Year 7 Mathematics | Victorian Curriculum 2.0
3D objects
Topic 11 | Measurement & Space | Answer key

Year 7 core - answers

Fluency

Naming, faces, edges, vertices

    1. Cube: F=6F = 6, E=12E = 12, V=8V = 8.
    2. Triangular prism: F=5F = 5, E=9E = 9, V=6V = 6.
    3. Square pyramid: F=5F = 5, E=8E = 8, V=5V = 5.
    4. Triangular pyramid (tetrahedron).
    5. Hexagonal prism.
    6. 55 rectangular faces (plus 22 pentagonal ends).
    7. 44 triangular faces.
Fluency

Nets

    1. 1111.
    2. 22 triangles and 33 rectangles.
    3. 66 rectangles: two 3×23 \times 2, two 3×43 \times 4, two 2×42 \times 4.
    4. A 44 cm square with four isosceles triangles attached to each side, each triangle with base 44 cm and slant height 55 cm.
    5. No. Six squares in a single row overlap when folded - they cannot form a closed cube.
Reasoning

Explain and spot the mistake

    1. Not true. A pyramid’s side-faces are always triangles, but the base can be any polygon (square, pentagon, etc.). “Triangular pyramid” is one particular type.
    2. A triangular prism has 55 faces: 22 triangular ends and 33 rectangular sides. Sam probably mixed faces with edges (which total 99) or counted the same face twice.
    3. Strictly no - a prism has a polygon base joined by flat rectangular sides. A cylinder has a circular base and a curved surface, not a polygon. (It is often informally called a circular prism because the volume formula V=base area×heightV = \text{base area} \times \text{height} still applies.)
    4. A top view (plan) or a side elevation would resolve the ambiguity, since those show lengths directly without the isometric distortion.
Problem solving

Real-world problems

    1. Surface area 20002000 cm^2; volume 48004800 cm^3. Method: SA =2(30×20+30×8+20×8)=2(600+240+160)=2000= 2(30 \times 20 + 30 \times 8 + 20 \times 8) = 2(600 + 240 + 160) = 2000; V=30×20×8V = 30 \times 20 \times 8.
    2. 22 equilateral triangles (side 66 cm) and 33 rectangles (66 cm by 2424 cm).
    3. 144144 cm^2. Method: base 82=648^2 = 64; four triangles each 12×8×5=20\tfrac{1}{2} \times 8 \times 5 = 20; total =64+4×20= 64 + 4 \times 20.
    4. 1212 m. Method: a cube has 1212 edges, each 11 m.

Extension - answers

Reasoning

Using the formulas

    1. V=14V = 14, E=21E = 21, F=9F = 9. Method: n=7n = 7; V=2nV = 2n, E=3nE = 3n, F=n+2F = n + 2.
    2. V=7V = 7, E=12E = 12, F=7F = 7. Method: n=6n = 6; V=n+1V = n + 1, E=2nE = 2n, F=n+1F = n + 1.
    3. F=12F = 12. Method: V−E+F=2V - E + F = 2.
    4. E=12E = 12. Method: 6−E+8=26 - E + 8 = 2, so E=12E = 12.
    5. Regular octahedron.
Year 7 Mathematics study companion | Answer key