Mathematics notes, chapter 6

Three-Dimensional Geometrical Shapes

Textbook Mathematics Form 2 (DLP), pages 98–119

What this chapter covers

Compare, contrast and classify prisms, pyramids, cylinders, cones and spheres, and describe their geometric properties (6.1)
Analyse nets of pyramids, prisms, cylinders and cones, and draw nets and build models (6.2)
Derive and use the formulae for the surface area of cubes, cuboids, pyramids, prisms, cylinders, cones and spheres (6.3)
Solve problems involving the surface area of three-dimensional shapes, including combined solids (6.3)
Derive the formulae for the volume of prisms, cylinders, pyramids and cones, and use the volume formulae including the sphere (6.4)
Solve problems involving the volume of three-dimensional shapes, including converting units (6.4)

Look at a tin of milk, a cone-shaped party hat, a football and a triangular gift box. They are all three-dimensional (3D) shapes: they have length, width and height, so they take up space. In this chapter you will describe their properties, draw their nets, and calculate their surface areas and volumes. In the Form 2 textbook, the Tun Mustapha Tower in Sabah (a cylinder) is the opening example.

The word "geometry" comes from the Greek geo (earth) and metria (measurement). Euclid, the "Father of Geometry", wrote the book Elements.

English	Bahasa Melayu
two-dimensional shape, three-dimensional shape	bentuk dua dimensi, bentuk tiga dimensi
geometrical characteristic	sifat geometri
net	bentangan
surface area	luas permukaan
volume	isi padu
subject of formula	perkara rumus
cross section	keratan rentas

6.1Geometric properties of three-dimensional shapes

A two-dimensional shape (a square, a triangle) has length and width only, so it has an area but no volume. A three-dimensional shape has length, width and height, so it has a surface area and a volume.

Five solids drawn in a row: a triangular prism, a square-based pyramid, a cylinder, a cone and a sphere — Prism, pyramid, cylinder, cone and sphere (textbook p. 101).

Shape	Geometrical characteristics
Prism	Two bases that are congruent, parallel polygons; flat rectangular side faces; the same (uniform) cross section all along its length.
Pyramid	One flat base that is a polygon; the other faces are triangles that meet at one point, the apex.
Cylinder	Two circular bases that are congruent and parallel; one curved surface.
Cone	One circular base; one apex; one curved surface joining the base and the apex.
Sphere	One curved surface; every point on the surface is the same distance from the centre.

Prisms and pyramids are named after their base: a triangular prism, a hexagonal prism, a square-based pyramid. Congruent means the same size and the same shape. A cube and a cuboid are special prisms (their cross section is a square or a rectangle).

Faces, edges and vertices

A polyhedron is a solid whose surfaces are all flat polygons. A cylinder, a cone and a sphere are non-polyhedra because they have curved surfaces. Count the parts of a polyhedron:

Solid	Faces (F)	Edges (E)	Vertices (V)
Cuboid	6	12	8
Triangular prism	5	9	6
Hexagonal prism	8	18	12
Square-based pyramid	5	8	5
Tetrahedron (triangular-based pyramid)	4	6	4

Example 1: name the solid

Which solid has one vertex, one circular base and one curved surface?

One apex (vertex) and one circular base: this is a cone.
A pyramid also has one vertex, but its base is a polygon and it has no curved surface.

A solid has 7 faces: one hexagonal base and six triangles. Name the solid and say how many vertices it has.

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A hexagonal-based pyramid. Vertices: 6 on the base + 1 apex = 7.

6.2Nets of three-dimensional shapes

A net is what you get when you open a 3D solid and lay all its surfaces flat as one 2D pattern. Folding the net along its edges rebuilds the solid.

One solid can have more than one net (a cube has 11 different nets), but all of its nets contain the same pieces.

Net of a cylinder: a rectangle with a circle above and a circle below — Net of a cylinder. The length of the rectangle is the circumference of the circle, 2πr.

Net of a cone: a sector with radius s and a circle of radius r — Net of a cone: one sector (the curved surface) and one circle (the base). The radius of the sector is the slant height s.

Net of a triangular prism: three rectangles in a row with a triangle above and below the middle one — Net of a triangular prism: 3 rectangles and 2 triangles.

Net of a square-based pyramid: a square with a triangle on each side — Net of a square-based pyramid: 1 square and 4 triangles.

Solid	Surfaces in the net
Cube	6 squares
Cuboid	6 rectangles (opposite faces are equal)
Triangular prism	2 triangles + 3 rectangles
Square-based pyramid	1 square + 4 triangles
Cylinder	2 circles + 1 rectangle
Cone	1 circle + 1 sector (curved surface)

To build a model: draw the net on 1 cm grid paper, add small tabs on some edges, cut it out, fold along the edges and stick the tabs with tape.

The rectangle in the net of a cylinder is 2πr long, not 2r. It wraps once round the circle, so its length is the circumference.

A net has 2 triangles and 3 rectangles. Which solid is it? A net has 1 circle and 1 sector. Which solid is it?

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A triangular prism. A cone.

6.3Surface area of three-dimensional shapes

The surface area of a solid is the total area of all its surfaces, that is, the total area of its net. It is measured in square units: mm², cm², m².

Solid	Surface area
Cube, side a	6 × a²
Cuboid, l × w × h	2(lw + lh + wh)
Prism	(2 × area of cross section) + (perimeter of cross section × length)
Pyramid with square base	area of base + 4 × area of a triangle
Closed cylinder	2πr² + 2πrh
Cone	πr² + πrs
Sphere	4πr²

Where the cylinder and cone formulae come from

Cylinder. The net is two circles and a rectangle. The rectangle is 2πr long (the circumference) and h wide (the height). So surface area = 2 × πr² + 2πr × h.

Cone. Cut the curved surface into many thin sectors and arrange them side by side. They almost form a rectangle with length πr (half the circumference) and width s (the slant height). So the curved surface area is πr × s = πrs, and the whole surface is πr² + πrs.

A cone with its radius r, height h and slant height s marked — The radius r, the height h and the slant height s form a right-angled triangle, so s² = r² + h².

In the cone formula use the slant height s, not the vertical height h. If you are given h, first find s = √(r² + h²).

Example 2 (like textbook p. 107): cube, cuboid and pyramid

Cube of side 4 cm: 6 × (4 × 4) = 96 cm².
Cuboid 7 cm × 4 cm × 4 cm: (4 × 7 × 4) + (2 × 4 × 4) = 112 + 32 = 144 cm².
Pyramid, square base 8 cm, each triangle has base 8 cm and slant height 5 cm: (8 × 8) + 4 × (½ × 8 × 5) = 64 + 80 = 144 cm².

Example 3 (textbook p. 108): cylinder and cone

Use π = 227.

Closed cylinder, r = 7 cm, h = 9 cm:
2πr² + 2πrh = 2 × 227 × 7² + 2 × 227 × 7 × 9 = 308 + 396 = 704 cm².
Cone, r = 3 cm, s = 5 cm:
πr² + πrs = 227 × 3² + 227 × 3 × 5 = 28.29 + 47.14 = 75.43 cm².

Surface area of a sphere

Surface area of a sphere = 4πr², where r is the radius.

Example 4 (textbook p. 109)

A sphere has radius 14 cm. Surface area = 4 × 227 × 14² = 2 464 cm².

Remember the radius is half the diameter. A football with diameter 22 cm has r = 11 cm. A solid hemisphere has a curved part 2πr² and a flat circle πr², so its total surface is 3πr².

Solving problems: combined shapes

A cube of side 5 cm with a square-based pyramid on top, height 6 cm, slant height s — Example 5: a cube with a pyramid on top (textbook p. 109).

Example 5 (textbook p. 109): cube and pyramid

A cube of side 5 cm has a pyramid on top. The height of the whole object is 11 cm. Find the surface area in m².

Understand: the pyramid sits on the top face of the cube, so that face is not on the surface. The object has 5 square faces and 4 triangles.
Height of pyramid = 11 − 5 = 6 cm. Half of the base = 2.5 cm.
Slant height s = √(6² + 2.5²) = √42.25 = 6.5 cm.
Surface area = 5 × (5 × 5) + 4 × (½ × 5 × 6.5) = 125 + 65 = 190 cm².
Convert: 1 m² = 10 000 cm², so 190 ÷ 10 000 = 0.019 m².

When two solids are joined, the faces that touch are hidden. Do not count them. Here the top face of the cube and the base of the pyramid are inside.

A closed cylinder has radius 3.5 cm and height 8 cm. Find its surface area (π = 227).

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2πr² = 2 × 227 × 3.5² = 77. 2πrh = 2 × 227 × 3.5 × 8 = 176. Total = 253 cm².

6.4Volume of three-dimensional shapes

The volume of a solid is the amount of space it occupies. It is measured in cubic units: mm³, cm³, m³. Also 1 litre = 1 000 cm³ and 1 m³ = 1 000 000 cm³.

Prisms and cylinders

A prism has the same cross section from one end to the other, so its volume is the area of the cross section multiplied by the length (or height). A cylinder is a prism-like solid with a circular cross section.

Volume of a prism = area of cross section × height
Volume of a cylinder = area of base × height = πr²h

A cuboid is a prism with a rectangle as cross section: V = l × w × h. If you cut a cuboid along a diagonal of its cross section you get two equal triangular prisms, so the volume of one triangular prism is ½ × l × w × h.

Pyramids and cones

Six identical pyramids fit exactly inside a cube whose base is the base of one pyramid and whose height is twice the height of the pyramid. So one pyramid is 16 of that cube, and since the cube is twice as tall as the pyramid, one pyramid is 13 of a prism with the same base and height. In the same way, three cones of sago fill a cylinder with the same base and height (textbook p. 112).

A cylinder and a cone drawn with the same base radius and the same height — Three cones fill the cylinder, so the volume of a cone is 13 of the volume of the cylinder.

Solid	Volume
Prism	area of cross section × height
Cuboid	l × w × h
Cylinder	πr²h
Pyramid	13 × area of base × height
Cone	13πr²h
Sphere	43πr³

Surface area uses square units, volume uses cubic units. Also, in the pyramid and cone formulae h is the vertical height, not the slant height.

Example 6 (like textbook pp. 112-113): prism, cylinder, cone and pyramid

Triangular prism: the cross section is a triangle with base 8 cm and height 3 cm, and the prism is 12 cm long. V = (½ × 8 × 3) × 12 = 144 cm³.
Cylinder: r = 3.5 cm, h = 12 cm. V = 227 × 3.5² × 12 = 462 cm³.
Cone: r = 7 cm, h = 12 cm. V = 13 × 227 × 7² × 12 = 616 cm³.
Pyramid: square base 4 cm × 4 cm, height 3 cm. V = 13 × 16 × 3 = 16 cm³.

Volume of a sphere and a hemisphere

Volume of a sphere = 43πr³. Volume of a hemisphere = 12 × 43πr³ = 23πr³.

Example 7 (textbook p. 114)

Sphere, r = 7 cm: V = 43 × 227 × 7³ = 1 437.33 cm³.
Hemisphere, r = 5 cm: V = 23 × 227 × 5³ = 261.90 cm³.

Solving problems

Example 8: a cylinder with a hemisphere on top

A water container is a cylinder of radius 7 cm and height 10 cm with a hemisphere of the same radius on top. How many litres does the whole container hold? (π = 227)

Cylinder: 227 × 7² × 10 = 1 540 cm³.
Hemisphere: 23 × 227 × 7³ = 718.67 cm³.
Total = 1 540 + 718.67 = 2 258.67 cm³.
Litres: 2 258.67 ÷ 1 000 = 2.26 litres (2 decimal places).

Example 9 (like textbook p. 116): cone inside a cylinder

A cylinder of radius 7 cm and height 15 cm is full of water. A solid cone of the same radius and height is lowered into it so the cone is fully under water. How much water remains?

Volume of cylinder = 227 × 7² × 15 = 2 310 cm³.
Volume of cone = 13 × 2 310 = 770 cm³.
Water left = 2 310 − 770 = 1 540 cm³, which is 23 of the cylinder.

If a question gives you a volume and asks for a missing length, make that length the subject of the formula. For a cylinder, h = V ÷ (πr²). Check your answer by putting it back into the formula.

Find the volume of a pyramid with a rectangular base 8 cm × 5 cm and a height of 6 cm. Then find the volume of a hemisphere with radius 3 cm (π = 227).

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Pyramid: 13 × 40 × 6 = 80 cm³. Hemisphere: 23 × 227 × 3³ = 56.57 cm³.

Summary

Solid	Surface area	Volume
Prism	2 × cross section + perimeter × length	cross section × height
Pyramid	base + triangles	13 × base × height
Cylinder	2πr² + 2πrh	πr²h
Cone	πr² + πrs	13πr²h
Sphere	4πr²	43πr³