1. Problem 12a: Show that it is possible for 2 squares and 3 equilateral triangles to meet at one point. - Each square has an internal angle of $90^\circ$. - Each equilateral triangle has an internal angle of $60^\circ$. - At one point, the sum of the angles meeting must be $360^\circ$. - Calculate the total angle: $2 \times 90^\circ + 3 \times 60^\circ = 180^\circ + 180^\circ = 360^\circ$. - Since the sum is exactly $360^\circ$, it is possible for 2 squares and 3 equilateral triangles to meet at one point. 2. Problem 12b: Draw a different way for 2 squares and 3 equilateral triangles with sides the same length to meet at one point. - Since the shapes have the same side length, you can arrange them in a different order around the point. - For example, alternate the shapes: triangle, square, triangle, square, triangle. - The angle sum remains $360^\circ$ because the number of each shape is the same. - This different sequence still satisfies the angle sum condition. 3. Problem 12c: Can you work out a third way for 2 squares and 3 equilateral triangles to meet at one point? Give a reason for your answer. - The total angle sum must be $360^\circ$. - The only way to have 2 squares and 3 equilateral triangles meeting at a point is to have their angles sum to $360^\circ$. - Since the angles are fixed, the only variations are in the order of shapes around the point. - Therefore, any third way is just a permutation of the order of the 2 squares and 3 triangles. - The reason is that the angle measures and counts fix the total sum, limiting the possible arrangements. 4. Problem 13a: Draw a tessellation of equilateral triangles. - Equilateral triangles tile the plane because their internal angle $60^\circ$ divides $360^\circ$ evenly. - Six triangles meet at each vertex. 5. Problem 13b: Draw a tessellation of regular hexagons. - Regular hexagons have internal angles of $120^\circ$. - Three hexagons meet at each vertex: $3 \times 120^\circ = 360^\circ$. - This allows a perfect tessellation. 6. Problem 13c: Explain why it is not possible to draw a tessellation of regular pentagons. - Regular pentagons have internal angles of $108^\circ$. - $360^\circ / 108^\circ \approx 3.33$, not an integer. - Therefore, pentagons cannot meet at a point without gaps or overlaps. - Hence, regular pentagons cannot tessellate the plane. 7. Problem 13d: Draw a tessellation of squares and equilateral triangles. Is there more than one way to do this? - Squares ($90^\circ$) and equilateral triangles ($60^\circ$) can combine to fill $360^\circ$ at vertices. - For example, two squares and one triangle: $2 \times 90^\circ + 60^\circ = 240^\circ$ (not $360^\circ$). - But combinations like 1 square and 2 triangles: $90^\circ + 2 \times 60^\circ = 210^\circ$ (not $360^\circ$). - The combination 2 squares and 3 triangles (from problem 12) works. - There are multiple ways to arrange these shapes to tessellate, so yes, more than one way exists. 8. Problem 13e: Draw a tessellation using regular octagons and squares. - Regular octagons have internal angles of $135^\circ$. - Squares have $90^\circ$. - At each vertex, one octagon and two squares meet: $135^\circ + 2 \times 90^\circ = 315^\circ$ (not $360^\circ$). - But the known tessellation is with one octagon and one square alternating around vertices. - Actually, the classic tessellation is with one octagon and one square alternating, repeated around the vertex. - The angle sum at vertex is $135^\circ + 90^\circ + 135^\circ + 90^\circ = 450^\circ$ which is too large. - The correct arrangement is one octagon and two squares meeting at a vertex with gaps filled by other shapes or adjusted arrangement. - The known semi-regular tessellation uses octagons and squares fitting perfectly. 9. Problem 13f: What other tessellations can you draw using regular polygons? - Regular triangles, squares, and hexagons tessellate alone. - Semi-regular tessellations combine these polygons in patterns where angles sum to $360^\circ$ at vertices. - Examples include combinations like triangle-hexagon, square-octagon, triangle-square-hexagon, etc. - The key is that the sum of the internal angles of polygons meeting at a vertex equals $360^\circ$. Final answers: - 12a: Yes, because $2 \times 90^\circ + 3 \times 60^\circ = 360^\circ$. - 12b: Different orderings of the same shapes around the point. - 12c: Only permutations of the order are possible due to fixed angles. - 13a: Equilateral triangles tessellate by six meeting at a point. - 13b: Regular hexagons tessellate by three meeting at a point. - 13c: Regular pentagons cannot tessellate because $108^\circ$ does not divide $360^\circ$ evenly. - 13d: Multiple tessellations with squares and triangles exist. - 13e: Octagons and squares tessellate in a semi-regular pattern. - 13f: Other tessellations include combinations of triangles, squares, hexagons, octagons, etc., where angles sum to $360^\circ$ at vertices.