In this unit students make and investigate various solids, including regular and semi-regular polyhedra, and cylinders and cones. They look for patterns in the numbers of faces, edges and vertices they see whether they can “discover” Euler’s famous formula. By truncating the vertices of the Platonic solids, they find a class of semi-regular polyhedra. They also generalise how to know if arrangements of polygons will create a closed polyhedron and the common properties of nets for prisms and pyramids.

- Construct models of polyhedra using construction materials, like geoshapes or polydrons.
- Use the terms faces, edges and vertices to describe models of polyhedral and look for relationships between these features.
- Anticipate the features of the solid created when a Platonic solid is truncated.
- Anticipate if an arrangement of regular polygons around a vertex will create a bounded polyhedron.
- Create nets for regular and semi-regular polyhedra using knowledge of the faces and symmetry.

A polyhedron is a three-dimensional solid object which consists of a collection of polygons that bound a space. That means that the space is fully enclosed by the polygons. The simplest polyhedra are created by joining regular polygons, such as equilateral triangles, squares and regular pentagons. This family of polyhedra are known as the Platonic solids, named after the Greek mathematician Plato (though actually proved by Euclid).

There are 5 platonic solids, the cube (6 squares, 3 meeting at each vertex), the tetrahedron (4 triangles, 3 meeting at each vertex), the octahedron (8 triangles, 4 meeting at each vertex), the dodecahedron (12 pentagons, 3 meeting at each vertex), and the icosahedron (20 triangles, 5 meeting at each vertex).

Terms commonly used to describe the attributes of polyhedra include:

Face: A single polygon in a solid figure

Edge: A line where two faces connect

Vertex: A point of intersection of edges – a corner

In the 1750’s Leonhard Euler discovered a famous relationship between these three properties. The number of vertices, plus the number of faces take away two equals the number of edges.

E = V + F - 2

This unit involves exploration with three dimensional shapes and would be ideal as a lead in to the unit *Building with Triangles*, a level four unit which goes on to look at a group of polyhedra, the Platonic solids, in more detail.

This unit builds on the Level Three unit called Polyhedra (3D Shapes). It extends students’ knowledge of polyhedra to include Euler’s theorem and the properties of shapes around a vertex for semi-regular solids.

Interlocking shapes

**Session One**

1. Show the students slide one of PowerPoint One, which has all five Platonic Solids.

*What do you notice about these solids?**What shapes make up each solid? (Platonic solids are made from one type of regular polygon)*

*What are the flat surfaces (polygons) that make up the solid called? (faces)**How many faces does each solid have? How did you count the faces systematically?**Look for edges and corners (vertices – singular is vertex). How many edges and vertices has each solid got?*

2. The icosahedron is the most complex Platonic solid. From slide two of PowerPoint One students might be able to count the number of vertices, 3 + 6 + 3 = 12. If possible, have a model of the icosahedron available (made from interlocking shapes).

Ask your students to imagine the model separated out.

*How many triangles will there be? (Twenty since icosa- is the prefix for twenty)**How many corners (vertices) will the twenty triangles have in total? (20 x 3 = 60)**How many of the sixty corners form each vertex of the icosahedron? (Refer your students to the model and systematically count the five triangles that surround each vertex)**If five triangle corners make one vertex, how many vertices must the icosahedron have? (60 **÷ **5 = 12)**Could the same kind of thinking help us to work out the number of edges? (The twenty triangles have 20 x 3 = 60 sides. Two sides are needed to make one edge. Since 60 **÷**2 = 30 the icosahedron must have 30 edges.)*

3. By referring only to slide one of PowerPoint One ask your students to make models of all five Platonic Solids using interlocking shapes. Watch to see that students understand that the arrangement of polygons around a vertex is consistent in each Platonic Solid (For example, a dodecahedron has three pentagons around every vertex). Ask the students to complete Copymaster One which systematically lists the number of faces, edges and vertices. Students are heavily guided on the copymaster to reinvent Euler’s famous theorem about networks (V + F = E + 2 where V equals the number of vertices, F equals the number of faces, and E equals the number of edges).

4. Slide three of PowerPoint One shows a postage stamp featuring Leonhard Euler, commonly regarded as one of the three greatest mathematicians of all time. Students may want to research how Euler came to invent network theory as a branch of mathematics.

#### Session Two

In this session students work with truncations of the Platonic Solids. A truncation occurs when the vertices are ‘cut off’. PowerPoint Two begins with the simplest truncation, that of a cube.

Look at slide one.

*These two solids are related yet they look so different. How are they related?*

Student might notice if the corners of a cube are cut off the result is the right-hand solid, called a truncated cube (or hexahedron). You can demonstrate truncation using a cube cut from a large potato of kumara. Where vertices are truncated new triangular faces are formed.

*How many faces, edges and vertices will the truncated cube have?*

*What shapes are the faces? Why are the faces those shapes.*

Expect your students to use structure as well as visualisation to answer the question. For example, there are eight vertices on a cube so changing each vertex into a triangular face adds eight new faces. That gives a total of 6 + 8 = 14 faces. The faces made from a vertex are triangles because three squares meet at each vertex of a cube. That also means that 8 x 3 = 24 new edges are added making 12 + 24 = 36 edges in total. Each of the eight vertices of a cube are replaced by three new vertices. That makes 8 x 3 = 24 vertices.

*A truncated cube has 14 faces, 24 vertices and 36 edges. Does it still fit Euler’s theorem?*

Show the students slide two of PowerPoint Two. Ask the students to name the Platonic Solids (Tetrahedron and Octahedron) and recall the properties of the solids, including numbers of faces, vertices and edges.

*Here is the challenge. Use the connecting shapes to build the truncations of these solids. Before you start, visualise what the truncated solid will look like. Then build it.*

Give your students ample time to construct the solids. Look for the following:

- Do your students apply structure to anticipate the result of truncation? For example, a tetrahedron has four vertices so four new faces will be formed by truncating each vertex. For the octahedron six new faces will be formed.
- Do your students predict the shape of the new faces from the number of triangles around each vertex of the original solid? The tetrahedron has three triangles around each vertex but an octahedron has four. Therefore, the new faces will be triangles for a tetrahedron and squares for an octahedron.
- Can your students anticipate how the original faces will be changed? Cutting off the vertices of triangles results in hexagons.

Students can check their models against the picture on slide three of PowerPoint Two. Draw their attention to the arrangement of shapes around each vertex. For the truncated tetrahedron, one triangle and two hexagons surround a vertex. For the truncated octahedron, the arrangement is one square and two hexagons.

Challenge your students further.*Here are the dodecahedron and icosahedron. Imagine these solids are truncated.**What shapes would the faces of the truncated solid be?**How many faces would there be?*

Let your students solve the problem in small groups. Access to the models they built in Session One will be helpful. To build a truncated dodecahedron requires 20 triangles and 12 decagons (ten sided polygons). Therefore, a model cannot be built with a normal set of connecting shapes. However, a truncated icosahedron can be built from 12 pentagons and 20 hexagons.

Slide four of PowerPoint Two has images of both truncated solids. Ask students to name the arrangement of shapes around each vertex. That arrangement is consistent for the whole solid.

*It is hard to visualise how many edges and vertices each solid has. You do have Euler’s theorem to help. Use your knowledge of the faces to work the numbers out.*

Look for your students to use the properties of shapes meeting at vertices and edges to solve the problem. They could organise the data in a table:

Solid |
Number of faces |
Number of edges |
Number of vertices |

Truncated Dodecahedron |
20 triangles + 12 decagons = 32 |
(20 x 3) + (12 x 10) ÷ 2 = 90 |
20 x 3 = 60 |

Truncated Icosahedron |
12 pentagons + 20 hexagons = 32 |
(12 x 5) + (20 x 6) ÷ 2 = 90 |
12 x 5 = 60 |

*Does Euler’s theorem hold for both solids?*

#### Session Three

In this session students create nets for the Platonic solids and possibly their truncations. Begin with this challenge.

*Use protractors, rulers and scissor to make and cut out an equilateral triangle, a square, a regular pentagon, a regular hexagon, a regular octagon and a regular decagon. Every side must be 5cm long. Use light cardboard.*

The purpose of making the shapes is to create templates to form nets with. Some students will need support with creating the polygons. Ideally students will use the sum of internal angles to work out the angle measures. That is a nice investigation but may interfere with the flow of this unit.

Number of sides |
Name of polygon |
Sum of internal angles |
Each internal angle |

3 |
Equilateral triangle |
180° |
60° |

4 |
Square |
360° |
90° |

5 |
Regular pentagon |
540° |
108° |

6 |
Regular hexagon |
720° |
120° |

7 |
Regular heptagon |
900° |
128.57° |

8 |
Regular octagon |
1080° |
135° |

9 |
Regular nonagon |
1260° |
140° |

10 |
Regular decagon |
1440° |
144° |

For students who do not know the pattern you might provide Copymaster Two which has the regular polygons on it. Students can measure the angles and use that information to create cut outs.

Once the students have made cardboard polygons they can use them to construct nets for the Platonic solids and the truncations. Models made from interlocking shapes can be ‘unpeeled’, if needs be, to reveal a net that will work. By tracing around the shapes students can create nets quickly.

The tetrahedron and cube are easy constructions. For the other three Platonic solids encourage your students to consider making the net for one half of the solid and joining two halves to make the complete net.

The halves can be joined to form the full net.

Of the truncated solids the tetrahedron and the cube are easiest. Encourage students to connect nets for the original Platonic solids with the nets for the truncated solids. For example:

A net for the truncated cube can be made by a similar process.

#### Session Four

In this session student find the properties of shapes surrounding a vertex that can be used to predict whether a polyhedron will be formed.

Set up the investigation as follows. Open up models of the Platonic solids to create nets.

For each net find a vertex where two sides will fold up to form an edge of the solid. The black dots give examples of such points. The ‘missing angle’ is known as the angle defect. Slide one of PowerPoint Three gives the case of the octahedron.

What is the angle defect, that is the angle that is missing from a full turn of 360°?

Students should see that four angles of 60° exist at the vertex. Since 4 x 60 = 240 and 360 – 240 = 120, the angle defect is 120°. Some students may see that two equilateral triangles could fill the angle space so the defect equals 2 x 60 = 120°. The PowerPoint slide introduces a protractor so the analytical answer can be checked by measurement.

*How many vertices does the octahedron have? (six)**There are six vertices where the angle defect is 120**°**.**What is the total of six defects of 120 degrees? (6 x 120 = 720**°**)*

Slide two looks at the cube. The defect angle is 90°. A cube has eight vertices, so the total of the defect angles is 8 x 90 = 720°.

*Maybe that is just co-incidence. Check out the total of the angle defects of the other Platonic solids.*

Let your students explore the other Platonic solids using the nets for support. Expect them to record the data systematically. Slide three shows the angle defects and multiplies each defect by the number of vertices for the relevant solid.

The students should notice that the sum of the angle defects is always 720°.

*How could we use this theorem to see if an arrangement of shapes will make a closed solid?**Could we then know how many vertices the solid will have?*

Slides four to six has some arrangements of shapes around one vertex. The notation represents the regular polygons around each vertex. For example, (8, 8, 3) represents two octagons and one triangle about a vertex. Discuss whether they believe each arrangement will work. On each slide the truncated solid that matches the arrangement appears on the last mouse click.

For each slide ask students to calculate the angle defect. If the arrangement works, then 720° is a multiple of the angle defect. With Slide four the angle defect is 60° since the angles present add to 90 + 60 + 90 + 60 = 300. 12 x 60 = 720 so the arrangement will create a closed solid and the number of vertices will be twelve.

The truncated cube has an angle defect of 30°, and 24 x 30 = 720, so the solid will have 24 vertices.

The truncated octahedron also has an angle defect of 30°, and 24 x 30 = 720, so the solid will have 24 vertices.

Slide seven provides a final challenge. The arrangement of polygons has an angle defect of 360 – (108 + 60 + 108 + 60) = 24°. Since 720 ÷ 24 = 30 the arrangement will produce a polyhedron called the icosa-dodecahedron that has 30 vertices. It has that name because the centre of each pentagonal face is the vertex of an icosahedron, and in the centre of each triangular face is the vertex of a dodecahedron. Challenge your students to anticipate how many of each shape, triangle and pentagon, will be needed (20 triangles and 12 pentagons) before they build it.

#### Session Five

In this session students explore the classes of solids known as prisms and pyramids. Assigning solids to these classes allows the students to generalise the structure of nets and volume formulae. In the process a cylinder can be regarded as the ‘limiting case’ of a prism and a cone as the ‘limiting case’ of a pyramid.

Make a triangular prism, a cuboid, and a hexagonal prism from the connecting shapes.

*What is the same about each solid?**What differences are there among the solids?*

Look for students to talk about the defining feature of prisms, consistent cross-section if the solid is cut parallel to the ‘end’ faces. These faces name the prism, for example a triangular prism has a triangular cross section.

Ask your students to sketch nets (flat patterns) for the prisms. The sketches can be confirmed by unpeeling the models you have. Standard nets for the prisms look like this.

*How do the similarities among the prisms show in their nets?**How do the differences show?*

Students should not that all three nets have rectangular faces, and the number of those faces matches the number of sides of the end faces. The endpoint faces are those that give the prism its cross-section.

Show the students Slide one of PowerPoint Four which is a cylinder.

*Is this solid a type of prism?*

Most definitions of a cylinder classify it as a curved surface. A prism is a type of polyhedron which means it is bounded (enclosed) by flat polygons. However, it is advantageous to consider the similarity of a cylinder to a type of prism. A cylinder has consistent cross-section (a circle) so the volume is worked out identically to a prism. Ask the students to sketch the net for a cylinder. The standard net looks like this:

The net has much in common with the other nets for prisms. The two end faces are circles and the rectangular faces are infinite so form a continuous whole. The length of the whole rectangle has to equal the circumference of the circles.

*How do you find the volume of a cuboid? (rectangular prism)*

Students are likely to suggest length x width x height. Rebuild the cuboid model and ask how many 1 cm^{3 }place value blocks will fit in. Highlight that multiplying two dimensions is like finding the area of the cross-section. Multiplying by the other dimension layers the cross-section.

*How do you think we could find the volume of the triangular and hexagonal prisms, and the cylinder?*

The same method applies. Find the area of an end face (triangle, hexagon, and circle) and multiply that area by the other dimension. Students may know the formula for the area of a circle a = πr^{2}, though that knowledge is not expected at Level Four. That means the volume of a cylinder involves the area of the circular face multiplied by height (v = πr^{2 }x h). You might test the formula out with a cylinder, e.g. a tin can. Measure the radius and height in centimetres and calculate the volume in cubic centimetres. Fill the container with water and measure to see if the capacity in millilitres matches the volume, since 1 cm^{3} = 1 mL.

Follow a similar process with pyramids. Begin with three pyramids, triangular, square and pentagonal-based. Ask what is in common with these solids and what is different.

Insert picture of models

Expects students to notice a base and triangular faces converging to an apex. Students should also note that the bases are different. Like prisms, pyramids are named by their bases, e.g. square based pyramid.

Ask students to sketch the nets for these solids. Most students will provide the ‘flower’ shaped net which is the standard template.

Students should be able to extend the idea to the net for a cone, though the curved surface of the solid is harder to visualise. Let students experiment to find out what pattern works. Notice that the flower shaped net does not work for the cone since the curved surface is a single part circle (sector). The arc length of the sector must match the circumference of the circle.

The volume of a pyramid is related to that of a prism. Students might look the formula up and discover that volume of a pyramid is one third of the surrounding prism. For example, the volume of a cylinder is given by v = πr^{2 }x h so the volume of a cone is given by v = ⅓πr^{2 }x h.

If you are ambitious, you might test the volume formula out by making a cone with card, lining the cone with plastic wrap and filling it with water.

Dear Family and Whānau,

This week we have been exploring polyhedral that are 3-dimensional shapes made from 2-dimensional shapes. Ask your child to explain how these solid shapes have faces, edges and vertices. For homework your child has been asked to either:

- find photographs of different polyhedra in the real world and create a poster page for their maths book;
- make a model of their favourite polyhedron from cardboard packets.