Exam drill: geometric themes

6.1 Ten questions, full solutions

These questions are written in the style of a chapter test. Fractals, solids and nets are mixed together, and each solution shows the formula, the substitution and the final statement. Attempt the question before reading the answer. Keep the working short: write the formula, substitute once, simplify and then say what the answer means. If a number looks surprising, test the same method on a smaller example before moving on, because the small case usually shows which count or factor went wrong.

Question 1. A Sierpinski carpet begins with area 1. Find the area remaining after 3 steps and the number of holes at that step.

Answer. The area factor is 8/9 per step, so the remaining area is (8/9)³ = 512/729. The holes at step 3 number (8³ − 1)/7 = (512 − 1)/7 = 73. The two answers describe different quantities: one is area, the other is a count of removed squares.

Question 2. A Sierpinski triangle has area 16 cm² at step 0. What area remains after 4 steps?

Answer. Each step keeps 3/4 of the area, so the area after 4 steps is 16 × (3/4)⁴ = 16 × 81/256 = 81/16 = 5.0625 cm². The shape still contains many small triangles even though its area is shrinking.

Question 3. For a Koch snowflake beginning with a side of 1 unit, find the number of sides and the perimeter at step 5.

Answer. S₅ = 3 × 4⁵ = 3 × 1,024 = 3,072 sides. P₅ = 3 × (4/3)⁵ = 3 × 1,024/243 = 1,024/81 ≈ 12.64 units. The perimeter keeps growing, but the figure remains within a bounded region.

Question 4. A prism has a pentagonal base. Find F, E and V by formula, and verify Euler's relation.

Answer. A prism with an n-sided base has F = n + 2, E = 3n and V = 2n. For n = 5, F = 7, E = 15 and V = 10. Euler check: 10 − 15 + 7 = 2.

Question 5. A convex polyhedron has 8 faces and 12 vertices. Find its number of edges and name a possible solid.

Answer. Euler's relation gives E = V + F − 2 = 12 + 8 − 2 = 18 edges. The truncated tetrahedron fits: it has 4 triangular faces and 4 hexagonal faces (8 faces), 12 vertices and 18 edges, and 12 − 18 + 8 = 2. A square pyramid has 5 faces, 8 edges and 5 vertices, so it does not fit.

Question 6. Explain why every net of a cube has perimeter 14 units.

Answer. A cube net contains six unit squares joined along five edges to stay in one piece without overlap. The six squares alone have perimeter 24 units, and each of the five shared edges deletes two units from the outer boundary, leaving 24 − 10 = 14 units. The argument works for each of the eleven different nets.

Question 7. A cuboid measures 8 cm by 5 cm by 2 cm. Find its surface area and the dimensions of the rectangles in a net.

Answer. Surface area = 2(8×5 + 8×2 + 5×2) = 2(40 + 16 + 10) = 132 cm². A net contains two rectangles of each size: 8×5, 8×2 and 5×2, because each pair of opposite faces must appear once.

Question 8. Two opposite vertices of a cuboid measuring 10 cm by 8 cm by 6 cm are joined by a path on the surface. Find the length of the shortest such path.

Answer. Unfold the cuboid in the three possible ways and apply the Baudhayana theorem. The three candidates are √((10+8)² + 6²) = √360, √((10+6)² + 8²) = √320 and √((8+6)² + 10²) = √296. The shortest is √296 = 2√74 ≈ 17.2 cm.

Question 9. A line segment of length 12 cm makes an angle of 30° with a plane. Find the length of its projection on the plane, and state when the projection equals the segment.

Answer. The projection length is p = l cos θ = 12 × cos 30° = 12 × (√3/2) = 6√3 ≈ 10.39 cm. The projection equals the full length when the segment lies parallel to the plane, that is, when θ = 0°.

Question 10. Describe the front, top and side views of an upright cylinder, and count the cubes in a stack with layers of 3 × 3, 2 × 2 and 1 cube.

Answer. The front view of an upright cylinder is a rectangle; the top view is a circle; the side view is a rectangle congruent to the front view. The cube stack has 9 + 4 + 1 = 14 cubes, counted layer by layer from the bottom up.