The pie chart can be constructed by carefully measuring these angles at the centre of a circle. The slices should be labelled, and an appropriate title given to the chart. The source of the data should also be stated.Try some yourselfQuestion 1A company carried out a survey, recording how staff in a particular office spent their working time. The table shows the average number of minutes spent in each hour on various activities.

Activity

Time taken on average in one hour/mins

Keyboarding

35

Answering telephone

12

Talking with colleagues

10

Other

3

The data is to be displayed as a pie chart. Work out the angle at the centre for each slice.Since one hour will be represented by 360° on the pie chart, 1 minute will be represented by 360° ÷ 60 = 6°.So the required angles on the chart are:How office staff spend their time (Source: Company survey)

Keyboarding

35 × 6° = 210°

Answering telephone

12 × 6° = 72°

Talking with colleagues

10 × 6° = 60°

Other activities

3 × 6° = 18°

Check: 210° + 72° + 60° + 18° = 360°.1.3.4 Vertically opposite anglesWhen two straight lines cross, they form four angles. In the diagram below, these angles are labelled α, β, θ and φ and referred to as alpha, beta, theta and phi. The angles opposite each other are equal. They are called vertically opposite angles. Here α and β are a pair of vertically opposite angles, as are θ and φ. Although such angles are called ‘vertically opposite’, they do not need to be vertically above and below each other!Activity 3: Vertically opposite anglesClick here for Activity 3.For two intersecting straight lines, vertically opposite angles are equal.We can show that vertically opposite angles are equal as follows:α and θ lie on a line.So, α + θ = 180°and α = 180° – θbut β and θ also lie on a line.So, β + θ = 180°and β = 180° – θ.Hence, α = β because they are both equal to 180° – θ.Try Some YourselfQuestion 1Find all the remaining angles in each of the diagrams below.(a) 130°, 50°, 130°.(b) 120°, 60°, 120°.(c) 90°, 90°, 90°.1.4 Parallel linesTwo straight lines that do not intersect, no matter how far they are extended, are said to be parallel. Arrows are used to indicate parallel lines.Look at the line l, which cuts two parallel lines m and n.The next two activities look at the angles formed by this line.Activity 4: Corresponding anglesClick here for Activity 4.Activity 5: Alternate anglesClick here for Activity 5.1.4.1 Corresponding anglesIf you trace the lines at one of the intersections in the diagram below and place them over the lines at the other intersection, you will find that the two sets of lines coincide exactly. The four angles at each intersection also coincide exactly: thus α = a, β = b, γ = c and δ = d.The pairs of angles that correspond to each other at such intersections are called corresponding angles.In the diagram below, α and a are corresponding angles: they are equal because m and n are parallel.When a line intersects two parallel lines, corresponding angles are equal.You can check these results using ‘Corresponding angles’ activity in the previous section.Example 5This diagram represents the type of arrangement that occurs in a garden trellis or a wine rack (r and s are parallel lines, indicated by the single arrowheads; l and m are also parallel, indicated by the double arrowheads). Calculate the angles α and β.Line l is parallel to line m, therefore α and the angle 60° are corresponding angles. So α = 60°.The angles 60° and θ are vertically opposite angles. So θ = 60°.Line r is parallel to line s, therefore θ and β are corresponding angles. So β = 60°.1.4.2 Alternate anglesOther pairs of equal angles can be identified in Example 5. These pairs of angles occur in a Z-shape, as indicated by the solid line in the diagram below. Such angles are called alternate angles.When a line intersects two parallel lines, alternate angles are equal.To prove this result consider the diagram below:ABC = DCE (corresponding angles)and DCE = FCB (vertically opposed angles)So, ABC = FCB (both equal to DCE).The other two angles are also equal and are also called alternate angles.It is important to realise that you can find the sizes of unknown angles in many shapes by using a combination of the angle properties that have been outlined. To recap:Vertically opposite angles are equal.Angles at a point add up to 360°.Angles on a straight line add up to 180°.Corresponding angles on parallel lines are equal.Alternate angles on parallel lines are equal.Example 6Find α and β in the following diagram.Line l is parallel to line m, therefore and are alternate angles. SoSimilarly, and are alternate angles. Butand henceThese properties of corresponding and alternate angles mean that the opposite angles in a parallelogram are also equal.Try some yourselfQuestion 1Find α and β in each of the diagrams below.(a) Now so But so Question 2(b)There are many ways of finding the sizes of these angles. This is only one of them:As it follows that But so Similarly, so But so Question 3(c) α = 45° (alternate angles).β = 55° (alternate angles).Question 4(d) α = = 30° (alternate angles).β = = 70° (corresponding angles).Question 5(e) As it follows that But β and are corresponding angles, so Whereas α and are alternate angles, so Question 62 This diagram shows part of some bannister rails. The handrail makes an angle of 40° with the horizontal. Calculate angles α, β, γ and δ.It is a good idea to sketch a diagram, adding some horizontal lines where necessary.Assume the lines marked are pairs of parallel lines. Then, since the handrail makes an angle of 40° with the horizontal,Question 73 The arrows on the diagram below indicate the idealised path (ABCD) of a snooker ball on a snooker table. Assume that the angles between the cushion (the edge of the snooker table) and the path of the ball before and after it impacts with the cushion are equal. Calculate the sizes of the angles marked a, b, c, d, e and f.2 Shapes and symmetry2.1 Geometric shapes – trianglesThis section deals with the simplest geometric shapes and their symmetries. All of the shapes are two-dimensional – hence they can be drawn accurately on paper.Simple geometric shapes are studied in mathematics partly because they are used in thousands of practical applications. For instance, triangles occur in bridges, pylons and, more mundanely, in folding chairs; rectangles occur in windows, cinema screens and sheets of paper; while circles are an essential part of wheels, gears and plates.By definition, triangles are shapes with three straight sides. However, there are various types of triangle:An equilateral triangle is a triangle with all three sides of equal length. The three angles are also all equal.An isosceles triangle is a triangle with two sides of equal length. The two angles opposite the equal sides are also equal to one another.A right-angled triangle is a triangle with one angle that is a right angle.A scalene triangle is a triangle with all the sides of different lengths. The angles are also all different.It is a general convention that equal sides are marked by drawing a short line, /, through them, and a right angle is marked by a square between the arms of the angle. If sides and angles are not marked, do not assume that they are equal, just because they look equal!Activity 6: Special trianglesClick here for Activity 6.2.2 Geometric shapes – quadrilateralsThe next few activities explore some properties of quadrilaterals and their diagonals.Activity 7: QuadrilateralsClick here for an activity on parallelograms.Now try the same activity for the following shapes:Click here for an activity on squares.Click here for an activity on rectangles.Click here for an activity on rhombi (that's the plural of ‘rhombus’).Click here for an activity on kites.Try some yourselfNow try to construct your own shapes from the diagonals here.Some of the properties of these shapes and their diagonals are summarised in the attached Quadrilateral table.Click ‘Document’ to open the Quadrilateral table.DocumentWhich ones did you spot?2.3 Geometric shapes – circlesAll circles are the same shape – they can only have different sizes.In a circle, all the points are the same distance from a point called the centre. The centre is often labelled with the letter O.The outside edge of a circle is called the circumference. A straight line from the centre to a point on the circumference is called a radius of the circle (the plural of radius is radii).A line with both ends on the circumference and passing through the centre is called a diameter. Any diameter cuts the circle into two halves called semicircles.In the circle below, the lines labelled OA, OB, OC, OD and OE are all radii, and AD and BE are diameters. The points A, B, C, D and E all lie on the circumference.Although the terms ‘radius’, ‘diameter’ and ‘circumference’ each denote a certain line, these words are also employed to mean the lengths of those lines. So it is common to say, for example, ‘Mark a point on the circumference’ and ‘The circumference of this circle is 7.3 cm’. It is obvious from the context whether the line itself or the length is being referred to.2.4 Drawing circlesDrawing circles freehand often produces very uncircle-like shapes! If you need a reasonable circle, you could draw round a circular object, but if you need to draw an accurate circle with a particular radius, you will need a pair of compasses and a ruler. Using the ruler, set the distance between the point of the compasses and the tip of the pencil at the desired radius; place the point on the paper at the position where you want the centre of the circle to be and carefully rotate the compasses on the point so that the pencil marks out the required circle.To draw a large circle, perhaps to create a circular flower bed, a similar set-up is needed. The essentials are a fixed central point (possibly a stake) and a means of ensuring a constant radius (possibly a string). To draw a circle on a computer or calculator screen, you may also need to fix the centre (maybe using coordinates) and the radius.It is often necessary to label diagrams of geometric figures, such as circles or triangles, in order to make it easier to refer to specific parts of the figure. Usually points are labelled as A, B, C, … and lines as AB, BC, …, or a, b, c, … and using combinations of the letters, such as ‘triangle ABC’ (‘Triangle ABC’ is often written as ‘ΔABC’.). It is rather laborious to read, but unfortunately is unavoidable.Note that, as in the case of words like ‘radius’ and ‘circumference’, AB may be used to mean the line from A to B or the length of the line itself.2.5 SymmetrySymmetry is a feature that has been used in the design of objects and patterns in many cultures throughout recorded history. From Greek vases and medieval windows to Victorian tiles and Native American decorations, symmetry has been seen as a way of achieving balance and beauty.Symmetry can be described mathematically, and is a useful concept when dealing with shapes.2.6 Line symmetryLook at the shapes below. The symmetry of the shape on the left and its relationship to the shape on the right can be thought of in two ways:Fold the left-hand shape along the central line. Then one side lies exactly on top of the other, and gives the shape on the right.Imagine a mirror placed along the central dotted line. The reflection in the mirror gives the other half of the shape.This type of symmetry is called line symmetry.Any isosceles triangle has line symmetry.The dashed lines represent lines of symmetry, and each shape is said to be symmetrical about this line.The following all have line symmetry:A shape can have more than one line of symmetry. Thus a rectangle has two lines of symmetry, an equilateral triangle has three lines of symmetry, and a square has four.A circle has an infinite number of lines of symmetry since it can be folded about any diameter. Only eight of the possible lines of symmetry are indicated below.Some shapes, such as a scalene triangle, have no lines of symmetry – it is not possible to fold the shape about a line so that the two halves fit exactly on top of one another.2.7 Rotational symmetryThere is another kind of symmetry which is often used in designs. It can be seen, for instance, in a car wheel trim.Look at the trim on the left. It does not have line symmetry but it has rotational symmetry. If the wheel is rotated through a quarter of a full turn, it will look exactly the same; likewise, if it is rotated through half a complete turn, or through three-quarters of a turn. There are four positions in which the wheel looks the same: hence the wheel is said to have rotational symmetry of order 4 or four-fold rotational symmetry.The wheel trim on the right has rotational symmetry of order 6. In this case there are six positions in which the trim will look exactly the same. These occur when the wheel is rotated through one-sixth of a complete turn, two-sixths of a turn, and so on, to five-sixths of a turn and finally a complete turn (when, of course, the wheel is back in its original position).The centre of the shape is the point about which the shape is rotated; it is called the centre of rotation.A shape does not have to be round to have rotational symmetry. The following shapes have rotational symmetry of orders 3 and 4, respectively.It is not difficult to create shapes with both line symmetry and rotational symmetry. The two designs below are examples.The design on the left has three lines of symmetry and rotational symmetry of order 3. The one on the right has four lines of symmetry and rotational symmetry of order 4.A shape with no rotational symmetry, like the one below, is sometimes said to have ‘rotational symmetry of order 1’. This is because it will only fit on top of itself in one position – after a complete turn.Try some yourselfQuestion 1Draw a line of symmetry on each of the shapes below.Each of the shapes has only one line of symmetry, so these are the only possible answers.Question 2Mark all of the lines of symmetry on these shapes. For each shape, state the total number of lines of symmetry.Question 3Which of these have rotational symmetry?(a) The dartboard has rotational symmetry.(b) The letter Z has rotational symmetry.(c) The letter K does not have rotational symmetry.Question 4Mark the centre of rotation on each of the shapes below. For each, state the order of rotational symmetry.Notice that (d) has no rotational symmetry and no centre of rotation.Question 5Describe the symmetry of each of these shapes. Mark all the lines of symmetry in each case. Also mark the centre of rotation, and state the order of rotational symmetry.2.8 The angles of a triangleThe sum of the angles of any triangle is 180°. This property can be demonstrated in several ways. One way is to draw a triangle on a piece of paper, mark each angle with a different symbol, and then cut out the angles and arrange them side by side, touching one another as illustrated.You can see why it is that the angles fit together in this way by looking at the triangle below. An extra line has been added parallel to the base. The angle of the triangle, , is equal to the angle β at the top (they are alternate angles), and similarly the angle of the triangle, , is equal to the angle γ at the top (they are also alternate angles). The three angles at the top (β, γ and the angle of the triangle, ) form a straight line of total angle 180°, and so the angles of the triangle must also add up to 180°.The sum of the angles of a triangle is 180°.The fact that the angles of a triangle add up to 180° is another angle property that enables you to find unknown angles.Example 7Find α, β and θ in the diagram below.First, look at the angles of ΔABD: and Then, by the angle sum property of triangles,SoandAs CDB is a straight line and α = 60°, it follows thatNow consider the angles of ΔADC: and ThereforeSo(Check for yourself that the angles of ΔABC also add up to 180°.)It is possible to deduce more information about the angles in certain special kinds of triangles.In a right-angled triangle, since one angle is a right angle (90°), the other two angles must add up to 90°. Thus, in the example below, α + β = 90°.In an equilateral triangle, all the angles are the same size. So each angle of an equilateral triangle must be 180° ÷ 3 = 60°.In an isosceles triangle, two sides are of equal length and the angles opposite those sides are equal. Therefore, α = β in the triangle below.Such angles are often called base angles.This means that there are only two different sizes of angle in an isosceles triangle: if the size of one angle is known, the sizes of the other two angles can easily be found. The next example shows how this is done.Example 8Find the unknown angles in these isosceles triangles, which represent parts of the roof supports of a house.(a) As α and 50° are the base angles, α = 50°. By the angle sum property of triangles,therefore(b) As γ and δ are the base angles, γ = δ. In this triangle,thereforeThe various angle properties can also be used to find the sum of the angles of a quadrilateral.Example 9The diagram below represents the four stages of a walk drawn on an Ordnance Survey map.The figure ABCD is a quadrilateral. Find θ and φ, and thus the sum of all the angles of the quadrilateral.From ΔABC,From ΔACD,Then the sum of all the angles of the quadrilateral isIn fact, you can find the sum of the four angles of a quadrilateral without calculating each angle as in Example 9. Look again at the quadrilateral: the dotted line splits it into two triangles, and the angles of these triangles together make up the angles of the quadrilateral. Each triangle has an angle sum of 180°, so the angle sum of the quadrilateral is 2 × 180° = 360°. This is true for any quadrilateral.The sum of the angles of a quadrilateral is 360°.Similarly, other polygons (that is, other shapes with straight sides) can be divided into triangles to find the sum of their angles.Try some yourselfQuestion 1Find the unknown angles in each of these diagrams, which represent part of the bracing structure supporting a marquee.(a) In ΔABC,therefore(b) In ΔFGH,thereforeAs EFG is a straight line,SoIn ΔEFH,Question 2Deduce the value of α in the triangle below.Lines XY and XZ are of equal length. This means that the triangle is isosceles, so the base angles and are equal. Then The third angle in the triangle is a right angle: .Because the three angles in a triangle must add up to 180°,HenceQuestion 3Find the unknown angles in the following isosceles triangles, which represent roof rafters.(a) As this is an isosceles triangle, α = β.SoTherefore(b) As this is an isosceles triangle, γ = δ.SoTherefore2.9 Similar and congruent shapesTwo shapes are said to be similar if they are the same shape but not necessarily the same size. In other words, one may be an enlargement of the other. They may also have different orientations, as in the drawing below.When a photograph is enlarged, the two images are similar.But if a photograph is stretched in only one direction, the resulting shape is not similar to the original.In effect, when two shapes are similar, one is a scaled up (or down) version of the other. Thus an accurate model and its original will be similar in this mathematical sense. If you measure the sides of the model, you will find that to produce the original, each side must be scaled up by the same amount. However, the angles remain the same in each version.The simplest scaled shapes are similar triangles. In two similar triangles, angles in equivalent positions must be the same size. This provides a way of identifying similar triangles.It is not necessary to calculate all the angles in two similar triangles. If two angles in one triangle match two angles in the other, then the third angle must also be the same in both, because in each case it will be 180° minus the sum of the other two angles.Examples of similar triangles are set out below.If two figures are the same shape and the same size, they are said to be congruent.Example 10This diagram shows, in simplified form, a wooden buttress supporting the wall of a medieval church.The angle between the ground and the buttress, , is 65°. By making appropriate assumptions, identify which triangles are similar. Calculate all the angles in the structure.Assume that the wall is vertical, and that the ground and BE are both horizontal. Also assume that BF and CE are at right angles to AD.Consider the angles of ΔACD: and ThenIt is easiest to see which triangles are similar if you look at them in pairs.In each diagram, the two triangles under consideration are emphasised by heavy lines.In (a), both the triangles that are outlined by heavy lines have the same angle at A, 25°, and both also have a right angle (at C and B, respectively). Therefore the third angle in the two triangles (at D and E) must also be the same. (You can confirm this by noticing that these are corresponding angles.) The size of these angles must be 180° − 25° − 90° = 65°.In (b), both triangles have the same angle at A, 25°, and they both have a right angle (at E and F, respectively). Then the third angle in each will be the same size, 65°.In (c), each triangle has a right angle (at E and F, respectively), and in the larger triangle is the same size as in the smaller triangle (they are corresponding angles). These corresponding angles are each 65°; hence the third angle must again be 25°.This gives six triangles, each with angles of 25°, 90° and 65°, and so all are similar.There is a seventh triangle that is also similar to the others, ΔBEC. This has a right angle, and its angle in ΔECD (they are alternate angles), and so is 25°. Its third angle must therefore be 65°.You may have met other examples of similar shapes, for example, when using scale diagrams. The scale plan of a house is similar to the actual layout of the house.Try some yourselfQuestion 1Which of these triangles are similar?Triangles a, c and g are similar since they have angles of 90°, 45° (and hence another angle of 45°).Triangles b and f are similar since they have angles of 90°, 60° (and hence another angle of 30°).Triangles d, e and h are similar since they have angles of 45° and 60° (and hence another angle of 75°).Question 2An aluminium ladder can be used in three different ways:The manufacturer says that in use, each segment of the ladder should make an angle of 20° with the vertical.For each diagram, add construction lines and labels so as to identify two similar triangles. Are any of the similar triangles also congruent?There are many alternative solutions.Here are some similar triangles which are identified by using the labels given in the diagram above:

domestic steps

ΔACD and ΔACB,

stair ladder

ΔSTR, ΔSTQ, ΔPQR, ΔSVU,

extension ladder

ΔFJH, ΔEIH, ΔEGF.

Some congruent triangles are

domestic steps

ΔACD and ΔACB,

stair ladder

ΔQST and ΔSTR.

Question 3This diagram shows the arrangement of the struts in a wall of a shed.The lines OABC and DE are each horizontal. The struts EA and DC are parallel.(a) Which of these are right angles?(b) Write down two angles that are equal to (c) Several of the triangles formed by the struts are similar (that is, they are the same shape). Write down all the triangles that are similar to ΔOAF.3(a) All four of the given angles are right angles.(b) (which is the same as ) is equal to . They are alternate angles. (which is the same as ) is equal to . This is because ΔBCD and ΔOAF are similar: each has a right angle, and and are corresponding angles.(c) There are four triangles that are similar to ΔOAF: they are ΔOBD, ΔDEF, ΔOCD and ΔBCD.3 Areas and volumes3.1 Areas of quadrilaterals and trianglesYou may like to add the area formulas in this section to your notes for future reference.The simplest areas to find are those of rectangles. The area of a rectangle is its length multiplied by its breadth. Sometimes the dimensions of a rectangle are referred to as the base and the height, instead of the length and the breadth. The area is then expressed as the base multiplied by the height.Area of a rectangle = length × breadth = base × heightA square is a special kind of rectangle in which the length is equal to the breadth. Hence its area is the length of one side multiplied by itself, or the length of one side squared.Area of a square = length × length = length^{2}For example, the area of a square mirror with sides 50 cm long is 50 cm × 50 cm = 2500 cm^{2}.Now consider parallelograms.Area of a parallelogram = base × heightIn the formula for the area of a parallelogram, the height is the perpendicular distance from the base to the opposite side. In order to avoid ambiguity it is sometimes called the perpendicular height rather than just the height. The height is not the length of the sloping side.At first sight, the formula for a parallelogram is quite surprising: it is the same formula as that for a rectangle. Imagine the bottom side of the parallelogram is fixed, but the top side slides along a line, as in the diagram below. The top and bottom of the parallelogram remain the same length and the same distance apart, while the other two sides lengthen or shrink. The shape always remains a parallelogram. (Notice that in one position, the parallelogram will become a rectangle – its sides will be at right angles to the base.)The area of the parallelogram stays the same as the parallelogram shifts: it is equal to the area of the rectangle (which, of course, is given by base × height). This is easy to see by looking at the next diagram. In this, the first figure consists of two identical triangles and a parallelogram. Imagine the left-hand triangle slides to the right: it will fit above the other triangle and leave a rectangle to the left. The second figure shows the same two triangles and the rectangle. Therefore the area of the parallelogram must be the same as the area of the rectangle.Next think about the areas of triangles. Any triangle can be seen as half of a parallelogram.So the area of a triangle is half the area of a parallelogram.Again, the height is the perpendicular height, which is now the distance from the base to the opposite corner, or vertex, of the triangle.This formula is true for any triangle, because any triangle will be half of a parallelogram even when the perpendicular height lies outside the triangle, as below.If a triangle does not have a side that is horizontal, it is not clear which side is ‘the base’. The beauty of the formula for the area is that it works no matter which side is called ‘the base’. Thus the area of the following triangle can be evaluated in three ways.You can often use what you know about the areas of rectangles and triangles to find the areas of more complex shapes.Example 11The lawn shown below is trapezium-shaped. Find its area.Divide the lawn into three parts – a rectangle and two triangles. Then combine the two triangles into one.SoExample 12Suppose a friend of yours decides to lay crazy paving in his garden which measures 7 m by 5 m, but he wants to leave two rectangular areas, each 2 m by 1 m, for flowerbeds. What area of crazy paving will be needed?The first thing to do when tackling a problem like this is to draw a diagram, and to include on it all the information that has been given.Note that, as the positions of the flowerbeds have not been specified, it does not matter where they are placed.From the diagram,area of garden = 7 m × 5 m = 35 m^{2},area of one flowerbed = 2 m × 1 m = 2 m^{2}.Therefore,Try some yourselfQuestion 1Find the area of each of these shapes.(a)(b) The trapezium can be split into a triangle and a rectangle:Question 2A girl is decorating a box by glueing wrapping paper on each face. She wants to put paper on the sides, the top and the bottom, and intends to cut out six pieces of paper and stick them on. Assuming no wastage, calculate what area of paper she will need.Therefore, the amount of material needed is 5250 cm^{2}.Question 3A rug measures 3 m by 2 m. It is to be laid on a wooden floor that is 5 m long and 4 m wide. The floorboards not covered by the rug are to be varnished.(a) What area of floor will need to be varnished?(b) A tin of varnish covers 2.5 m^{2}. How many tins will be required?So six tins will have to be purchased.Question 4This diagram represents the end wall of a bungalow; the wall contains two windows. The wall is to be treated with a special protective paint. In order to decide how much paint is required, the owner wants to know the area of the wall. Divide the wall up into simple shapes and then find the total area.The end wall of the bungalow, minus the windows, can be divided into simple shapes as shown.The dimensions of the windows, in metres, are 2.2 m by 1.45 m and 1.25 m by 0.88 m, respectively.Question 5The diagram below shows the dimensions of a frame tent. Calculate the amount of canvas needed to make the tent, ignoring the door which is made of different material.Area of one side of tent = 2 m × 5 m = 10 m^{2}.Area of door = 0.8 m × 1.75 m = 1.4 m^{2}.Total area of canvas= ( 2 × area of one side of sloping roof )+ ( 2 × area of side of tent )+ ( 2 × area of front/back of tent )− ( area of door )= ( 2 × 10.6 m^{2} ) + ( 2 × 10 m^{2} )+ ( 2 × 8.25 m^{2} ) − 1.4 m^{2}= 56.3 m^{2}.So 56.3 m^{2} of canvas are needed.(In practice, the amount needed will depend upon the width of the canvas and on how many joins there are. It is likely that at least 60 m^{2} will be needed.)3.2 Areas of circlesThere are two very famous formulas for circles:circumference of a circle = × diameterandarea of a circle = × radius^{2}. is the Greek letter for ‘p’ and it has the name ‘pi’. Its value is approximately 3.14. Most calculators have a key for which you can use when carrying out calculations.Try measuring the circumference and diameter of some circular objects such as tins, bottles or bowls. For each object, divide the circumference by the diameter. You should find that your answer is always just over 3. In fact the ratio is the constant . Therefore:Circumference of a circle = × diameterSince the diameter is twice the radius, this formula can be written ascircumference = × 2 ×radius = 2 × radius.The formula for the area of a circle can be explained, as outlined below.The circle here has been divided into equal ‘slices’ or sectors. The eight sectors can then be cut out and rearranged into the shape shown: this shape has the same area as the circle.You can see that the total distance from A to B along the ‘bumps’ is the same as half the circumference of the circle, that is × 2 × radius = × radius. Also the length OA is the same as the radius of the circle.Imagine dividing the circle into more and more sectors and rearranging them as described above. For example, dividing the circle into 16 equal sectors gives the following shape, whose area is still the same as that of the circle.Again the total distance from A to B along the bumps is × radius, and the length of OA is the same as the radius.Notice how the rearranged shape is beginning to look more like a rectangle. The more sectors, the straighter AB will become and the more perpendicular OA will be. Eventually it will not be possible to distinguish the rearranged shape from a rectangle. The area of this rectangle will be the same as that of the circle, and its sides will have the lengths × radius (for AB) and radius (for OA). So the following formula can be deduced:Area of a circle = × (radius)^{2}Example 13A circular flowerbed is situated in the centre of a traffic roundabout. The radius of the flowerbed is 10 m. Find its circumference and its area.Circumference = 2 × radius = 2 × 10 m = 20 m ≈ 62.8 mExample 14A circular pond has a diameter of 7 m. A 1 m wide gravel path is to be laid around the pond. What is the area of the path?The diagram shows the area of the path.This area can be found by calculating the area of the path and pond together, and then subtracting the area of the pond.So, Area of path = Area of path and pond – Area of pond.The pond and the path form a circle of diameter 1m + 1m + 7m = 9m. A circle of diameter 9m has a radius of 4.5 m.So, the area of the path and pond = × 4.5^{2} m^{2}.The pond has diameter 7 m, so its radius is 7 m ÷ 2 = 3.5 m.Hence, the area of the pond = × 3.5^{2} m^{2}.So, the area of the path = × 4.5^{2} m^{2} − × 3.5^{2} m^{2} 25 m^{2}.Try some yourselfQuestion 1Find the area of a circle of (a) radius 8 cm, and (b) radius 15 m.(a)(b)Question 2Calculate the areas of the following shapes:(a) Area = 10 m × 6 m = 60 m^{2}.(b) Area = (1.1 m)^{2} ≈ 3.80 m^{2}.(c) Area = × 10m × 5m = 25m^{2}.Question 3Use your answers to the previous question to find the area of turf needed for the proposed lawn shown below, which has a circular flowerbed in the middle. Round your answer to the nearest square metre.Add together the areas of the rectangle and the triangle from Question 2, and subtract the area of the circle to findQuestion 4A manufacturer produces a patio kit consisting of 28 paving slabs which are shaped so that they fit together to form rings, as shown. The outside edge of each ring is a circle, and all three circles have the same centre. Circles with the same centre are called concentric circles. The slabs are of three sizes, one for each ring of the patio. All of the slabs in a particular ring are identical.The radii of the three circles are 0.4 m, 0.8 m and 1.2 m.Making appropriate assumptions, calculate which of the three types of slab is the heaviest and which the lightest.In the following calculation, full calculator accuracy numbers are indicated by three dots. For example the full calculator accuracy value for is written as 3.141 ... .There are four slabs in this circle, so each slab will have aThe surface area of the eight slabs in the inner ring is calculated by subtracting the central circle from the circle with radius 0.8 m:There are eight slabs in the inner ring, so each slab will have aUse a similar method to find the surface area of the outer ring of slabs:There are sixteen slabs in the inner ring, so each slab will have a

Type of slab

Surface area/m^{2}

Central slab

0.1257

Lightest

Inner ring slab

0.1885

Heaviest

Outer ring slab

0.1571

Assuming that the slabs are of equal thickness, and are made of the same material, the weights of the slabs will be proportional to the surface areas. So the results show that the lightest slabs are in the central circle and the heaviest in the inner ring.HintTry breaking the problem in question 4 down into steps and considering what you know about what you want to find out. You know the radii of the circles, so can find the area of each circle. That should help you to find the area of each ‘ring’ of slabs. Then count how may slabs are in each ring and use that to work out the area of each slab.3.3 VolumesWhat is a volume? The word usually refers to the amount of three-dimensional space that an object occupies. It is commonly measured in cubic centimetres (cm^{3}) or cubic metres (m^{3}).A closely related idea is capacity; this is used to specify the volume of liquid or gas that a container can actually hold. You might refer to the volume of a brick and the capacity of a jug – but not vice versa. Note that a container with a particular volume will not necessarily have the same amount of capacity. For example, a toilet cistern will have a smaller capacity than its total volume because the overflow pipe makes the volume above the pipe outlet unusable. Some units are used only for capacity – examples are litre, gallon and pint; cubic centimetres and cubic metres can be used for either capacity or volume.One of the simplest solid shapes is a cube; it has six identical square faces.Volume of a cube = length × length × length = (length) ^{3}A cuboid (or rectangular box) has 6 rectangular faces as shown below.Volume of a rectangular box = length × breadth × heightThe length × breadth is the area of the bottom (or top) of the box, so an alternative formula isvolume of box = area of base × height.The volume formula can also be written asvolume = area of end face × lengthorvolume = area of front face × breadth3.4 Cylinders and shapes with a uniform cross-sectionAn important idea when calculating volumes of simple shapes is that of a cross-section. In the case of the rectangular box considered above, it is possible to slice through the box horizontally so that the sliced area is exactly the same as the area of the base or top; in other words, the areas of the horizontal cross-sections are equal.Likewise, you could slice through the box vertically in either of two different directions, producing cross-sections that are the same as either the end faces or the front and back faces.For objects that have a constant cross-sectional area, there is a very useful formula for the volume.Volume = cross-sectional area × length(length is measured at right angles to the cross section)Notice that this fits with the formula for the volume of a rectangular box.Bear in mind that many objects can only be sliced in one direction to produce a constant cross-sectional area. This bread bin is one example.Another example is a cylinder. The formula at the top of the page can be used to find the volume of a cylinder because a cylinder has a constant cross-sectional area if it is sliced parallel to the circular face – the cross-sectional area is the area of the circle that forms the base of the cylinder, that is × (radius)^{2}. The following formula can then be deduced.Volume of cylinder = × (radius)^{2} × heightExample 15Find the volumes of these objects.(a) For this object,cross-sectional area = 8 cm × 8 cm = 64 cm^{2},thereforevolume = 16 cm × 64 cm^{2} = 1024 cm^{3}.(b) For this object,thereforevolume ≅ 157 m^{2} × 100 m = 15 700 m^{3}.Try some yourselfQuestion 1Find the volumes of these objects.(a)Sovolume = 50.265 cm^{2} × 10 cm = 502.65 cm^{3}.Thus the volume is 503 cm^{3} (to the nearest cubic centimetre).(If you used the approximate value of 3.14 for , you will have got a cross-sectional area of 50.24 cm^{2} and a volume of 502.4 cm^{3}.)(b)Sovolume = 37.5 m^{2} × 10 m = 375 m^{3}.Question 2Two car manufacturers both claim that their models have an engine capacity of 2 litres. The two models have four-cylinder, four-stroke engines.The table below shows the details of the four cylinders.

Car model

Cylinder diameter (bore)/mm

Cylinder height (stroke)/mm

Number of cylinders

A

86

86

4

B

92

75

4

By working out the total volume of the four cylinders for each model in cm^{3}, find out if the manufacturers’ claims are true.(Hint: 1 litre = 1000 cm^{3}.)Car A has four cylinders, each with a radius of 4.3 cm and a height of 8.6 cm. The volume of one cylinder is calculated by using the formulaSo, the four cylinders will haveCar B has four cylinders, each with a radius of 4.6 cm and a height of 7.5 cm. From the same formula, the four cylinders will haveTherefore, both engines have a cubic capacity very close to 2000 cm^{3}. They are both said to have two-litre engines. Hence the claims of both manufacturers are true.Question 3The guttering pictured here has a semicircular cross-section. Find the volume of water that the guttering will hold when full.The cross-section of the guttering is a semicircle of radius 0.05 m. SoThen, since the length of the guttering is 12m,Therefore the guttering will hold about 0.047 m^{3} of water.3.5 Scaling areas and volumesIn OpenLearn course Diagrams, graphs and charts you saw how a scale is used on plans of houses and other structures. The scale makes it possible to take a length on the plan and calculate the corresponding length in reality. The scale can also be used to convert between areas on the plan and real areas. Moreover, if a three-dimensional scale model is made, it is possible to use the scale to convert between volumes in the model and the real volumes.Example 16Dudley's and June's hobby is constructing dolls' houses. They decide to make a model of their own house, using a scale in which 1 cm on the model represents 20 cm on their real house.They are making the curtains for the model. The window in the real dining room measures 240 cm by 120 cm. What is the area of the real window and the area of the window in the model? How many times greater is the real area?The real window has an area of 240 cm × 120 cm = 28 800 cm^{2}. It might be easier to think of this in square metres, that isTo find the dimensions of the window in the model, divide the real lengths by 20 as the scale is 1 cm to 20 cm:or in square metres,Now find the number of times that the area of the real window exceeds the area of the window in the model:working in square centimetres 28 800 ÷ 72 = 400,orworking in square metres 2.88 ÷ 0.0072 = 400.As the real lengths are 20 times greater than those on the model, the areas are 20^{2} (= 400) times greater.This example has demonstrated a general result:To scale areas, multiply or divide by the scale squared.Example 17Dudley and June have a cold-water tank in their loft, which has a capacity of 250 litres. If they make a scale model of the tank, what will its capacity be?Just as areas must be multiplied or divided by the scale squared, so volumes (and capacities) must be multiplied or divided by the cube of the scale. Here the capacity of the real tank must be divided by 20^{3} (= 8000). Thereforecapacity of model tank = 250 litres ÷ 8000 = 0.03125 litres.As there are 1000 cm^{3} in one litre,capacity = 0.03125 × 1000 cm^{3} = 31.25 cm^{3}.A check on this value can be made by considering the volume of the real water tank. If it is assumed that the full tank holds exactly 250 litres, the volume of the tank would be at least 250 × 1000 cm^{3} = 250 000 cm^{3}.The question does not give the dimensions of the real tank, but to produce this volume, the dimensions might perhaps be 50 cm by 50 cm by 100 cm. (Note that 50 × 50 × 100 = 250 000.) The dimensions of the model of such a tank would beSovolume of model tank = 31.25 cm^{3}.This example illustrates a general result:To scale volumes, multiply or divide by the scale cubed.Try some yourselfQuestion 1Calculate the area of a carpet in a model house if the real carpet has an area of 22 m^{2}. On the scale used, 1 cm represents 0.25 m.(a) Since 1 cm in the model represents 25 cm in real life, areas must be scaled by 25 × 25 (the area scale is 25^{2} because the length scale is 25). SoThis can be converted to square centimetres by multiplying by 100 × 100:Alternatively, you could convert to cm^{2} first:SoQuestion 2A model steam engine that runs in a park is built to a scale such that 1 cm represents 0.2 m. On the model there is space in the tender for m^{3} of coal. What volume of coal could be carried in the real engine's tender?Since 1 cm in the model represents 20 cm in real life, volumes must be scaled by 20 × 20 × 20.So the volume of the tender in real life must beThus the volume of coal that could be carried in the real engine's tender is 40 m^{3}.4 OpenMark quizNow try the quiz and see if there are any areas you need to work on.ConclusionThis free course provided an introduction to studying Mathematics. It took you through a series of exercises designed to develop your approach to study and learning at a distance and helped to improve your confidence as an independent learner.Keep on learningStudy another free courseThere are more than 800 courses on OpenLearn for you to choose from on a range of subjects. Find out more about all our free courses.Take your studies furtherFind out more about studying with The Open University by visiting our online prospectus. If you are new to university study, you may be interested in our Access Courses or Certificates.What’s new from OpenLearn?Sign up to our newsletter or view a sample.For reference, full URLs to pages listed above:OpenLearn – www.open.edu/openlearn/free-coursesVisiting our online prospectus – www.open.ac.uk/coursesAccess Courses – www.open.ac.uk/courses/do-it/accessCertificates – www.open.ac.uk/courses/certificates-heNewsletter – www.open.edu/openlearn/about-openlearn/subscribe-the-openlearn-newsletterAdapted from the works of Steve Jurvetson: [Details correct as of 8th July 2008]Course image: Daniel Oines in Flickr made available under Creative Commons Attribution 2.0 Licence.All other material contained within this course originated at The Open University.Don't miss out:If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University - www.open.edu/openlearn/free-coursesDiscussion2016091000