1. **State the problem:** We have a polygon ABCDEF with given conditions: BCD is a straight line, CA is parallel to DF, angles ABD, BDE, and DEF are right angles, and lengths AB = 4 m, DE = 11 m, EF = 4 m. We need to show that angle FDE = x° equals 20.0° correct to 3 significant figures. 2. **Show the formula and rules:** Since ABD, BDE, and DEF are right angles, triangle DEF is right angled at E. We can use trigonometric ratios in triangle DEF to find angle x = FDE. 3. **Intermediate work:** In right triangle DEF, with right angle at E: - DE = 11 m (adjacent to angle x) - EF = 4 m (opposite to angle x) Using tangent: $$\tan(x) = \frac{\text{opposite}}{\text{adjacent}} = \frac{EF}{DE} = \frac{4}{11}$$ Calculate angle x: $$x = \tan^{-1}\left(\frac{4}{11}\right)$$ Using a calculator: $$x \approx 20.0^\circ$$ (to 3 significant figures) 4. **Answer for (a):** $$x = 20.0^\circ$$ --- 5. **Given that x = y, calculate AC:** Since BCD is a straight line and angles ABD and BDE are right angles, triangle ABC is right angled at B and triangle BDE is right angled at D. Using the fact that CA is parallel to DF and x = y, triangles ABC and DEF are similar. The ratio of corresponding sides is: $$\frac{AB}{DE} = \frac{4}{11}$$ Since AC corresponds to EF in the similar triangles: $$\frac{AC}{EF} = \frac{AB}{DE}$$ Therefore: $$AC = EF \times \frac{AB}{DE} = 4 \times \frac{4}{11} = \frac{16}{11} \approx 1.45\,m$$ 6. **Calculate the area of ACDF:** ACDF is a trapezium with parallel sides AC and DF, and perpendicular distance between them is 7 m. Length of DF is: $$DF = DE + EF = 11 + 4 = 15\,m$$ Area of trapezium: $$\text{Area} = \frac{1}{2} \times (AC + DF) \times \text{distance} = \frac{1}{2} \times (1.45 + 15) \times 7$$ Calculate: $$= \frac{1}{2} \times 16.45 \times 7 = 8.225 \times 7 = 57.575\,m^2$$ Rounded to 3 significant figures: $$57.6\,m^2$$ **Final answers:** - (a) $x = 20.0^\circ$ - (b) $AC \approx 1.45\,m$ - Area of ACDF $\approx 57.6\,m^2$