To solve the problem of finding the areas of the given quadrilaterals, we will use the formula for the area of a trapezoid:
\[
\text{Area of a Trapezoid} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
Let's solve each problem step by step.
---
Problem 1:
-
Dimensions:
- Base 1 = 9 cm
- Base 2 = 11 cm
- Height = 5 cm
\[
\text{Area} = \frac{1}{2} \times (9 + 11) \times 5 = \frac{1}{2} \times 20 \times 5 = 50 \, \text{cm}^2
\]
Answer: \( 50 \, \text{cm}^2 \)
---
Problem 2:
-
Dimensions:
- Base 1 = 13 in
- Base 2 = 10 in
- Height = 8 in
\[
\text{Area} = \frac{1}{2} \times (13 + 10) \times 8 = \frac{1}{2} \times 23 \times 8 = 92 \, \text{in}^2
\]
Answer: \( 92 \, \text{in}^2 \)
---
Problem 3:
-
Dimensions:
- Base 1 = 17 m
- Base 2 = 8 m
- Height = 4 m
\[
\text{Area} = \frac{1}{2} \times (17 + 8) \times 4 = \frac{1}{2} \times 25 \times 4 = 50 \, \text{m}^2
\]
Answer: \( 50 \, \text{m}^2 \)
---
Problem 4:
-
Dimensions:
- Base 1 = 11 km
- Base 2 = 0 km (since it's a triangle)
- Height = 5 km
This is actually a triangle, not a trapezoid. The formula for the area of a triangle is:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
\[
\text{Area} = \frac{1}{2} \times 11 \times 5 = 27.5 \, \text{km}^2
\]
Answer: \( 27.5 \, \text{km}^2 \)
---
Problem 5:
-
Dimensions:
- Base 1 = 16 yd
- Base 2 = 6 yd
- Height = 7 yd
\[
\text{Area} = \frac{1}{2} \times (16 + 6) \times 7 = \frac{1}{2} \times 22 \times 7 = 77 \, \text{yd}^2
\]
Answer: \( 77 \, \text{yd}^2 \)
---
Problem 6:
-
Dimensions:
- Base 1 = 13 cm
- Base 2 = 13 cm (since it's a parallelogram)
- Height = 8 cm
For a parallelogram, the area is:
\[
\text{Area} = \text{Base} \times \text{Height}
\]
\[
\text{Area} = 13 \times 8 = 104 \, \text{cm}^2
\]
Answer: \( 104 \, \text{cm}^2 \)
---
Problem 7:
-
Dimensions:
- Base 1 = 9 mm
- Base 2 = 4 mm
- Height = 2 mm
\[
\text{Area} = \frac{1}{2} \times (9 + 4) \times 2 = \frac{1}{2} \times 13 \times 2 = 13 \, \text{mm}^2
\]
Answer: \( 13 \, \text{mm}^2 \)
---
Problem 8:
-
Dimensions:
- Base 1 = 6 yd
- Base 2 = 2 yd
- Height = 19 yd
\[
\text{Area} = \frac{1}{2} \times (6 + 2) \times 19 = \frac{1}{2} \times 8 \times 19 = 76 \, \text{yd}^2
\]
Answer: \( 76 \, \text{yd}^2 \)
---
Final Answers:
1. \( 50 \, \text{cm}^2 \)
2. \( 92 \, \text{in}^2 \)
3. \( 50 \, \text{m}^2 \)
4. \( 27.5 \, \text{km}^2 \)
5. \( 77 \, \text{yd}^2 \)
6. \( 104 \, \text{cm}^2 \)
7. \( 13 \, \text{mm}^2 \)
8. \( 76 \, \text{yd}^2 \)
\boxed{50, 92, 50, 27.5, 77, 104, 13, 76}
Parent Tip: Review the logic above to help your child master the concept of quadrilateral area worksheet.
