Trapezium Area 48D1D4

1. **State the problem:** We have trapezium ABCD with one line of symmetry, where $AD=12$ cm, $DC=47$ cm, and $\angle ADC=60^\circ$. We need to find the area of trapezium ABCD. 2. **Understand the trapezium and symmetry:** Since ABCD has one line of symmetry, the top side $AB$ is parallel to $DC$ and centered above it. The trapezium is isosceles, so $AD=BC=12$ cm. 3. **Find the height of the trapezium:** The height $h$ is the perpendicular distance between the parallel sides $AB$ and $DC$. Using triangle $ADC$, drop a perpendicular from $A$ to $DC$ at point $E$. Since $\angle ADC=60^\circ$ and $AD=12$ cm, the height is: $$h = AD \times \sin 60^\circ = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.3923 \text{ cm}$$ 4. **Find the length of $DE$:** $$DE = AD \times \cos 60^\circ = 12 \times \frac{1}{2} = 6 \text{ cm}$$ 5. **Find the length of $AB$:** Since $DC=47$ cm and the trapezium is symmetric, the top base $AB$ is shorter by twice $DE$: $$AB = DC - 2 \times DE = 47 - 2 \times 6 = 47 - 12 = 35 \text{ cm}$$ 6. **Calculate the area of trapezium:** The area formula for trapezium is: $$\text{Area} = \frac{(AB + DC)}{2} \times h$$ Substitute values: $$\text{Area} = \frac{(35 + 47)}{2} \times 10.3923 = \frac{82}{2} \times 10.3923 = 41 \times 10.3923 = 426.0583 \text{ cm}^2$$ 7. **Round to 3 significant figures:** $$\boxed{426 \text{ cm}^2}$$