To solve the problem, we need to calculate the areas of the given triangles and trapezoids using their respective formulas. Let's go through each figure step by step.
Formula Recap:
1.
Area of a Triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
2.
Area of a Trapezoid:
\[
\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}
\]
---
Figure 1: Trapezoid
-
Dimensions:
- Parallel sides: \(10 \, \text{cm}\) and \(16 \, \text{cm}\)
- Height: \(9.0 \, \text{cm}\)
\[
\text{Area} = \frac{1}{2} \times (10 + 16) \times 9.0 = \frac{1}{2} \times 26 \times 9.0 = 13 \times 9.0 = 117 \, \text{cm}^2
\]
---
Figure 2: Triangle
-
Dimensions:
- Base: \(9 \, \text{yd}\)
- Height: \(8 \, \text{yd}\)
\[
\text{Area} = \frac{1}{2} \times 9 \times 8 = \frac{1}{2} \times 72 = 36 \, \text{yd}^2
\]
---
Figure 3: Trapezoid
-
Dimensions:
- Parallel sides: \(12 \, \text{yd}\) and \(22 \, \text{yd}\)
- Height: \(15 \, \text{yd}\)
\[
\text{Area} = \frac{1}{2} \times (12 + 22) \times 15 = \frac{1}{2} \times 34 \times 15 = 17 \times 15 = 255 \, \text{yd}^2
\]
---
Figure 4: Trapezoid
-
Dimensions:
- Parallel sides: \(6 \, \text{yd}\) and \(21.5 \, \text{yd}\)
- Height: \(9 \, \text{yd}\)
\[
\text{Area} = \frac{1}{2} \times (6 + 21.5) \times 9 = \frac{1}{2} \times 27.5 \times 9 = 13.75 \times 9 = 123.75 \, \text{yd}^2
\]
---
Figure 5: Triangle
-
Dimensions:
- Base: \(70 \, \text{mm}\)
- Height: \(50 \, \text{mm}\)
\[
\text{Area} = \frac{1}{2} \times 70 \times 50 = \frac{1}{2} \times 3500 = 1750 \, \text{mm}^2
\]
---
Figure 6: Trapezoid
-
Dimensions:
- Parallel sides: \(170 \, \text{yd}\) and \(240 \, \text{yd}\)
- Height: \(150 \, \text{yd}\)
\[
\text{Area} = \frac{1}{2} \times (170 + 240) \times 150 = \frac{1}{2} \times 410 \times 150 = 205 \times 150 = 30750 \, \text{yd}^2
\]
---
Figure 7: Trapezoid
-
Dimensions:
- Parallel sides: \(22.5 \, \text{in}\) and \(24.5 \, \text{in}\)
- Height: \(12.5 \, \text{in}\)
\[
\text{Area} = \frac{1}{2} \times (22.5 + 24.5) \times 12.5 = \frac{1}{2} \times 47 \times 12.5 = 23.5 \times 12.5 = 293.75 \, \text{in}^2
\]
---
Figure 8: Trapezoid
-
Dimensions:
- Parallel sides: \(11.5 \, \text{cm}\) and \(21.5 \, \text{cm}\)
- Height: \(7 \, \text{cm}\)
\[
\text{Area} = \frac{1}{2} \times (11.5 + 21.5) \times 7 = \frac{1}{2} \times 33 \times 7 = 16.5 \times 7 = 115.5 \, \text{cm}^2
\]
---
Final Answers:
\[
\boxed{
\begin{array}{ll}
1. & 117 \, \text{cm}^2 \\
2. & 36 \, \text{yd}^2 \\
3. & 255 \, \text{yd}^2 \\
4. & 123.75 \, \text{yd}^2 \\
5. & 1750 \, \text{mm}^2 \\
6. & 30750 \, \text{yd}^2 \\
7. & 293.75 \, \text{in}^2 \\
8. & 115.5 \, \text{cm}^2 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of area of rectangles and triangles worksheet.
