Edge โ a line where two faces meet.
Vertex โ a corner where edges meet. Plural: vertices.
| Solid | Faces | Edges | Vertices |
|---|---|---|---|
| Cube | $6$ (all squares) | $12$ | $8$ |
| Cuboid | $6$ (rectangles) | $12$ | $8$ |
| Triangular prism | $5$ ($2$ triangles, $3$ rectangles) | $9$ | $6$ |
| Square-based pyramid | $5$ ($1$ square, $4$ triangles) | $8$ | $5$ |
| Tetrahedron (triangular pyramid) | $4$ (all triangles) | $6$ | $4$ |
| Hexagonal prism | $8$ | $18$ | $12$ |
| Cylinder | $3$ ($2$ flat circles, $1$ curved) | $2$ (curved) | $0$ |
| Cone | $2$ ($1$ flat circle, $1$ curved) | $1$ (curved) | $1$ (the apex) |
| Sphere | $1$ (curved) | $0$ | $0$ |
A polyhedron has $12$ faces and $30$ edges. How many vertices does it have?
(This is a dodecahedron.)
A triangular prism has a cross-section that is a right-angled triangle with legs $6$ cm and $8$ cm. The prism is $15$ cm long. Find its volume.
| Solid | Description | Volume |
|---|---|---|
| Pyramid | A flat base with triangular faces meeting at an apex | $\tfrac{1}{3} \times$ base area $\times$ height |
| Cone | A circular base tapering to an apex | $\tfrac{1}{3}\pi r^2 h$ |
| Sphere | Every point the same distance from the centre | $\tfrac{4}{3}\pi r^3$ |
| Hemisphere | Half a sphere | $\tfrac{2}{3}\pi r^3$ |
A cone has base radius $9$ cm and vertical height $12$ cm. Find its slant height and its volume.
A net is the 2D shape you get by unfolding a solid and laying it flat. It shows every face at its true size, which makes nets the easiest route to a surface area.
| Solid | Net consists of |
|---|---|
| Cube | $6$ identical squares |
| Cuboid | $3$ pairs of matching rectangles |
| Triangular prism | $2$ triangles $+$ $3$ rectangles |
| Square-based pyramid | $1$ square $+$ $4$ triangles |
| Cylinder | $2$ circles $+$ $1$ rectangle of width $2\pi r$ |
| Cone | $1$ circle $+$ $1$ sector of radius $l$ |
A cuboid measures $5$ cm by $3$ cm by $2$ cm. Use its net to find the surface area.
A plane of symmetry is a flat surface that slices a solid into two mirror-image halves.
| Solid | Planes of symmetry |
|---|---|
| Cube | $9$ |
| Cuboid (all edges different) | $3$ |
| Triangular prism (equilateral cross-section) | $4$ |
| Cylinder | Infinitely many (plus one horizontal) |
| Cone | Infinitely many (all vertical, through the apex) |
| Sphere | Infinitely many |
| Square-based pyramid | $4$ |
Explain why a cuboid with all three edge lengths different has exactly $3$ planes of symmetry.
So there are exactly $3$. A cube has more ($9$) because its diagonal slices do produce matching halves.
Face, edge, vertex
Surface, line where faces meet, corner.
Cube / cuboid
$6$ faces, $12$ edges, $8$ vertices.
Euler
$F + V - E = 2$ for solids with flat faces.
Prism
Constant cross-section; $V = $ area $\times$ length.
Pyramid / cone
$V = \tfrac{1}{3} \times$ base area $\times$ height.
Sphere
$V = \tfrac{4}{3}\pi r^3$, surface $= 4\pi r^2$.
Cone heights
$l^2 = r^2 + h^2$ links slant and vertical height.
Nets
Unfold the solid; add the face areas for the surface area.
Cylinder net
$2$ circles plus a rectangle of width $2\pi r$.
State the number of faces, edges and vertices of a square-based pyramid.
โถ Show solution
Faces: $5$ โ one square base and four triangles.
Edges: $8$ โ four around the base and four sloping up.
Vertices: $5$ โ four at the base and one apex.
Check with Euler: $5 + 5 - 8 = 2$ โ
A polyhedron has $8$ vertices and $12$ edges. How many faces does it have?
โถ Show solution
$F + V - E = 2$
$F + 8 - 12 = 2$
$F = 6$ (it could be a cube or cuboid).
Explain why a cone is not a prism.
โถ Show solution
A prism must have the same cross-section all the way through.
Slicing a cone parallel to its base gives circles, but they get smaller and smaller towards the apex.
The cross-section changes, so a cone is not a prism.
A hexagonal prism has how many (a) faces, (b) edges, (c) vertices? Verify with Euler's formula.
โถ Show solution
(a) $2$ hexagonal ends $+ 6$ rectangles $= 8$ faces.
(b) $6$ edges on each hexagon $+ 6$ joining edges $= 18$ edges.
(c) $6$ corners on each hexagon $= 12$ vertices.
Euler: $8 + 12 - 18 = 2$ โ
A cone has radius $5$ cm and slant height $13$ cm. Find its vertical height.
โถ Show solution
$l^2 = r^2 + h^2$, so $h^2 = l^2 - r^2$.
$h^2 = 169 - 25 = 144$
$h = 12$ cm
A prism has a trapezium cross-section with parallel sides $7$ cm and $11$ cm and perpendicular height $4$ cm. The prism is $20$ cm long. Find its volume.
โถ Show solution
Cross-section area $= \tfrac{1}{2}(7 + 11) \times 4 = \tfrac{1}{2} \times 18 \times 4 = 36\text{ cm}^2$
$V = 36 \times 20 = 720\text{ cm}^3$
Describe the net of a triangular prism whose cross-section is an equilateral triangle of side $6$ cm and whose length is $10$ cm.
โถ Show solution
The net consists of:
โข Two equilateral triangles of side $6$ cm (the two ends).
โข Three rectangles, each $6$ cm by $10$ cm (the three sloping/flat faces), joined edge to edge in a row.
The three rectangles form a single strip $18$ cm by $10$ cm, with a triangle attached to the top and bottom of the middle rectangle.
How many planes of symmetry does (a) a cube have, (b) a cylinder have?
โถ Show solution
(a) A cube has $9$: three parallel to the faces (slicing it in half each way) and six through pairs of opposite edges (the diagonal slices).
(b) A cylinder has infinitely many: any vertical plane through the central axis, plus one horizontal plane halfway up.
A cuboid measures $8$ cm by $6$ cm by $5$ cm. Find (a) its volume, (b) its surface area, (c) the length of a space diagonal (corner to opposite corner), to 1 d.p.
โถ Show solution
(a) $V = 8 \times 6 \times 5 = 240\text{ cm}^3$
(b) $2(8\times6) + 2(8\times5) + 2(6\times5) = 96 + 80 + 60 = 236\text{ cm}^2$
(c) Use Pythagoras twice. Base diagonal$^2 = 8^2 + 6^2 = 100$.
Space diagonal$^2 = 100 + 5^2 = 125$
Space diagonal $= \sqrt{125} = 11.2$ cm (1 d.p.)
A solid is made from a cylinder of radius $4$ cm and height $10$ cm with a cone of the same radius and vertical height $3$ cm on top.
(a) Find the total volume, in terms of $\pi$. (b) Find the slant height of the cone. (c) Find the total surface area of the solid, to 1 d.p.
โถ Show solution
(a) Cylinder: $\pi r^2 h = \pi \times 16 \times 10 = 160\pi$
Cone: $\tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times \pi \times 16 \times 3 = 16\pi$
Total $= 160\pi + 16\pi = \mathbf{176\pi\text{ cm}^3}$
(b) $l^2 = 4^2 + 3^2 = 25$, so $l = 5$ cm.
(c) The surfaces are: the flat circular base, the curved cylinder, and the curved cone. (The top of the cylinder is hidden by the cone.)
Base circle: $\pi r^2 = 16\pi$
Cylinder curved surface: $2\pi r h = 2\pi \times 4 \times 10 = 80\pi$
Cone curved surface: $\pi r l = \pi \times 4 \times 5 = 20\pi$
Total $= 16\pi + 80\pi + 20\pi = 116\pi = \mathbf{364.4\text{ cm}^2}$ (1 d.p.)