1. **Problem Statement:** Find the area and perimeter of two trapezoids with given dimensions. --- ### a. Trapezoid with bases 14 cm and 20.3 cm, height 12 cm 2. **Formulas:** - Area of trapezoid: $$A = \frac{(b_1 + b_2)}{2} \times h$$ where $b_1$ and $b_2$ are the lengths of the two bases, and $h$ is the height. - Perimeter: sum of all sides. 3. **Calculate area:** $$A = \frac{(14 + 20.3)}{2} \times 12$$ $$= \frac{34.3}{2} \times 12$$ $$= 17.15 \times 12$$ $$= 205.8\text{ cm}^2$$ 4. **Calculate perimeter:** We know two bases: 14 cm and 20.3 cm, and height 12 cm (which is perpendicular). To find the non-parallel side length, use the Pythagorean theorem: The difference between bases is $20.3 - 14 = 6.3$ cm. The slant side length $s = \sqrt{12^2 + 6.3^2} = \sqrt{144 + 39.69} = \sqrt{183.69} \approx 13.55$ cm. Perimeter $P = 14 + 20.3 + 12 + 13.55 = 59.85$ cm. --- ### b. Trapezoid with bases 17.8 m and 7.5 m, height 4 m, other sides 6 m and 7 m 5. **Calculate area:** $$A = \frac{(17.8 + 7.5)}{2} \times 4$$ $$= \frac{25.3}{2} \times 4$$ $$= 12.65 \times 4$$ $$= 50.6\text{ m}^2$$ 6. **Calculate perimeter:** Sum all sides: $17.8 + 7.5 + 6 + 7 = 38.3$ m. --- **Final answers:** - a) Area = $205.8\text{ cm}^2$, Perimeter = $59.85$ cm - b) Area = $50.6\text{ m}^2$, Perimeter = $38.3$ m