1. **Problem statement:** ABCD is a four-sided shape but two of the sides cross. 2. **Part a:** Explain why the sum of the angles at A, B, C, and D must be less than 360°. 3. When two sides cross, the figure is a *complex quadrilateral* or a *self-intersecting polygon*. 4. In such a figure, the interior angles at the vertices do not form a simple closed loop, so the sum of the angles at A, B, C, and D is less than the sum of interior angles of a simple quadrilateral (which is 360°). 5. This happens because the crossing creates overlapping regions, effectively reducing the total angle sum at the vertices. 6. **Part b:** Find the sum of the angles at A, B, C, and D and give a reason. 7. The sum of the interior angles of any quadrilateral (simple or complex) is given by the formula: $$\text{Sum of interior angles} = (n-2) \times 180^\circ$$ where $n=4$ for a quadrilateral. 8. So, for a quadrilateral: $$\text{Sum} = (4-2) \times 180^\circ = 2 \times 180^\circ = 360^\circ$$ 9. However, for a self-intersecting quadrilateral, the sum of the angles at the vertices is less than 360° because the figure folds over itself. 10. To find the exact sum, consider the exterior angles or use the fact that the sum of the exterior angles of any polygon is 360°. 11. Since the sides cross, the sum of the interior angles at A, B, C, and D is less than 360°, but the sum of the exterior angles remains 360°. 12. Therefore, the sum of the angles at A, B, C, and D is less than 360°, consistent with the crossing sides. **Final answers:** - a) The sum of the angles at A, B, C, and D is less than 360° because the figure is self-intersecting, causing overlapping and reducing the total interior angle sum. - b) The sum of the angles at A, B, C, and D is less than 360°, as explained by the nature of the crossed sides and the polygon's geometry.