Isosceles Trapezium Area 713A76

1. **State the problem:** We have trapezium ABCD with one line of symmetry, where $DC = 47$ cm, $AD = 12$ cm, and $\angle ADC = 60^\circ$. We need to find the area of the trapezium. 2. **Understand the trapezium:** Since ABCD has one line of symmetry and $AD = BC$, the trapezium is isosceles with $AB \parallel DC$. 3. **Draw height from A and B to DC:** Let the height be $h$. Drop perpendiculars from $A$ and $B$ to $DC$ at points $E$ and $F$ respectively. Because of symmetry, $E$ and $F$ divide $DC$ into three parts: $DE$, $EF$, and $FC$, with $DE = FC$. 4. **Calculate height $h$ using triangle $ADC$:** In triangle $ADC$, $AD = 12$ cm, $\angle ADC = 60^\circ$, and $DC = 47$ cm. The height $h$ is the perpendicular from $A$ to $DC$, so: $$h = AD \times \sin 60^\circ = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.392$$ 5. **Calculate base segment $DE$:** The horizontal distance from $D$ to $E$ is: $$DE = AD \times \cos 60^\circ = 12 \times \frac{1}{2} = 6$$ 6. **Calculate length of $AB$:** Since $DC = 47$ cm and $DE = FC = 6$ cm, the length of $AB$ is: $$AB = DC - DE - FC = 47 - 6 - 6 = 35$$ 7. **Calculate area of trapezium:** Area formula: $$\text{Area} = \frac{(AB + DC)}{2} \times h = \frac{(35 + 47)}{2} \times 10.392 = \frac{82}{2} \times 10.392 = 41 \times 10.392 = 425.672$$ 8. **Final answer rounded to 3 significant figures:** $$\boxed{426}$$ cm$^2$