Area of Trapezoids Worksheets - Free Printable
Educational worksheet: Area of Trapezoids Worksheets. Download and print for classroom or home learning activities.
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Step-by-step solution for: Area of Trapezoids Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Area of Trapezoids Worksheets
To solve the problem of finding the areas of trapezoids and parallelograms, we will use the following formulas:
1. Area of a Trapezoid:
\[
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
2. Area of a Parallelogram:
\[
\text{Area} = \text{Base} \times \text{Height}
\]
Let's solve each problem step by step.
---
- Bases: \(3 \, \text{m}\) and \(6 \, \text{m}\)
- Height: \(6.42 \, \text{m}\)
\[
\text{Area} = \frac{1}{2} \times (3 + 6) \times 6.42 = \frac{1}{2} \times 9 \times 6.42 = 4.5 \times 6.42 = 28.89 \, \text{m}^2
\]
Answer: \(28.89 \, \text{m}^2\)
---
- Bases: \(4 \, \text{in}\) and \(7 \, \text{in}\)
- Height: \(3.52 \, \text{in}\)
\[
\text{Area} = \frac{1}{2} \times (4 + 7) \times 3.52 = \frac{1}{2} \times 11 \times 3.52 = 5.5 \times 3.52 = 19.36 \, \text{in}^2
\]
Answer: \(19.36 \, \text{in}^2\)
---
- Bases: \(5 \, \text{m}\) and \(2.57 \, \text{m}\)
- Height: \(3 \, \text{m}\)
\[
\text{Area} = \frac{1}{2} \times (5 + 2.57) \times 3 = \frac{1}{2} \times 7.57 \times 3 = 3.785 \times 3 = 11.355 \, \text{m}^2
\]
Answer: \(11.355 \, \text{m}^2\)
---
- Base: \(6 \, \text{m}\)
- Height: \(6.47 \, \text{m}\)
\[
\text{Area} = 6 \times 6.47 = 38.82 \, \text{m}^2
\]
Answer: \(38.82 \, \text{m}^2\)
---
- Base: \(4 \, \text{in}\)
- Height: \(4.37 \, \text{in}\)
\[
\text{Area} = 4 \times 4.37 = 17.48 \, \text{in}^2
\]
Answer: \(17.48 \, \text{in}^2\)
---
- Base: \(13 \, \text{ft}\)
- Height: \(10 \, \text{ft}\)
\[
\text{Area} = 13 \times 10 = 130 \, \text{ft}^2
\]
Answer: \(130 \, \text{ft}^2\)
---
- Base: \(15 \, \text{in}\)
- Height: \(7.58 \, \text{in}\)
\[
\text{Area} = 15 \times 7.58 = 113.7 \, \text{in}^2
\]
Answer: \(113.7 \, \text{in}^2\)
---
- Bases: \(6 \, \text{cm}\) and \(5 \, \text{cm}\)
- Height: \(6.38 \, \text{cm}\)
\[
\text{Area} = \frac{1}{2} \times (6 + 5) \times 6.38 = \frac{1}{2} \times 11 \times 6.38 = 5.5 \times 6.38 = 35.09 \, \text{cm}^2
\]
Answer: \(35.09 \, \text{cm}^2\)
---
- Base: \(6 \, \text{cm}\)
- Height: \(6.65 \, \text{cm}\)
\[
\text{Area} = 6 \times 6.65 = 39.9 \, \text{cm}^2
\]
Answer: \(39.9 \, \text{cm}^2\)
---
- Bases: \(5 \, \text{yd}\) and \(3 \, \text{yd}\)
- Height: \(2.82 \, \text{yd}\)
\[
\text{Area} = \frac{1}{2} \times (5 + 3) \times 2.82 = \frac{1}{2} \times 8 \times 2.82 = 4 \times 2.82 = 11.28 \, \text{yd}^2
\]
Answer: \(11.28 \, \text{yd}^2\)
---
\[
\boxed{
\begin{array}{ll}
1. & 28.89 \, \text{m}^2 \\
2. & 19.36 \, \text{in}^2 \\
3. & 11.355 \, \text{m}^2 \\
4. & 38.82 \, \text{m}^2 \\
5. & 17.48 \, \text{in}^2 \\
6. & 130 \, \text{ft}^2 \\
7. & 113.7 \, \text{in}^2 \\
8. & 35.09 \, \text{cm}^2 \\
9. & 39.9 \, \text{cm}^2 \\
10. & 11.28 \, \text{yd}^2 \\
\end{array}
}
\]
1. Area of a Trapezoid:
\[
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
2. Area of a Parallelogram:
\[
\text{Area} = \text{Base} \times \text{Height}
\]
Let's solve each problem step by step.
---
Problem 1: Trapezoid
- Bases: \(3 \, \text{m}\) and \(6 \, \text{m}\)
- Height: \(6.42 \, \text{m}\)
\[
\text{Area} = \frac{1}{2} \times (3 + 6) \times 6.42 = \frac{1}{2} \times 9 \times 6.42 = 4.5 \times 6.42 = 28.89 \, \text{m}^2
\]
Answer: \(28.89 \, \text{m}^2\)
---
Problem 2: Trapezoid
- Bases: \(4 \, \text{in}\) and \(7 \, \text{in}\)
- Height: \(3.52 \, \text{in}\)
\[
\text{Area} = \frac{1}{2} \times (4 + 7) \times 3.52 = \frac{1}{2} \times 11 \times 3.52 = 5.5 \times 3.52 = 19.36 \, \text{in}^2
\]
Answer: \(19.36 \, \text{in}^2\)
---
Problem 3: Trapezoid
- Bases: \(5 \, \text{m}\) and \(2.57 \, \text{m}\)
- Height: \(3 \, \text{m}\)
\[
\text{Area} = \frac{1}{2} \times (5 + 2.57) \times 3 = \frac{1}{2} \times 7.57 \times 3 = 3.785 \times 3 = 11.355 \, \text{m}^2
\]
Answer: \(11.355 \, \text{m}^2\)
---
Problem 4: Parallelogram
- Base: \(6 \, \text{m}\)
- Height: \(6.47 \, \text{m}\)
\[
\text{Area} = 6 \times 6.47 = 38.82 \, \text{m}^2
\]
Answer: \(38.82 \, \text{m}^2\)
---
Problem 5: Parallelogram
- Base: \(4 \, \text{in}\)
- Height: \(4.37 \, \text{in}\)
\[
\text{Area} = 4 \times 4.37 = 17.48 \, \text{in}^2
\]
Answer: \(17.48 \, \text{in}^2\)
---
Problem 6: Parallelogram
- Base: \(13 \, \text{ft}\)
- Height: \(10 \, \text{ft}\)
\[
\text{Area} = 13 \times 10 = 130 \, \text{ft}^2
\]
Answer: \(130 \, \text{ft}^2\)
---
Problem 7: Parallelogram
- Base: \(15 \, \text{in}\)
- Height: \(7.58 \, \text{in}\)
\[
\text{Area} = 15 \times 7.58 = 113.7 \, \text{in}^2
\]
Answer: \(113.7 \, \text{in}^2\)
---
Problem 8: Trapezoid
- Bases: \(6 \, \text{cm}\) and \(5 \, \text{cm}\)
- Height: \(6.38 \, \text{cm}\)
\[
\text{Area} = \frac{1}{2} \times (6 + 5) \times 6.38 = \frac{1}{2} \times 11 \times 6.38 = 5.5 \times 6.38 = 35.09 \, \text{cm}^2
\]
Answer: \(35.09 \, \text{cm}^2\)
---
Problem 9: Parallelogram
- Base: \(6 \, \text{cm}\)
- Height: \(6.65 \, \text{cm}\)
\[
\text{Area} = 6 \times 6.65 = 39.9 \, \text{cm}^2
\]
Answer: \(39.9 \, \text{cm}^2\)
---
Problem 10: Trapezoid
- Bases: \(5 \, \text{yd}\) and \(3 \, \text{yd}\)
- Height: \(2.82 \, \text{yd}\)
\[
\text{Area} = \frac{1}{2} \times (5 + 3) \times 2.82 = \frac{1}{2} \times 8 \times 2.82 = 4 \times 2.82 = 11.28 \, \text{yd}^2
\]
Answer: \(11.28 \, \text{yd}^2\)
---
Final Answers
\[
\boxed{
\begin{array}{ll}
1. & 28.89 \, \text{m}^2 \\
2. & 19.36 \, \text{in}^2 \\
3. & 11.355 \, \text{m}^2 \\
4. & 38.82 \, \text{m}^2 \\
5. & 17.48 \, \text{in}^2 \\
6. & 130 \, \text{ft}^2 \\
7. & 113.7 \, \text{in}^2 \\
8. & 35.09 \, \text{cm}^2 \\
9. & 39.9 \, \text{cm}^2 \\
10. & 11.28 \, \text{yd}^2 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of area of triangle and parallelogram worksheet.