Mixed practice: fractals and solids

5.1 Working between the two themes

The chapter has two halves: fractals, where a pattern repeats at smaller scales, and solids, where faces, edges and vertices describe the shape. This page mixes the two. The formulas are the ones collected on the revision page, and the questions use fresh data, so the page bridges the exercises and the final revision.

For each question, write the formula first, substitute the numbers, and then say in one sentence what the answer means. That habit catches most slips in this chapter.

5.1.1 Fractal formulas

Question 1. A Sierpinski carpet starts from a square of area 1. Find the number of remaining squares, the number of holes and the area of the remaining region after 4 steps.

Answer. The remaining squares multiply by 8 at every step, so R₄ = 8⁴ = 4,096. The holes follow Hₙ = (8ⁿ − 1)/7, so H₄ = (4,096 − 1)/7 = 4,095/7 = 585. The area keeps only 8/9 of the previous area at each step, so the area after 4 steps is (8/9)⁴ = 4,096/6,561 ≈ 0.624.

Question 2. For a Sierpinski gasket built from a triangle of side 1 unit and area 1 square unit, find the number of small triangles, the side of each small triangle, the total length of the boundary and the remaining area after 5 steps.

Answer. The count is 3⁵ = 243 triangles. Each step halves the side, so the side after 5 steps is (1/2)⁵ = 1/32 unit. Every small triangle contributes three boundary edges, so the boundary length is 3 × 243 × 1/32 = 729/32 ≈ 22.8 units. The area keeps 3/4 at each step, so the remaining area is (3/4)⁵ = 243/1,024 ≈ 0.237 square units.

Question 3. A Koch snowflake starts from an equilateral triangle of side 1 unit. Find the number of sides and the perimeter after 4 steps, and explain what happens to the perimeter as the step number grows.

Answer. The number of sides is 3 × 4⁴ = 768. The perimeter is 3 × (4/3)⁴ = 3 × 256/81 = 256/27 ≈ 9.48 units. At every step each side is replaced by four sides of length one third of the old side, so the perimeter is multiplied by 4/3. Since 4/3 is greater than 1, the perimeter grows without bound, while the snowflake stays inside a finite region.

Question 4. A Sierpinski carpet step has 4,681 holes. Which step is it, and how many squares remain?

Answer. Solve (8ⁿ − 1)/7 = 4,681. Multiplying by 7 gives 8ⁿ − 1 = 32,767, so 8ⁿ = 32,768 = 8⁵ and n = 5. The number of remaining squares is 8⁵ = 32,768. The step is the fifth one.

5.1.2 Euler's relation for solids

For a convex solid with no holes, the numbers of vertices V, edges E and faces F satisfy V − E + F = 2. The relation is a check on any count of a solid, and it can find a missing value when the other two are known.

Question 5. Verify Euler's relation for a cube, a regular tetrahedron and a regular octahedron, and use it to find the edges of a solid with 12 faces and 20 vertices.

Answer. A cube has V = 8, E = 12 and F = 6, and 8 − 12 + 6 = 2. A regular tetrahedron has V = 4, E = 6 and F = 4, and 4 − 6 + 4 = 2. A regular octahedron has V = 6, E = 12 and F = 8, and 6 − 12 + 8 = 2. For the solid with 12 faces and 20 vertices, E = V + F − 2 = 20 + 12 − 2 = 30, which is the regular dodecahedron: twelve pentagonal faces, twenty vertices and thirty edges.

Question 6. A prism has 15 edges. Find the number of sides of its base, then its faces and vertices, and check Euler's relation.

Answer. A prism with an n-sided base has E = 3n, so 3n = 15 gives n = 5. It has F = n + 2 = 7 faces and V = 2n = 10 vertices. Euler's relation: 10 − 15 + 7 = 2, so the counts are consistent.

Question 7. A pyramid has 12 faces. Find its base, its edges and its vertices, and check Euler's relation.

Answer. A pyramid with an n-sided base has F = n + 1, so n = 11. It has E = 2n = 22 edges and V = n + 1 = 12 vertices. Euler's relation: 12 − 22 + 12 = 2, so the counts agree.

5.1.3 Nets and folding

A net is a flat pattern that folds into a solid. Every face of the solid appears once in the net, and when the net is folded two matching edges meet.

Question 8. A cuboid measures 7 cm by 4 cm by 2 cm. Find its surface area, and state one thing every correct net of this cuboid must contain.

Answer. The surface area is 2(7×4 + 7×2 + 4×2) = 2(28 + 14 + 8) = 100 cm². Every correct net must contain six rectangles: two of each of the sizes 7×4, 7×2 and 4×2, arranged so that the matching edges meet when folded.

Question 9. All eleven nets of a cube are made of six unit squares joined edge to edge. Show that the perimeter of every such net is 14 units.

Answer. Six separate squares have total perimeter 6 × 4 = 24 units. In a cube net the squares are joined along five internal edges, and each shared edge removes two units from the outside boundary. So the perimeter is 24 − 2 × 5 = 14 units.

Question 10. A square pyramid has a base of side 6 cm and each triangular face has slant height 5 cm. Find the area of its net.

Answer. The net contains the square base of area 6 × 6 = 36 cm² and four congruent triangles, each of area ½ × 6 × 5 = 15 cm². The total area of the net is 36 + 4 × 15 = 96 cm².