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Introduction

Geometry (Greek: geo = earth, metria = measure) arose as the field of knowledge dealing with spatial relationships. It was one of the two fields of pre-modern mathematics, the other being the study of numbers. In modern times, geometric concepts have become very complex and abstract and are barely recognizable as the descendants of early geometry. Geometry is often split into Euclidean geometry and analytical geometry. Euclidean geometry is covered in this chapter.

Research Project : History of Geometry

Work in pairs or groups and investigate the history of the foundation of geometry. Describe the various stages of development and how the following cultures used geometry to improve their lives. This list should serve as a guideline and provide the minimum requirement, there are many other people who contributed to the foundation of geometry.

  1. Ancient Indian geometry (c. 3000 - 500 B.C.)
    1. Harappan geometry
    2. Vedic geometry
  2. Classical Greek geometry (c. 600 - 300 B.C.)
    1. Thales and Pythagoras
    2. Plato
  3. Hellenistic geometry (c. 300 B.C - 500 C.E.)
    1. Euclid
    2. Archimedes

Quadrilaterals

In this section we will look at the properties of some special quadrilaterals. We will then use these properties to solve geometrical problems. It should be noted that although all the properties of a figure are given, we only need one unique property of the quadrilateral to prove that it is that quadrilateral. For example, if we have a quadrilateral with two pairs of opposite sides parallel, then that quadrilateral is a parallelogram. We can then prove the other properties of the quadrilateral using what we have learnt about parallel lines and triangles.

Trapezium

A trapezium is a quadrilateral with one pair of parallel opposite sides. It may also be called a trapezoid. A special type of trapezium is the isosceles trapezium, where one pair of opposite sides is parallel, the other pair of sides is equal in length and the angles at the ends of each parallel side are equal. An isosceles trapezium has one line of symmetry and its diagonals are equal in length.

Note: The term trapezoid is predominantly used in North America and refers to what we call a trapezium. Rather confusingly, they use the term 'trapezium' to refer to a general irregular quadrilateral, that is a quadrilateral with no parallel sides!

Figure 1: Examples of trapeziums.
Figure 1 (MG10C13_040.png)

Parallelogram

A trapezium with both sets of opposite sides parallel is called a parallelogram. A summary of the properties of a parallelogram is:

  • Both pairs of opposite sides are parallel.
  • Both pairs of opposite sides are equal in length.
  • Both pairs of opposite angles are equal.
  • Both diagonals bisect each other (i.e. they cut each other in half).
Figure 2: An example of a parallelogram.
Figure 2 (MG10C13_041.png)

Rectangle

A rectangle is a parallelogram that has all four angles equal to 90∘90∘. A summary of the properties of a rectangle is:

  • Both pairs of opposite sides are parallel.
  • Both pairs of opposite sides are of equal length.
  • Both diagonals bisect each other.
  • Diagonals are equal in length.
  • All angles at the corners are right angles.
Figure 3: Example of a rectangle.
Figure 3 (MG10C13_042.png)

Rhombus

A rhombus is a parallelogram that has all four sides of equal length. A summary of the properties of a rhombus is:

  • Both pairs of opposite sides are parallel.
  • All sides are equal in length.
  • Both pairs of opposite angles are equal.
  • Both diagonals bisect each other at 90∘90∘.
  • Diagonals of a rhombus bisect both pairs of opposite angles.
Figure 4: An example of a rhombus. A rhombus is a parallelogram with all sides equal.
Figure 4 (MG10C13_043.png)

Square

A square is a rhombus that has all four angles equal to 90∘∘.

A summary of the properties of a square is:

  • Both pairs of opposite sides are parallel.
  • All sides are equal in length.
  • All angles are equal to 90∘90∘.
  • Both pairs of opposite angles are equal.
  • Both diagonals bisect each other at 90∘90∘.
  • Diagonals are equal in length.
  • Diagonals bisect both pairs of opposite angles (ie. all 45∘45∘).
Figure 5: An example of a square. A square is a rhombus with all angles equal to 90∘∘.
Figure 5 (MG10C13_044.png)

Kite

A kite is a quadrilateral with two pairs of adjacent sides equal.

A summary of the properties of a kite is:

  • Two pairs of adjacent sides are equal in length.
  • One pair of opposite angles are equal where the angles are between unequal sides.
  • One diagonal bisects the other diagonal and one diagonal bisects one pair of opposite angles.
  • Diagonals intersect at right-angles.
Figure 6: An example of a kite.
Figure 6 (MG10C13_045.png)

Rectangles are a special case (or a subset) of parallelograms. Rectangles are parallelograms that have all angles equal to 90. Squares are a special case (or subset) of rectangles. Squares are rectangles that have all sides equal in length. So all squares are parallelograms and rectangles. So if you are asked to prove that a quadrilateral is a parallelogram, it is enough to show that both pairs of opposite sides are parallel. But if you are asked to prove that a quadrilateral is a square, then you must also show that the angles are all right angles and the sides are equal in length.

Polygons

Polygons are all around us. A stop sign is in the shape of an octagon, an eight-sided polygon. The honeycomb of a beehive consists of hexagonal cells. The top of a desk is a rectangle. Note that although in the first two of these cases the sides of the polygon are all the same, but this need not be the case. Polygons with all sides and angles the same are called 'regular', while those with some sides or angles that are different are called 'irregular'. Although we often work with irregular triangles and quadrilaterals, once one gets up to polygons with greater than four sides, the more interesting ones are often the regular ones.

In this section, you will learn about similar polygons.

Similarity of Polygons

Discussion : Similar Triangles

Fill in the table using the diagram and then answer the questions that follow.

Table 1
AB DE AB DE =...cm...cm=......cm...cm=... A^A^=...∘∘ D^D^...∘∘
BC EF BC EF =...cm...cm=......cm...cm=... B^B^=...∘∘ E^E^=...∘∘
AC DF AC DF =...cm...cm=......cm...cm=... C^C^...∘∘ F^F^=...∘∘

Figure 7
Figure 7 (MG10C14_010.png)

  1. What can you say about the numbers you calculated for: AB DE AB DE , BC EF BC EF , AC DF AC DF ?
  2. What can you say about A^A^ and D^D^?
  3. What can you say about B^B^ and E^E^?
  4. What can you say about C^C^ and F^F^?

If two polygons are similar, one is an enlargement of the other. This means that the two polygons will have the same angles and their sides will be in the same proportion.

We use the symbol |||||| to mean is similar to.

Definition 1: Similar Polygons

Two polygons are similar if:

  1. their corresponding angles are equal, and
  2. the ratios of corresponding sides are equal.

Exercise 1: Similarity of Polygons

Show that the following two polygons are similar.

Figure 8
Figure 8 (MG10C14_011.png)

Solution
  1. Step 1. Determine what is required :

    We are required to show that the pair of polygons is similar. We can do this by showing that the ratio of corresponding sides is equal and by showing that corresponding angles are equal.

  2. Step 2. Corresponding angles :

    We are given the angles. So, we can show that corresponding angles are equal.

  3. Step 3. Show that corresponding angles are equal :

    All angles are given to be 90∘∘ and

    A ^ = E ^ B ^ = F ^ C ^ = G ^ D ^ = H ^ A ^ = E ^ B ^ = F ^ C ^ = G ^ D ^ = H ^
    (1)
  4. Step 4. Show that corresponding sides have equal ratios :

    We first need to see which sides correspond. The rectangles have two equal long sides and two equal short sides. We need to compare the ratio of the long side lengths of the two different rectangles as well as the ratio of the short side lenghts.

    Long sides, large rectangle values over small rectangle values:

    Ratio = 2 L L = 2 Ratio = 2 L L = 2
    (2)

    Short sides, large rectangle values over small rectangle values:

    Ratio = L 1 2 L = 1 1 2 = 2 Ratio = L 1 2 L = 1 1 2 = 2
    (3)

    The ratios of the corresponding sides are equal, 2 in this case.

  5. Step 5. Final answer :

    Since corresponding angles are equal and the ratios of the corresponding sides are equal the polygons ABCD and EFGH are similar.

Tip:

All squares are similar.

Exercise 2: Similarity of Polygons

If two pentagons ABCDE and GHJKL are similar, determine the lengths of the sides and angles labelled with letters:

Figure 9
Figure 9 (MG10C14_012.png)

Solution
  1. Step 1. Determine what is given :

    We are given that ABCDE and GHJKL are similar. This means that:

    AB GH = BC HJ = CD JK = DE KL = EA LG AB GH = BC HJ = CD JK = DE KL = EA LG
    (4)

    and

    A ^ = G ^ B ^ = H ^ C ^ = J ^ D ^ = K ^ E ^ = L ^ A ^ = G ^ B ^ = H ^ C ^ = J ^ D ^ = K ^ E ^ = L ^
    (5)
  2. Step 2. Determine what is required :

    We are required to determine the

    1. lengths: aa, bb, cc and dd, and
    2. angles: ee, ff and gg.
  3. Step 3. Decide how to approach the problem :

    The corresponding angles are equal, so no calculation is needed. We are given one pair of sides DCDC and KJKJ that correspond. DCKJ=4,53=1,5DCKJ=4,53=1,5 so we know that all sides of KJHGLKJHGL are 1,5 times smaller than ABCDEABCDE.

  4. Step 4. Calculate lengths :
    a 2 = 1 , 5 ∴ a = 2 × 1 , 5 = 3 b 1 , 5 = 1 , 5 ∴ b = 1 , 5 × 1 , 5 = 2 , 25 6 c = 1 , 5 ∴ c = 6 ÷ 1 , 5 = 4 d = 3 1 , 5 ∴ d = 2 a 2 = 1 , 5 ∴ a = 2 × 1 , 5 = 3 b 1 , 5 = 1 , 5 ∴ b = 1 , 5 × 1 , 5 = 2 , 25 6 c = 1 , 5 ∴ c = 6 ÷ 1 , 5 = 4 d = 3 1 , 5 ∴ d = 2
    (6)
  5. Step 5. Calculate angles :
    e = 92 ∘ ( corresponds to H ) f = 120 ∘ ( corresponds to D ) g = 40 ∘ ( corresponds to E ) e = 92 ∘ ( corresponds to H ) f = 120 ∘ ( corresponds to D ) g = 40 ∘ ( corresponds to E )
    (7)
  6. Step 6. Write the final answer :
    a = 3 b = 2 , 25 c = 4 d = 2 e = 92 ∘ f = 120 ∘ g = 40 ∘ a = 3 b = 2 , 25 c = 4 d = 2 e = 92 ∘ f = 120 ∘ g = 40 ∘
    (8)

Activity: Similarity of Equilateral Triangles

Working in pairs, show that all equilateral triangles are similar.

Polygons-mixed

  1. Find the values of the unknowns in each case. Give reasons.
    Figure 10
    Figure 10 (mg10c14_2.png)
    Click here for the solution
  2. Find the angles and lengths marked with letters in the following figures:
    Figure 11
    Figure 11 (MG10C14_014.png)
    Click here for the solution

Investigation: Defining polygons

Investigate the different ways of defining polygons. Polygons that you should pay special attention to are:

  • Isoceles, equilateral and right-angled triangle
  • Kites, parallelograms, rectangles, rhombuses (or 'rhombi'), squares and trapeziums

Things to consider are how these figures have been defined in this book and what alternative definitions exist. For example, a triangle is a three-sided polygon or a figure having three sides and three angles. Triangles can be classified using either their sides or their angles. Could you also classify quadrilaterals in this way? What other names exist for these figures? For example, quadrilaterals can also be called tetragons.

Proofs and conjectures in geometry

You have seen how to use geometry and the properties of polygons to help you find unknown lengths and angles in various quadrilaterals and polygons. We will now extend this work to proving some of the properties and to solving riders. A conjecture is the mathematicians way of saying I believe that this is true, but I have no proof. The following worked examples will help make this clearer.

Exercise 3: Proofs - 1

Given quadrilateral ABCD, with AB∥CDAB∥CD and AD∥BCAD∥BC, prove that B A ^ D = B C ^ A B A ^ D= B C ^ A and A B ^ C = A D ^ C A B ^ C= A D ^ C.

Solution

  1. Step 1. Draw a diagram: We draw the following diagram and construct the diagonals.
    Figure 12
    Figure 12 (geomproof1.png)
  2. Step 2. Write down what you are given and what you need: Given: AB∥CDAB∥CD and AD∥BCAD∥BC. We need to prove A=CA=C and B=DB=D. In the formal language of maths we say that we are required to prove (RTP) B A ^ D = B C ^ A B A ^ D= B C ^ A and A B ^ C = A D ^ C A B ^ C= A D ^ C.
  3. Step 3. Solve the problem:
    B A ^ C = A C ^ D ( corresponding angles ) D A ^ C = B C ^ A ( corresponding angles ) B A ^ D = B C ^ A B A ^ C = A C ^ D ( corresponding angles ) D A ^ C = B C ^ A ( corresponding angles ) B A ^ D = B C ^ A
    (9)
    Similarly we find that:
    A B ^ C = A D ^ C A B ^ C= A D ^ C
    (10)

Proofs - 2 4

In parallelogram ABCD, the bisectors of the angles (AW, BX, CY and DZ) have been constructed:

Figure 13
Figure 13 (geomproof2.png)
You are also given that AB=CDAB=CD, AD=BCAD=BC, AB∥CDAB∥CD, AD∥BCAD∥BC, A ^ = C ^ A ^ = C ^ , and B ^ = D ^ B ^ = D ^ . Prove that MNOP is a parallelogram.

Solution

  1. Step 1. Write down what you are given and what you need to prove: Given: AB=CDAB=CD, AD=BCAD=BC, AB∥CDAB∥CD, AD∥BCAD∥BC, A ^ = C ^ A ^ = C ^ , and B ^ = D ^ B ^ = D ^ . RTP: MNOP is a parallelogram.
  2. Step 2. Solve the problem:
    In ▵ADW and ▵CBY D A ^ W = B C ^ Y ( given ) A D ^ C = A B ^ C ( given ) AD = BC(given) ∴▵ADW = ▵ CBY(AAS) ∴DW=BY In ▵ADW and ▵CBY D A ^ W = B C ^ Y ( given ) A D ^ C = A B ^ C ( given ) AD = BC(given) ∴▵ADW = ▵ CBY(AAS) ∴DW=BY
    (11)
    In ▵ABX and ▵CDZ D C ^ Z = B A ^ X ( given ) Z D ^ C = X B ^ A ( given ) DC = AB(given) ∴▵ABX ≡ ▵ CDZ(AAS) ∴AX=CZ In ▵ABX and ▵CDZ D C ^ Z = B A ^ X ( given ) Z D ^ C = X B ^ A ( given ) DC = AB(given) ∴▵ABX ≡ ▵ CDZ(AAS) ∴AX=CZ
    (12)
    In ▵XAM and ▵ZCO X A ^ M = Z C ^ O ( given ) A X ^ M = C Z ^ O ( proven above ) AX = CZ(proven above) ∴▵XAM ≡ ▵ COZ(AAS) ∴ A O ^ C = A M ^ X In ▵XAM and ▵ZCO X A ^ M = Z C ^ O ( given ) A X ^ M = C Z ^ O ( proven above ) AX = CZ(proven above) ∴▵XAM ≡ ▵ COZ(AAS) ∴ A O ^ C = A M ^ X
    (13)
    A M ^ X = P M ^ N(vert. opp. ∠'s) C O ^ Z = N O ^ P(vert. opp. ∠'s) ∴ P M ^ N = N O ^ P A M ^ X = P M ^ N(vert. opp. ∠'s) C O ^ Z = N O ^ P(vert. opp. ∠'s) ∴ P M ^ N = N O ^ P
    (14)
    In ▵BYN and ▵DWP Y B ^ N = W D ^ P ( given ) B Y ^ N = W D ^ P ( proven above ) DW = BY(proven above) ∴▵YBN ≡ ▵ WDP(AAS) ∴ B N ^ Y = D P ^ W In ▵BYN and ▵DWP Y B ^ N = W D ^ P ( given ) B Y ^ N = W D ^ P ( proven above ) DW = BY(proven above) ∴▵YBN ≡ ▵ WDP(AAS) ∴ B N ^ Y = D P ^ W
    (15)
    D P ^ W = M P ^ O(vert. opp. ∠'s) B N ^ Y = O N ^ M(vert. opp. ∠'s) ∴ M P ^ O = O N ^ M D P ^ W = M P ^ O(vert. opp. ∠'s) B N ^ Y = O N ^ M(vert. opp. ∠'s) ∴ M P ^ O = O N ^ M
    (16)

    ∴∴ MNOP is a parallelogram (both pairs opp. ∠∠'s ==, and therefore both pairs opp. sides parallel too)

Warning:

It is very important to note that a single counter example disproves a conjecture. Also numerous specific supporting examples do not prove a conjecture.

Measurement

Areas of Polygons

  1. Area of triangle: 12×12× base ×× perpendicular height
    Figure 14
    Figure 14 (MG10C13_047.png)
  2. Area of trapezium: 12×12× (sum of ∥∥ (parallel) sides) ×× perpendicular height
    Figure 15
    Figure 15 (MG10C13_048.png)
  3. Area of parallelogram and rhombus: base ×× perpendicular height
    Figure 16
    Figure 16 (MG10C13_049.png)
  4. Area of rectangle: length ×× breadth
    Figure 17
    Figure 17 (MG10C13_050.png)
  5. Area of square: length of side ×× length of side
    Figure 18
    Figure 18 (MG10C13_051.png)
  6. Area of circle: ππ x radius22
    Figure 19
    Figure 19 (MG10C13_052.png)

Figure 20
Khan Academy video on area and perimeter

Figure 21
Khan Academy video on area of a circle

Exercise 5: Finding areas

Find the area of the following figure:

Figure 22
Figure 22 (area1.png)

Solution
  1. Step 1. Find the height: We first need to find the height, BE, of the parallelogram. We can use Pythagoras to do this:
    BE2 = AB2−AE2 BE2 = 52−32 BE2 = 16 BE = 4BE2 = AB2−AE2 BE2 = 52−32 BE2 = 16 BE = 4
    (17)
  2. Step 2. Apply the formula: We apply the formula for the area of a parallelogram to find the area:
    Area = h × b = 4 × 7 = 28 Area = h × b = 4 × 7 = 28
    (18)

Polygons

  1. For each case below, say whether the statement is true or false. For false statements, give a counter-example to prove it:
    1. All squares are rectangles
    2. All rectangles are squares
    3. All pentagons are similar
    4. All equilateral triangles are similar
    5. All pentagons are congruent
    6. All equilateral triangles are congruent
    Click here for the solution
  2. Find the areas of each of the given figures - remember area is measured in square units (cm22, m22, mm22).
    Figure 23
    Figure 23 (MG10C13_064.png)
    Click here for the solution

Right prisms and cylinders

In this section we study how to calculate the surface areas and volumes of right prisms and cylinders. A right prism is a polygon that has been stretched out into a tube so that the height of the tube is perpendicular to the base (the definition is motivated by the fact that the angle between base and side form a right angle). A square prism has a base that is a square and a triangular prism has a base that is a triangle.

Figure 24: Examples of a right square prism, a right triangular prism and a cylinder.
Figure 24 (MG10C14_001.png)

It is relatively simple to calculate the surface areas and volumes of prisms.

Surface Area

The term surface area refers to the total area of the exposed or outside surfaces of a prism. This is easier to understand if you imagine the prism as a solid object.

If you examine the prisms in Figure 24, you will see that each face of a prism is a simple polygon. For example, the triangular prism has two faces that are triangles and three faces that are rectangles. Therefore, in order to calculate the surface area of a prism you simply have to calculate the area of each face and add it up. In the case of a cylinder the top and bottom faces are circles, while the curved surface flattens into a rectangle.

Surface Area of Prisms

Calculate the area of each face and add the areas together to get the surface area. To do this you need to determine the correct shape of each and every face of the prism and then for each one determine the surface area. The sum of the surface areas of all the faces will give you the total surface area of the prism.

Discussion : surface areas

In pairs, study the following prisms and the adjacent image showing the various surfaces that make up the prism. Explain to your partner, how each relates to the other.

Figure 25
Figure 25 (MG10C14_002.png)

Activity: Surface areas

Find (or take one yourself) a picture of a building that does not have a well defined shape (i.e. is not simply a rectangle). For example a castle with towers, or a house with gable windows or a porch. Assume you have to paint the outside of the building. How much paint would you need? Think about what you have learnt about surface area and the area of polygons. Can you find regular polygons on your picture and use those to find the surface area?

Surface areas

  1. Calculate the surface area in each of the following:
    Figure 26
    Figure 26 (MG10C14_003.png)
    Click here for the solution
  2. If a litre of paint covers an area of 2m22m2, how much paint does a painter need to cover:
    1. A rectangular swimming pool with dimensions 4m×3m×2,5m4m×3m×2,5m, inside walls and floor only.
    2. The inside walls and floor of a circular reservoir with diameter 4m4m and height 2,5m2,5m
    Figure 27
    Figure 27 (MG10C14_004.png)
    Click here for the solution

Volume

The volume of a right prism is calculated by multiplying the area of the base by the height. So, for a square prism of side length aa and height hh the volume is a×a×h=a2ha×a×h=a2h.

Volume of Prisms

Calculate the area of the base and multiply by the height to get the volume of a prism.

Exercise 6

Find the surface area and volume for the a square prism of height 4cm4cm and base length, 3cm3cm.

Figure 28
Figure 28 (squareprism.png)

Solution
  1. Step 1. Find the surface area: We use the formula for the surface area of a prism:
    S.A. = 2 2 L × b + b × h = 2 2 3 × 4 + 3 × 4 = 72cm 2 S.A. = 2 2 L × b + b × h = 2 2 3 × 4 + 3 × 4 = 72cm 2
    (19)
  2. Step 2. Find the volume: To find the volume of the prism, we find the area of the base and multiply it by the height:
    V = l 2 × h = 3 2 × 4 = 36cm 3 V = l 2 × h = 3 2 × 4 = 36cm 3
    (20)

Volume

  1. Write down the formula for each of the following volumes:
    Figure 29
    Figure 29 (MG10C14_005.png)
    Click here for the solution
  2. Calculate the following volumes:
    Figure 30
    Figure 30 (MG10C14_006.png)
    Click here for the solution
  3. A cube is a special prism that has all edges equal. This means that each face is a square. An example of a cube is a die. Show that for a cube with side length aa, the surface area is 6a26a2 and the volume is a3a3.
    Figure 31
    Figure 31 (MG10C14_007.png)
    Click here for the solution

Now, what happens to the surface area if one dimension is multiplied by a constant? For example, how does the surface area change when the height of a rectangular prism is divided by 2?

Figure 32: Rectangular prisms
Figure 32 (MG10C14_008.png)

Table 2
Surface area=2(l×h+l×b+b×h)Surface area=2(l×h+l×b+b×h) Surface area=2(l×12h+l×b+b×12h)Surface area=2(l×12h+l×b+b×12h)
Volume=l×b×hVolume=l×b×h Volume=l×b×12h =12×l×b×hVolume=l×b×12h =12×l×b×h

Exercise 7: Scaling the dimensions of a prism

The size of a prism is specified by the length of its sides. The prism in the diagram has sides of lengths LL, bb and hh.

Figure 33
Figure 33 (MG10C14_009.png)

  1. Consider enlarging all sides of the prism by a constant factor xx, where x>1x>1. Calculate the volume and surface area of the enlarged prism as a function of the factor xx and the volume of the original volume.
  2. In the same way as above now consider the case, where 0<x<10<x<1. Now calculate the reduction factor in the volume and the surface area.
Solution
  1. Step 1. Identify :

    The volume of a prism is given by: V=L×b×hV=L×b×h

    The surface area of the prism is given by: A=2×(L×b+L×h+b×h)A=2×(L×b+L×h+b×h)

  2. Step 2. Rescale :

    If all the sides of the prism get rescaled, the new sides will be:

    L ' = x × L b ' = x × b h ' = x × h L ' = x × L b ' = x × b h ' = x × h
    (21)

    The new volume will then be given by:

    V ' = L ' × b ' × h ' = x × L × x × b × x × h = x 3 × L × b × h = x 3 × V V ' = L ' × b ' × h ' = x × L × x × b × x × h = x 3 × L × b × h = x 3 × V
    (22)

    The new surface area of the prism will be given by:

    A ' = 2 × ( L ' × b ' + L ' × h ' + b ' × h ' ) = 2 × ( x × L × x × b + x × L × x × h + x × b × x × h ) = x 2 × 2 × ( L × b + L × h + b × h ) = x 2 × A A ' = 2 × ( L ' × b ' + L ' × h ' + b ' × h ' ) = 2 × ( x × L × x × b + x × L × x × h + x × b × x × h ) = x 2 × 2 × ( L × b + L × h + b × h ) = x 2 × A
    (23)
  3. Step 3. Interpreting the above results :
    1. We found above that the new volume is given by: V'=x3×VV'=x3×V Since x>1x>1, the volume of the prism will be increased by a factor of x3x3. The surface area of the rescaled prism was given by: A'=x2×AA'=x2×A Again, since x>1x>1, the surface area will be increased by a factor of x2x2. Surface areas which are two dimensional increase with the square of the factor while volumes, which are three dimensional, increase with the cube of the factor.
    2. The answer here is based on the same ideas as above. In analogy, since here 0<x<10<x<1, the volume will be reduced by a factor of x3x3 and the surface area will be decreased by a factor of x2x2

When the length of one of the sides is multiplied by a constant the effect is to multiply the original volume by that constant, as for the example in Figure 32.

Right Pyramids, Right Cones and Spheres

A pyramid is a geometric solid that has a polygon base and the base is joined to a point, called the apex. Two examples of pyramids are shown in the left-most and centre figures in (Reference). The right-most figure has an apex which is joined to a circular base and this type of geometric solid is called a cone. Cones are similar to pyramids except that their bases are circles instead of polygons.

Figure 34: Examples of a square pyramid, a triangular pyramid and a cone.
Figure 34 (MG11C16_001.png)

Surface Area of a Pyramid

Figure 35
Khan academy video on solid geometry volumes

The surface area of a pyramid is calculated by adding the area of each face together.

Exercise 8: Surface Area

If a cone has a height of hh and a base of radius rr, show that the surface area is πr2+πrr2+h2πr2+πrr2+h2.

Solution

  1. Step 1. Draw a picture :

    Figure 36
    Figure 36 (MG11C16_002.png)

  2. Step 2. Identify the faces that make up the cone :

    The cone has two faces: the base and the walls. The base is a circle of radius rr and the walls can be opened out to a sector of a circle.

    Figure 37
    Figure 37 (MG11C16_003.png)

    This curved surface can be cut into many thin triangles with height close to aa (aa is called the slant height). The area of these triangles will add up to 12×12×base××height(of a small triangle) which is 12×2πr×a=πra12×2πr×a=πra

  3. Step 3. Calculate aa :

    aa can be calculated by using the Theorem of Pythagoras. Therefore:

    a = r 2 + h 2 a = r 2 + h 2
    (24)
  4. Step 4. Calculate the area of the circular base :
    A b = π r 2 A b = π r 2
    (25)
  5. Step 5. Calculate the area of the curved walls :
    A w = π r a = π r r 2 + h 2 A w = π r a = π r r 2 + h 2
    (26)
  6. Step 6. Calculate surface area A :
    A = A b + A w = π r 2 + π r r 2 + h 2 A = A b + A w = π r 2 + π r r 2 + h 2
    (27)

Volume of a Pyramid: The volume of a pyramid is found by:

V = 1 3 A · h V = 1 3 A · h
(28)

where AA is the area of the base and hh is the height.

A cone is like a pyramid, so the volume of a cone is given by:

V = 1 3 π r 2 h . V = 1 3 π r 2 h .
(29)

A square pyramid has volume

V = 1 3 a 2 h V = 1 3 a 2 h
(30)

where aa is the side length of the square base.

Exercise 9: Volume of a Pyramid

What is the volume of a square pyramid, 3cm high with a side length of 2cm?

Figure 38
Figure 38 (MG11C16_004.png)

Solution

  1. Step 1. Determine the correct formula :

    The volume of a pyramid is

    V = 1 3 A · h , V = 1 3 A · h ,
    (31)

    where AA is the area of the base and hh is the height of the pyramid. For a square base this means

    V = 1 3 a · a · h V = 1 3 a · a · h
    (32)

    where aa is the length of the side of the square base.

  2. Step 2. Substitute the given values :
    = 1 3 · 2 · 2 · 3 = 1 3 · 12 = 4 c m 3 = 1 3 · 2 · 2 · 3 = 1 3 · 12 = 4 c m 3
    (33)

We accept the following formulae for volume and surface area of a sphere (ball).

Surface area = 4 π r 2 Volume = 4 3 π r 3 Surface area = 4 π r 2 Volume = 4 3 π r 3
(34)

Exercise 10

A triangular pyramid is placed on top of a triangular prism. The prism has an equilateral triangle of side length 20 cm as a base, and has a height of 42 cm. The pyramid has a height of 12 cm.

  1. Find the total volume of the object.
  2. Find the area of each face of the pyramid.
  3. Find the total surface area of the object.
Figure 39
Figure 39 (MG11C16_006.png)

Solution

  1. Step 1. Find the volume of the triangular prism: We use the formula for the volume of a triangular prism:
    V = 12bh2 = 1220422 = 17640 V = 12bh2 = 1220422 = 17640
    (35)
  2. Step 2. Find the volume of the triangular pyramid: We use the formula for the volume of a triangular pyramid:
    V = 16bh2 = 1620422 = 5880 V = 16bh2 = 1620422 = 5880
    (36)
  3. Step 3. Find the volume: We note that we can simply add the volumes of each of the two shapes. So we obtain: 17640+5880=2352017640+5880=23520. This is the answer to part a.
  4. Step 4. Find the area of the faces of the pyramid: We note that we have four triangles that make up the pyramid. So the area of each face is simply:
    Area = 12bh = 122042 = 420 Area = 12bh = 122042 = 420
    (37)
    This is the answer to part b.
  5. Step 5. Find the area of the pyramid: The total area of the pyramid is simply 4×420=16804×420=1680
  6. Step 6. Find the area of the prism: The surface area of the prism is:
    Surface area = b×h+2×H×S+H×b = 20×20+2×12×20+12×20 = 1120 Surface area = b×h+2×H×S+H×b = 20×20+2×12×20+12×20 = 1120
    (38)
  7. Step 7. Find the total surface area: To find the total surface area, we must subtract the area of one face of the pyramid from the area of the prism. We must also subtract the area of one of the triangular faces of the prism. Doing this gives the total surface area as: 1120-420+1680-420=19601120-420+1680-420=1960 This is the answer to part c.

Surface Area and Volume

  1. Calculate the volumes and surface areas of the following solids: *Hint for (e): find the perpendicular height using Pythagoras.
    Figure 40
    Figure 40 (MG11C16_005.png)
    Click here for the solution
  2. Water covers approximately 71% of the Earth's surface. Taking the radius of the Earth to be 6378 km, what is the total area of land (area not covered by water)?
    Click here for the solution

Transformations - Enrichment, not in CAPS

In this section you will learn about how the co-ordinates of a point change when the point is moved horizontally and vertically on the Cartesian plane. You will also learn about what happens to the co-ordinates of a point when it is reflected on the xx-axis, yy-axis and the line y=xy=x.

Translation of a Point

When something is moved in a straight line, we say that it is translated. You will recall that the term 'translation' was also mentioned in the section on functions to refer to what happens when an entire function is shifted in a straight line. What happens to the co-ordinates of a point that is translated horizontally or vertically?

Discussion : Translation of a Point Vertically

Complete the table, by filling in the co-ordinates of the points shown in the figure.

Figure 41
Figure 41 (MG10C14_022.png)

Table 3
Point xx co-ordinate yy co-ordinate
A    
B    
C    
D    
E    
F    
G    

What do you notice about the xx co-ordinates? What do you notice about the yy co-ordinates? What would happen to the co-ordinates of point A, if it was moved to the position of point G?

When a point is moved vertically up or down on the Cartesian plane, the xx co-ordinate of the point remains the same, but the yy co-ordinate changes by the amount that the point was moved up or down.

For example, in Item 19 Point A is moved 4 units upwards to the position marked by G. The new xx co-ordinate of point A is the same (xx=1), but the new yy co-ordinate is shifted in the positive yy direction 4 units and becomes yy=-2+4=2. The new co-ordinates of point A are therefore G(1;2). Similarly, for point B that is moved downwards by 5 units, the xx co-ordinate is the same (x=-2,5x=-2,5), but the yy co-ordinate is shifted in the negative yy-direction by 5 units. The new yy co-ordinate is therefore yy=2,5 -5=-2,5.

Figure 42: Point A is moved 4 units upwards to the position marked by G. Point B is moved 5 units downwards to the position marked by H.
Figure 42 (MG10C14_023.png)

Tip:

If a point is shifted upwards, the new yy co-ordinate is given by adding the shift to the old yy co-ordinate. If a point is shifted downwards, the new yy co-ordinate is given by subtracting the shift from the old yy co-ordinate.

Discussion : Translation of a Point Horizontally

Complete the table, by filling in the co-ordinates of the points shown in the figure.

Figure 43
Figure 43 (MG10C14_024.png)

Table 4
Point xx co-ordinate yy co-ordinate
A    
B    
C    
D    
E    
F    
G    

What do you notice about the xx co-ordinates? What do you notice about the yy co-ordinates?

What would happen to the co-ordinates of point A, if it was moved to the position of point G?

When a point is moved horizontally left or right on the Cartesian plane, the yy co-ordinate of the point remains the same, but the xx co-ordinate changes by the amount that the point was moved left or right.

For example, in Figure 44 Point A is moved 4 units right to the position marked by G. The new yy co-ordinate of point A is the same (yy=1), but the new xx co-ordinate is shifted in the positive xx direction 4 units and becomes xx=-2+4=2. The new co-ordinate of point A at G is therefore (2;1). Similarly, for point B that is moved left by 5 units, the yy co-ordinate is the same (y=-2,5y=-2,5), but the xx co-ordinate is shifted in the negative xx-direction by 5 units. The new xx co-ordinate is therefore xx=2,5 -5=-2,5. The new co-ordinates of point B at H is therefore (-2,5;1).

Figure 44: Point A is moved 4 units to the right to the position marked by G. Point B is moved 5 units to the left to the position marked by H.
Figure 44 (MG10C14_025.png)

Tip:

If a point is shifted to the right, the new xx co-ordinate is given by adding the shift to the old xx co-ordinate. If a point is shifted to the left, the new xx co-ordinate is given by subtracting the shift from the old xx co-ordinate.

Reflection of a Point

When you stand in front of a mirror your reflection is located the same distance (dd) behind the mirror as you are standing in front of the mirror.

Figure 45
Figure 45 (MG10C14_026.png)

We can apply the same idea to a point that is reflected on the xx-axis, the yy-axis and the line y=xy=x.

Reflection on the xx-axis

If a point is reflected on the xx-axis, then the reflection must be the same distance below the xx-axis as the point is above the xx-axis and vice-versa, as though it were a mirror image.

Figure 46: Points A and B are reflected on the xx-axis. The original points are shown with •• and the reflected points are shown with ∘∘.
Figure 46 (MG10C14_027.png)
Tip:
When a point is reflected about the xx-axis, only the yy co-ordinate of the point changes.
Exercise 11: Reflection on the xx-axis

Find the co-ordinates of the reflection of the point P, if P is reflected on the xx-axis. The co-ordinates of P are (5;10).

Solution
  1. Step 1. Determine what is given and what is required :

    We are given the point P with co-ordinates (5;10) and need to find the co-ordinates of the point if it is reflected on the xx-axis.

  2. Step 2. Determine how to approach the problem :

    The point P is above the xx-axis, therefore its reflection will be the same distance below the xx-axis as the point P is above the xx-axis. Therefore, yy=-10.

    For a reflection on the xx-axis, the xx co-ordinate remains unchanged. Therefore, xx=5.

  3. Step 3. Write the final answer :

    The co-ordinates of the reflected point are (5;-10).

Reflection on the yy-axis

If a point is reflected on the yy-axis, then the reflection must be the same distance to the left of the yy-axis as the point is to the right of the yy-axis and vice-versa.

Figure 47: Points A and B are reflected on the yy-axis. The original points are shown with •• and the reflected points are shown with ∘∘.
Figure 47 (MG10C14_028.png)
Tip:
When a point is reflected on the yy-axis, only the xx co-ordinate of the point changes. The yy co-ordinate remains unchanged.
Exercise 12: Reflection on the yy-axis

Find the co-ordinates of the reflection of the point Q, if Q is reflected on the yy-axis. The co-ordinates of Q are (15;5).

Solution
  1. Step 1. Determine what is given and what is required :

    We are given the point Q with co-ordinates (15;5) and need to find the co-ordinates of the point if it is reflected on the yy-axis.

  2. Step 2. Determine how to approach the problem :

    The point Q is to the right of the yy-axis, therefore its reflection will be the same distance to the left of the yy-axis as the point Q is to the right of the yy-axis. Therefore, xx=-15.

    For a reflection on the yy-axis, the yy co-ordinate remains unchanged. Therefore, yy=5.

  3. Step 3. Write the final answer :

    The co-ordinates of the reflected point are (-15;5).

Reflection on the line y=xy=x

The final type of reflection you will learn about is the reflection of a point on the line y=xy=x.

Casestudy : Reflection of a point on the line y=xy=x

Figure 48
Figure 48 (MG10C14_029.png)

Study the information given and complete the following table:

Table 5
  Point Reflection
A (2;1) (1;2)
B (-112112;-2) (-2;-11212)
C (-1;1)  
D (2;-3)  

What can you deduce about the co-ordinates of points that are reflected about the line y=xy=x?

The xx and yy co-ordinates of points that are reflected on the line y=xy=x are swapped around, or interchanged. This means that the xx co-ordinate of the original point becomes the yy co-ordinate of the reflected point and the yy co-ordinate of the original point becomes the xx co-ordinate of the reflected point.

Figure 49: Points A and B are reflected on the line y=xy=x. The original points are shown with •• and the reflected points are shown with ∘∘.
Figure 49 (MG10C14_030.png)
Tip:
The xx and yy co-ordinates of points that are reflected on the line y=xy=x are interchanged.
Exercise 13: Reflection on the line y=xy=x

Find the co-ordinates of the reflection of the point R, if R is reflected on the line y=xy=x. The co-ordinates of R are (-5;5).

Solution
  1. Step 1. Determine what is given and what is required :

    We are given the point R with co-ordinates (-5;5) and need to find the co-ordinates of the point if it is reflected on the line y=xy=x.

  2. Step 2. Determine how to approach the problem :

    The xx co-ordinate of the reflected point is the yy co-ordinate of the original point. Therefore, xx=5.

    The yy co-ordinate of the reflected point is the xx co-ordinate of the original point. Therefore, yy=-5.

  3. Step 3. Write the final answer :

    The co-ordinates of the reflected point are (5;-5).

Rules of Translation

A quick way to write a translation is to use a 'rule of translation'. For example (x;y)→(x+a;y+b)(x;y)→(x+a;y+b) means translate point (x;y) by moving a units horizontally and b units vertically.

So if we translate (1;2) by the rule (x;y)→(x+3;y-1)(x;y)→(x+3;y-1) it becomes (4;1). We have moved 3 units right and 1 unit down.

Translating a Region

To translate a region, we translate each point in the region.

Example

Region A has been translated to region B by the rule: (x;y)→(x+4;y+2)(x;y)→(x+4;y+2)

Figure 50
Figure 50 (MG10C14_031.png)

Discussion : Rules of Transformations

Work with a friend and decide which item from column 1 matches each description in column 2.

Table 6
Column 1 Column 2
1.(x;y)→(x;y-3)(x;y)→(x;y-3) A. a reflection on x-y line
2. ( x ; y ) → ( x - 3 ; y ) ( x ; y ) → ( x - 3 ; y ) B. a reflection on the x axis
3. ( x ; y ) → ( x ; - y ) ( x ; y ) → ( x ; - y ) C. a shift of 3 units left
4. ( x ; y ) → ( - x ; y ) ( x ; y ) → ( - x ; y ) D. a shift of 3 units down
5. ( x ; y ) → ( y ; x ) ( x ; y ) → ( y ; x ) E. a reflection on the y-axis
Transformations
  1. Describe the translations in each of the following using the rule (x;y)→→ (...;...)
    Figure 51
    Figure 51 (mg10c14_4.png)
    1. From A to B
    2. From C to J
    3. From F to H
    4. From I to J
    5. From K to L
    6. From J to E
    7. From G to H
    Click here for the solution
  2. A is the point (4;1). Plot each of the following points under the given transformations. Give the co-ordinates of the points you have plotted.
    1. B is the reflection of A in the x-axis.
    2. C is the reflection of A in the y-axis.
    3. D is the reflection of B in the line x=0.
    4. E is the reflection of C is the line y=0.
    5. F is the reflection of A in the line y= x
    Click here for the solution
  3. In the diagram, B, C and D are images of polygon A. In each case, the transformation that has been applied to obtain the image involves a reflection and a translation of A. Write down the letter of each image and describe the transformation applied to A in order to obtain the image.
    Figure 52
    Figure 52 (mg10c14_5.png)
    Click here for the solution
Investigation : Calculation of Volume, Surface Area and scale factors of objects
  1. Look around the house or school and find a can or a tin of any kind (e.g. beans, soup, cooldrink, paint etc.)
  2. Measure the height of the tin and the diameter of its top or bottom.
  3. Write down the values you measured on the diagram below:
    Figure 53
    Figure 53 (MG10C14_034.png)
  4. Using your measurements, calculate the following (in cm22, rounded off to 2 decimal places):
    1. the area of the side of the tin (i.e. the rectangle)
    2. the area of the top and bottom of the tin (i.e. the circles)
    3. the total surface area of the tin
  5. If the tin metal costs 0,17 cents/cm22, how much does it cost to make the tin?
  6. Find the volume of your tin (in cm33, rounded off to 2 decimal places).
  7. What is the volume of the tin given on its label?
  8. Compare the volume you calculated with the value given on the label. How much air is contained in the tin when it contains the product (i.e. cooldrink, soup etc.)
  9. Why do you think space is left for air in the tin?
  10. If you wanted to double the volume of the tin, but keep the radius the same, by how much would you need to increase the height?
  11. If the height of the tin is kept the same, but now the radius is doubled, by what scale factor will the:
    1. area of the side surface of the tin increase?
    2. area of the bottom/top of the tin increase?

Summary

  • The properties of kites, rhombuses, parallelograms, squares, rectangles and trapeziums was covered. These figures are all known as quadrilaterals
  • You should know the formulae for surface area of rectangular and triangular prisms as well as cylinders
  • The volume of a right prism is calculated by multiplying the area of the base by the height. So, for a square prism of side length aa and height hh the volume is a×a×h=a2ha×a×h=a2h.
  • Similarity of polygons: Two polygons are similar if:
    • their corresponding angles are equal
    • the ratios of corresponding sides are equal
    . All squares are similar

End of Chapter Exercises

  1. Assess whether the following statements are true or false. If the statement is false, explain why:
    1. A trapezium is a quadrilateral with two pairs of parallel opposite sides.
    2. Both diagonals of a parallelogram bisect each other.
    3. A rectangle is a parallelogram that has all four corner angles equal to 60°.
    4. The four sides of a rhombus have different lengths.
    5. The diagonals of a kite intersect at right angles.
    6. Two polygons are similar if only their corresponding angles are equal.
    Click here for the solution
  2. Calculate the area of each of the following shapes:
    Figure 54
    Figure 54 (MG10C14_041.png)
    Click here for the solution
  3. Calculate the surface area and volume of each of the following objects (assume that all faces/surfaces are solid – e.g. surface area of cylinder will include circular areas at top and bottom):
    Figure 55
    Figure 55 (MG10C14_042.png)
    Click here for the solution
  4. Calculate the surface area and volume of each of the following objects (assume that all faces/surfaces are solid):
    Figure 56
    Figure 56 (MG10C14_043.png)
    Click here for the solution
  5. Using the rules given, identify the type of transformation and draw the image of the shapes.
    1. (x;y)→→(x+3;y-3)
      Figure 57
      Figure 57 (MG10C14_037.png)
    2. (x;y)→→(x-4;y)
      Figure 58
      Figure 58 (MG10C14_038.png)
    3. (x;y)→→(y;x)
      Figure 59
      Figure 59 (MG10C14_039.png)
    4. (x;y)→→(-x;-y)
      Figure 60
      Figure 60 (MG10C14_040.png)
    Click here for the solution
  6. PQRS is a quadrilateral with points P(0; −3) ; Q(−2;5) ; R(3;2) and S(3;–2) in the Cartesian plane.
    1. Find the length of QR.
    2. Find the gradient of PS.
    3. Find the midpoint of PR.
    4. Is PQRS a parallelogram? Give reasons for your answer.
    Click here for the solution
  7. A(–2;3) and B(2;6) are points in the Cartesian plane. C(a;b) is the midpoint of AB. Find the values of a and b.
    Click here for the solution
  8. Consider: Triangle ABC with vertices A (1; 3) B (4; 1) and C (6; 4):
    1. Sketch triangle ABC on the Cartesian plane.
    2. Show that ABC is an isoceles triangle.
    3. Determine the co-ordinates of M, the midpoint of AC.
    4. Determine the gradient of AB.
    5. Show that the following points are collinear: A, B and D(7;-1)
    Click here for the solution
  9. In the diagram, A is the point (-6;1) and B is the point (0;3)
    Figure 61
    Figure 61 (MG10C14_5.png)
    1. Find the equation of line AB
    2. Calculate the length of AB
    3. A’ is the image of A and B’ is the image of B. Both these images are obtain by applying the transformation: (x;y)→→(x-4;y-1). Give the coordinates of both A’ and B’
    4. Find the equation of A’B’
    5. Calculate the length of A’B’
    6. Can you state with certainty that AA'B'B is a parallelogram? Justify your answer.
    Click here for the solution
  10. The vertices of triangle PQR have co-ordinates as shown in the diagram.
    Figure 62
    Figure 62 (mg10c14_6.png)
    1. Give the co-ordinates of P', Q' and R', the images of P, Q and R when P, Q and R are reflected in the line y=x.
    2. Determine the area of triangle PQR.
    Click here for the solution
  11. Which of the following claims are true? Give a counter-example for those that are incorrect.
    1. All equilateral triangles are similar.
    2. All regular quadrilaterals are similar.
    3. In any ▵ABC▵ABC
      with ∠ABC=90∘∠ABC=90∘ we have AB3+BC3=CA3AB3+BC3=CA3.
    4. All right-angled isosceles triangles with perimeter 10 cm are congruent.
    5. All rectangles with the same area are similar.
    Click here for the solution
  12. For each pair of figures state whether they are similar or not. Give reasons.
    Figure 63
    Figure 63 (MG10C13_053.png)
    Click here for the solution

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