Parallelogram

Given:
- Base = 12 in
- Height = 9 in

Formula for Area of a Parallelogram:
\[ \text{Area} = \text{Base} \times \text{Height} \]

Solution:
\[ \text{Area} = 12 \times 9 \]
\[ \text{Area} = 108 \, \text{in}^2 \]

Answer:
\[ \boxed{108 \, \text{in}^2} \]

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Problem 6: Area of a Triangle

Given:
- Base = 14 ft
- Height = 6.7 ft

Formula for Area of a Triangle:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]

Solution:
\[ \text{Area} = \frac{1}{2} \times 14 \times 6.7 \]
\[ \text{Area} = 7 \times 6.7 \]
\[ \text{Area} = 46.9 \, \text{ft}^2 \]

Answer:
\[ \boxed{46.9 \, \text{ft}^2} \]

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Problem 7: Area of a Trapezoid

Given:
- Bases: \( b_1 = 6 \, \text{cm} \), \( b_2 = 8 \, \text{cm} \)
- Height = 5 cm

Formula for Area of a Trapezoid:
\[ \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \]

Solution:
\[ \text{Area} = \frac{1}{2} \times (6 + 8) \times 5 \]
\[ \text{Area} = \frac{1}{2} \times 14 \times 5 \]
\[ \text{Area} = 7 \times 5 \]
\[ \text{Area} = 35 \, \text{cm}^2 \]

Answer:
\[ \boxed{35 \, \text{cm}^2} \]

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Problem 8: Perimeter of a Trapezoid

Given:
- Sides: 9 m, 10 m, 11 m, 12 m

Formula for Perimeter of a Trapezoid:
\[ \text{Perimeter} = \text{Sum of all sides} \]

Solution:
\[ \text{Perimeter} = 9 + 10 + 11 + 12 \]
\[ \text{Perimeter} = 42 \, \text{m} \]

Answer:
\[ \boxed{42 \, \text{m}} \]

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Final Answers:

1. \( \boxed{42 \, \text{cm}} \)
2. \( \boxed{4.75 \, \text{m}^2} \)
3. \( \boxed{10.6 \, \text{m}} \)
4. \( \boxed{70 \, \text{mm}} \)
5. \( \boxed{108 \, \text{in}^2} \)
6. \( \boxed{46.9 \, \text{ft}^2} \)
7. \( \boxed{35 \, \text{cm}^2} \)
8. \( \boxed{42 \, \text{m}} \)

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If you have any further questions or need additional clarification, feel free to ask!