new century maths 10: stage 5.2/5.3 - chapter 5

Wordbank

Z

deduction

The formal process of proving something logically and reaching a conclusion by reasoning from a general principle to a specific result.

Z

parallelogram

A quadrilateral with both pairs of opposite sides parallel.

Z

rectangle

A parallelogram with one angle a right angle.

Z

exterior angle

The angle formed when one of the sides of a polygon is extended.

05_NCM10EX2SB_TXT.fm Page 180 Monday, September 12, 2005 3:45 PM

CHAPTER 5

deductive geometrydeductive geometry

Think!

A square is a rhombus, but a rhombus is not necessarily a square. Why not? Can a square be considered as a rectangle?

SPACE AND GEOMETRY

In this chapter you will:

Z

recognise the types of triangles and quadrilaterals and the properties of their sides, angles and diagonals

Z

apply simple deductive reasoning to solving numerical and non-numerical problems

Z

construct and write geometrical arguments to prove a geometrical result, giving reasons at each step of the argument

Z

apply angle sum results to find unknown angles in convex polygons

Z

write formal proofs of the congruence of triangles, preserving the matching order of the vertices

Z

prove properties of special triangles and quadrilaterals from the formal definitions of the shapes

Z

prove and apply theorems related to triangles and quadrilaterals

Z

prove and apply tests for quadrilaterals

Z

calculate unknown sides in a pair of similar triangles

Z

determine what information is needed to establish that two triangles are similar

Z

write formal proofs of the similarity of triangles, preserving the matching order of the vertices, identifyng the similarity factor when appropriate

Z

solve Euclidian geometry problems

Z

prove and apply further theorems using similarity

Z

hence

Following from the previous result, another word for therefore.

Z

deductive geometry

When geometrical results and problems are proved or solved by a step-by-step process or argument. Reasons are given at each step of the argument.

Z

test for quadrilaterals

A property of a quadrilateral that proves that it is a particular type of quadrilateral. For example, if opposite angles in a quadrilateral are equal, then it

must

be a parallelogram.

Z

similarity test for triangles

One of four tests (SSS, SAS, AA, RHS) that prove that two triangles are similar.

Z

converse

The reverse of a geometrical rule or theorem, written in a back-to-front way. For example, the converse of If lines are parallel, corresponding angles are equal is If corresponding angles are equal, then the lines are parallel.


182

N EW CENTURY MATHS 10 : S TAGES 5 .2/5 .3

1 Find the values of the pronumerals in the following diagrams. Give reasons for your answers.a b c

d e f

2 Find the value of each pronumeral, giving reasons for each step.a b c

d e f

3 Find the values of the pronumerals in the following diagrams. Give reasons for your answers.a b c

d e f

m 63x

47

82 y78

135

k42

w

38127

4a

5a37 152

127

d

88

4m

66

2h

a

72104

m

k

81

w

p

h

71

63 m

35124

3h5h

2h

2y

137

83

3y

3k105

a

16327

34

Start upWorksheet

5-01Brainstarters 5

Skillsheet 5-01

Angles and parallel lines


D EDUCT I VE GEOMETRY

183

CHAPTER 5

Deduction in geometry

Many results in geometry can be shown or demonstrated by construction and measurement. For example, we can draw a triangle and measure the angles to show or demonstrate that the angle sum of a triangle is 180

. However this does not prove that the angle sum of

any

triangle is 180

.

To prove this and other geometrical results we use a process called

deduction

, in which a specic result is proved by reasoning logically from a general principle or known fact.When setting out proofs, reasons must be given for each statement or each step of the argument. The reasons are usually in brackets after the statements and should state the names of the triangles, angles, or lines referred to.Important results which have been proved and are then used as a rule or general principle are called

theorems

.

4 Which of these pairs of lines are parallel? Give reasons.a b c

6769

132

48

78

103

Euclid and Euclidian geometryEuclid was a Greek mathematician who lived in Alexandria during the third century BC. Euclids most famous written work was the thirteen-volume Elements, which he probably wrote before the age of 25. This work presented a systematic treatment of all known geometry, and actually covered most of the mathematics known at the time. There are 465 different propositions in Elements (Pythagoras theorem being number 47 in the rst volume).Geometry using the methods from Euclids Elements is called Euclidian geometry and the results and proofs that will be considered in the chapter are close to those of Euclid.Other geometries were developed by Gauss, Lobachevsky, Rieman and others in the early nineteenth century. Select one of these geometries and briey describe how it is different from Euclidean geometry.

Just for the record

Example 1

Find the values of the pronumerals in the following diagrams:a b

k

55

d

46

A B C

DE


184

N EW CENTURY MATHS 10 : S TAGES 5 .2/5 .3

Properties of triangles and quadrilaterals

Triangles

Denitions of special triangles

Properties

The properties of the various types of triangles can be summarised as follows.

Scalene

no equal sides no equal angles no symmetry

Isosceles

two equal sides two equal angles one axis of symmetry no rotational symmetry

Equilateral

three equal sides three equal angles (60

each) three axes of symmetry has rotational symmetry

Solutiona k + 55 + 55 = 180 (angle sum of isosceles triangle)

k + 110 = 180k = 70

b EDC = ECD = (180 46) 2 (angle sum of isosceles ECD)EDC = 67EBC = 67 (opposite angles of a parallelogram are equal)d + 67 = 180 (angle sum of a straight line)

d = 113

Skillsheet 5-02

Two-dimensional shapes

An acute-angled triangle is a triangle with all angles acute. An obtuse-angled triangle is a triangle with an obtuse angle. A right-angled triangle is a triangle with a right angle. A scalene triangle is a triangle with no two sides equal in length. An isosceles triangle is a triangle with two sides equal in length. An equilateral triangle is a triangle with all sides equal in length.


D EDUCT I VE GEOMETRY

185

CHAPTER 5

Quadrilaterals

A quadrilateral is a four-sided plane gure. A quadrilateral may be either

convex

or

non-convex

(

concave

).A convex quadrilateral has no interior angles greater than 180

.

A convex quadrilateral A non-convex quadrilateral

(all angles are less than 180

) (one interior angle is greater than 180

)

The diagonals of a

convex

quadrilateral lie

inside

the quadrilateral.

Diagonals

AC

and

BD

lie inside the The diagonal

MP

does not lie inside the convex quadrilateral

ABCD

. non-convex quadrilateral

LMNP

.

In the work that follows, convex quadrilaterals will be simply referred to as quadrilaterals.

Denitions of special quadrilaterals

Worksheet5-02

Naming quadrilaterals

Worksheet5-03

Investigating the geometrical constructions

Skillsheet 5-03

Geometrical constructions

A

B

C

D

L

M

N

P

A trapezium is a quadrilateral with at least one pair of opposite sides parallel.A parallelogram is a quadrilateral with both pairs of opposite sides parallel.A rectangle is a parallelogram with one angle a right angle.A square is a rectangle with two adjacent sides equal in length.A rhombus is a parallelogram with two adjacent sides equal in length.A kite is a convex quadrilateral with two pairs of equal adjacent sides.


186 NEW CENTURY MATHS 10 : S TAGES 5 .2/5 .3

SummaryThe properties of the various types of quadrilaterals can be summarised as follows:Trapezium one pair of opposite sides parallel

Parallelogram opposite sides parallel opposite sides equal opposite angles equal diagonals bisect each other has rotational symmetry

Rectangle opposite sides parallel opposite sides equal all angles are right angles diagonals are equal in length diagonals bisect each other two axes of symmetry has rotational symmetry

Square opposite sides parallel all sides equal all angles are right angles diagonals equal in length diagonals bisect each other at right angles diagonals bisect the angles of the square four axes of symmetry has rotational symmetry

Rhombus opposite sides parallel all sides equal opposite angles equal diagonals bisect each other at right angles diagonals bisect the angles of the rhombus two axes of symmetry has rotational symmetry

Kite two pairs of equal adjacent sides one pair of opposite angles equal one axis of symmetry diagonals intersect at right angles

Working mathematicallyReasoning: Is a square a rhombus?Looking at the denitions of the quadrilaterals above, we see that a parallelogram can also be classied as a trapezium since it has at least one pair of opposite sides parallel. This means that trapeziums are inclusive of parallelograms. Similarly parallelograms are inclusive of rectangles and rhombuses and these are inclusive of squares. This can be represented by a Venn diagram as shown on the next page.

Worksheet 5-02



DEDUCT I VE GEOMETRY 187 CHAPTER 5

1 a Why is a rectangle a special type of parallelogram but a parallelogram is not always a rectangle?

b How can you use the Venn diagram to answer part a?

2 The kite is not included in the Venn diagram. Where would you put kites on the diagram?

squaresrectangles

parallelograms

trapeziums

quadrilaterals

rhombuses

Example 2

Prove that the exterior opposite angles of a parallelogram are equal.SolutionTo prove FBC = FDEProof:

FBC = A (corresponding angles, AD || BF)and FDE = A (corresponding angles, AB || DF)

FBC = FDE Exterior opposite angles of a parallelogram equal.

A

D FE

B C

Geometric constructions: BisectorsThe perpendicular bisectors of the sides and angles of a triangle have some interesting properties. You can do these constructions using compasses and a ruler, but it is more fun to use a dynamic geometry program such as Geometers Sketchpad or Cabri Geometry.

Construction 1Step 1: Draw a triangle and label it ABC.Step 2: Construct the perpendicular bisectors of sides AB and

BC. (In Geometers Sketchpad you will need to nd the midpoint of the side rst.)

Step 3: Construct the point of intersection of the perpendicular bisectors and label it O.

Step 4: Construct a circle with centre O and a radius OA.Step 5: Construct the perpendicular bisector of side AC.1 What do you notice about the circle you have drawn?2 Do all three bisectors meet at point O?3 Do point O and the circles change in a signicant way when you drag a vertex or side of

ABC?

C

A

B

O

Using technology Skillsheet 5-03Geometrical constructions

Skillsheet 5-04

Starting Geometers Sketchpad

Skillsheet 5-05

Starting Cabri Geometry

Geometry 5-01

Side bisectors

STAGE5.3



4 The circle is called the circumcircle, and point O is called the circumcentre of the triangle. What does circum mean?

5 Can you explain why this construction produces the circumcentre?

Construction 2Step 1: Draw another triangle and label it PQR.Step 2: Construct the bisectors of PQR and RPQ.Step 3: Construct the point of intersection of the two angle

bisectors and label it O.Step 4: Use the drawing tool to draw a circle, with centre O,

just touching QR.Step 5: Construct the bisector QRP.

6 What do you notice about the circle you have drawn?7 Do all three bisectors meet at point O?8 Do point O and the circles change in a signicant way when you drag a vertex or side of

PQR?9 The circle is called the in-circle, and point O is called the in-centre of the triangle. Explain.

10 Can you explain why this construction produces the in-centre?

Extension11 Find the radius of the in-circle of the isosceles triangle with sides 10 cm, 10 cm and 4 cm.

O

Q

P

R

Geometry 5-02

Angle bisectors

1 Calculate the size of the angle indicated by the pronumeral in each of the following. (Give reasons for your answer.)a b c

d e f

y

80m

36h

x

a d3d

Exercise 5-01Example 1



g h i

2 Find the value of the pronumeral in each of the following. (Give reasons for your answers.)a b c

d e f

3 Find the value of the pronumeral in each of the following, giving reasons for each step:a b c

d e f

g h i

4 a Find the value of p, q and w in the diagram on the right. (Give reasons.)

b What type of triangle is XYZ? (Give reasons for your answer.)

y

70

p130

h

42

y

53

k

127

w

55

h

71

f 65

a

134

t

40m

y

58 d

74120

m

k

wh

115 c

43

a

w

78

p

34

A B

CD

X h

32

78 w

56

65

X

p

YZ

wq

70110



5 State whether each of the following is true (T) or false (F):a A rectangle is a parallelogram.b An equilateral triangle has three axes of symmetry.c The diagonals of a rhombus are equal.d A scalene triangle has two equal sides.e A square has four axes of symmetry.f Opposite sides of a trapezium are equal.g The angles opposite the equal sides in an isosceles triangle are equal.h Opposite angles of a parallelogram are equal.i The diagonals of a square, a rhombus and a rectangle bisect each other at right angles.j A scalene triangle has no equal sides but one axis of symmetry.k A rectangle and a rhombus both have only two axes of symmetry.l An equilateral triangle is also an isosceles triangle.m The diagonals of a parallelogram are equal.n The angles of a square and a rectangle are equal to 90.

6 KL = ML and MN = MP. Find the value of x, giving reasons for each step.

7 Segment YP bisects XYW, segment WP bisects TWY and YX || WT.Prove that YPW = 90.

8 BDF is an isosceles triangle and BF || CE.Prove that CDE is isosceles.

9 Segment NK bisects HKL. Find the size of NHK, giving reasons.

x

K

L M N

P

Y

P

X

W T

F E

C

B

D

K

H

N

L M

93

147

Example 2

STAGE5.3



10 AC || DE, BC = BE and CD = CE.a Find k, giving reasons.b Find m, giving reasons.

11 Find the value of x and y. (Give reasons for your answers.)a b c

d e f

12 In the gure on the right, AC || ED, AE || BD and BE || CD. Also, CB = CD.Prove that ABE is an isosceles triangle.

13 In the gure on the right, LP bisects KLN and LM = LN.Prove LP || MN.

14 In the gure on the right, AC = BC and DC = EC.Prove AB || DE.

15 UXY is an equilateral triangle and WX = XU.Prove that WUY is a right angle.

km

D E

BAC

42

U

Z X

V

Y

W

y

xB

A

C Dx

y

H G

E

F40

x y

H

K

I

J

108

y xD C B

A

E33

67xy 20 54

y

x

S R Q

P

T

E

A B C

D

P

K L M

N

B

A

CD

E

U

XW Y



Properties of convex polygonsThe general name for any plane gure bounded by straight sides is a polygon.A convex polygon has no interior angles greater than 180. All the diagonals of a convex polygon lie inside the polygon.

In the work that follows, convex polygons will be simply referred to as polygons.

Regular polygonsA regular polygon has all sides equal and all angles equal. The rst eight regular polygons are:

16 In the gure on the right, WY || PQ.Prove that the angle sum of PQT is 180.

17 Segment AC bisects FAB and AD bisects HAB.Prove CAD = 90.

W T

P Q

Y

F

C B

D

A H

Worksheet 5-04

Angles in polygons

Worksheet 5-05

Find the missing angle

Convex polygon Non-convex polygon

3 sides 4 sides 5 sides 6 sides

7 sides 8 sides 9 sides 10 sides

Equilateral triangle Square Pentagon Hexagon

Heptagon Octagon Nonagon Decagon

STAGE5.3



Canberra: The national capitalCanberra is located in the northern part of the ACT, 300 km south-west of Sydney and 650 km north-east of Melbourne.The city was designed by an American architect, Walter Burley Grifn, and construction began in 1913. The centre of Canberra is based on an equilateral triangle, bounded by the sides (Commonwealth Avenue, Kings Avenue and Constitution Avenue). The smaller Parliamentary triangle is bounded by Commonwealth Avenue, Kings Avenue and King Edward Terrace. The axis of symmetry of the triangle runs from Parliament House across Lake Burley Grifn directly along Anzac Parade to the Australian War Memorial.What other geometrical features can you see in Canberras design?

Just for the record



Angle sum of a convex polygon

The size of each angle of a regular polygon with n sides is given by the following formula.

Exterior angle sum of a convex polygon

The angle sum of a convex polygon with n sides is given by:Angle sum = (n 2) 180

Example 3

Find the angle sum of a polygon with 16 sides.SolutionAngle sum = (16 2) 180

= 14 180= 2520

One angle = where n is the number of sides

=

angle sumn

------------------------,

(n 2) 180n

----------------------------------

Example 4

1 Find the size of an angle of a regular octagon.Solution

2 The sum of the angles of a regular polygon is 3960.a How many sides does the polygon have? b Find the size of each angle.

Solution

Angle sum = (8 2) 180 or= 6 180= 1080

Each angle =

= 135

Each angle = = 135

a (n 2) 180 = 3960n 2 =

n 2 = 22n = 24

Number of sides is 24.

b Each angle =

= 165

10808

--------------

(8 2) 1808

----------------------------------

3960180------------

396024

------------

The sum of the exterior angles of a convex polygon is 360.



Example 5

1 The exterior angle of a regular polygon is 18. How many sides does the polygon have?

SolutionLet n be the number of sides of the polygon.Sum of exterior angles = n 18 = 360

18n = 360

n =

= 20 The polygon has 20 sides.

2 Each interior angle of a regular polygon is 140. How many sides does the polygon have?

Solution

Method 1:Exterior angle = 180 140 (since exterior angle + interior angle = 180)

= 40Number of exterior angles = 360 40

= 9 The polygon has 9 sides.

Method 2:

= 140

= n 140

(n 2) 180 = 140nSo: 180n 360 = 140n

180n = 140n + 360180n 140n = 140n + 360 140n

40n = 360 n = 9

36018---------

(n 2) 180n

--------------------------------

n(n 2) 180

n--------------------------------

1

1

Exterior angle sum of a convex polygonStep 1: Use your drawing program to copy the diagram on the right.Step 2: Measure the exterior angles indicated and calculate

their sum.Step 3: Repeat Steps 1 and 2 for a quadrilateral, a pentagon

and a hexagon. What is the sum of the exterior angles of a convex polygon?

Using technology Geometry 5-03Exterior angles of a polygon



1 Use the rule, Angle sum = (n 2) 180, to nd the angle sum of a polygon with:a 20 sides b 10 sides c 15 sidesd 24 sides e 30 sides f 12 sides

2 Find the number of sides for a polygon that has an angle sum of:a 4680 b 1800 c 900d 6840 e 2160 f 1260

3 Find the size of each interior angle of a regular:a pentagon b nonagon c dodecagon (12 sides)

4 Find the size of each interior angle of a regular polygon which has:a 30 sides b 45 sides c 7 sides d 60 sides

5 Find the number of sides of a regular polygon which has each of its exterior angles equal to:a 20 b 15 c 45 d 8 e 30

6 Find the size of each exterior angle of a regular:a hexagon b 12-sided polygon c 30-sided polygon

7 Find the number of sides of a regular polygon which has each of its interior angles equal to:a 120 b 165 c 172 d 135 e 156

8 The interior angle of a regular polygon is 11 times the size of an exterior angle. How many sides has the polygon?


Example 4

Example 5

Converting fractions and decimals to percentagesTo convert a fraction or decimal to a percentage, multiply it by 100%.1 Examine these examples:

a = 100% = = 2 20% = 40%

b = 100% = = 18 4% = 72%

c = 100% = = = 3 20% = 60%

d = 100% = = = 7 4% = 28%

2 Now convert these fractions to percentages:

a b c d e

f g h i j

k l m n o

25---

25---

25--- 100%

1

20

1825------

1825------

1825------ 100%

1

4

2440------

2440------

2440------ 100%

5

3 35--- 100%

1

20

2175------

2175------

2175------ 100%

3

4 213------ 4%

1

7

710------

3350------

2760------

2225------

2432------

3040------

6075------

45---

1120------

2880------

1550------

1620------

5460------

1840------

1325------

Skillbank 5A

SkillTest 5-01

Converting fractions and decimals to percentages



Proving congruent trianglesCongruent gures have exactly the same size and shape.The symbol for is congruent to is .In congruent gures, matching sides are equal and matching angles are equal.There are four sets of conditions that can be used to determine if two triangles are congruent. These are known as the four tests for congruent triangles.

3 Examine these examples:a 0.41 = 0.41 100% = 0.41 = 41% b 0.08 = 0.08 100% = 0.08 = 8%c 0.9 = 0.9 100% = 0.90 = 90% d 0.375 = 0.375 100% = 0.375 = 37.5%

4 Now convert these decimals to percentages:a 0.25 b 0.68 c 0.17 d 0.6 e 0.1f 0.333 g 0.59 h 0.702 i 0.84 j 0.7k 0.428 l 0.055 m 0.91 n 0.7825 o 0.314

Worksheet5-06

Congruent triangles proofs

1 Side, Side, Side (SSS)If three sides of one triangle are respectively equal to three sides of another triangle, then the two triangles are congruent.

2 Side, Angle, Side (SAS)If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the two triangles are congruent.

3 Angle, Angle, Side (AAS)If two angles and one side of one triangle are respectively equal to two angles and the matching side of another triangle, then the two triangles are congruent.

4 Right angle, Hypotenuse, Side (RHS)If the hypotenuse and a second side of one right-angled triangle are respectively equal to the hypotenuse and a second side of another right-angled triangle, then the two triangles are congruent.

Geometry 5-04

Congruent triangles



Example 6

Which congruence test (SSS, SAS, AAS, or RHS) can be used to prove that each of these pairs of triangles are congruent?a b

Solutiona AAS b SAS

In the diagram on the right, W = V and TX WV.Prove:a WXT VXTb X is the midpoint of WV.Solutiona In WXT and VXT

W = V (given)WXT = VXT (TX WV)TX is common WXT VXT (AAS)

b WX = VX (matching sides of congruent triangles) X is the midpoint of WV

3535

48

48

30 mm

30 mm

Example 7

T

XW V

1 Test whether each of the following pairs of triangles are congruent. If they are, which test (SSS, SAS, AAS or RHS) would prove them congruent?a b c

d e f

4 cm4 cm

10 cm

10 cm

8 cm 8 cm

5 cm

5 cm

40

40

70

70

9 cm

9 cm

7 cm

7 cm

15 cm15 cm


STAGE5.3



g h i

2 For these pairs of congruent triangles, nd the value of the pronumerals:a b

c d

e f

3 a AB = CB and EB = DB. b QT WT, PW WT and QW = PT.Prove ABE CBD. Prove QTW PWT.

70

70

60

60

8 cm8 cm

55

5510 cm

10 cm

40

40

58

58

31 mmd mm

50 50

25 k

w

25

12 cm

y cm33

a

p cm9 cm

35 mm

35 mm

30 mm

30 mm53 mm

53 mm109

p11 cm9 cm

9 cmk cm

60 60

For Questions 3 and 4 following, identify matching vertices before writing the proof.

A

E

BD

C

P W

QT

Example 7



c CH || EG, DH || FG and CH = EG. d YX bisects UXW and UX = WX.Prove CDH EFG. Prove UXY WXY.

e ABCD is a square and AY = CX. f LM = NP and LP = NM.Prove ABY CBX. Prove LMP NPM.

g O is the centre of the circles. h FGH is an isosceles triangle, so Prove AOB COD. FH = FG. HN FG and GM FH.

Prove FHN FGM.

4 a PQR is isosceles and QA = RB. b TP = XP and AP = CP. Prove: Prove:i PQA PRB i TAP XCP

ii PAB is isosceles. ii TA || XC.

C D E F

GH U

X Y

W

D X C

Y

A B

N

L

P

M

BC

DA

O

H

M

F

N

G

Q A B R

P T A

XC

P

STAGE5.3



Proving properties of triangles and quadrilateralsThe congruence tests can be used not only to solve numerical problems but also to establish properties of triangles and special quadrilaterals.

c AB || CD and AD || CB. Prove:i ABD CDB

ii AD = CB and AB = CD.

d A and B are the centres of two circles which intersect at C and D. Prove:i ADB ACB

ii AB bisects DAC.D C

A B

D

C

BA

e HEF = GFE and EH = FG. Prove:i HEF GFE

ii EHF = FGE.

f O is the centre of the circle and LT = MN. Prove:i LOT MON

ii LOT MON.

HK

E F

GT

O

L

N

M

g O is the centre of the circle and OD CE. Prove:i OCD OED

ii OD bisects CE.

h AB = AD and CB = CD. Prove:i ABC ADC

ii BCA = DCAiii BCY DCYiv BY = DY.

O

DC E

A C

D

B

Y



Using the word henceWhen the word hence is used, it means use or follow on from the previous result.For example, if a question states:a Prove ABC DBCb Hence prove A = D,then part b follows from what was proved in part a and the nal statement in the proof would be:

A = D (matching angles of congruent triangles)

Example 8

ABCD is a rectangle.a Prove that ABD BAC.b Hence show that the diagonals of a rectangle are equal.

Solutiona In ABD and BAC:

AD = BC (opposite sides of a rectangle)AB is common.DAB = CBA = 90 (angles in a rectangle) ABD BAC (SAS)

b BD = AC (matching sides of congruent triangles) The diagonals of a rectangle are equal.

D C

A B

Properties of triangles and quadrilateralsIsosceles triangle 1Step 1: Use your drawing program to construct isosceles triangle ABC, so that

AB = AC.Step 2: Measure B and C.

1 What can you say about the angles opposite the equal sides?Step 3: Construct a line from A through T, the midpoint of BC.Step 4: Measure ATB and ATC.

2 What can you say about AT and BC?Step 5: Measure BAT and CAT.

3 Does AT bisect BAC?

Isosceles triangle 2Step 1: Use your drawing program to construct isosceles triangle BEW, so that

BE = BW.Step 2: Construct the perpendicular bisector of EW.

4 Which point does the perpendicular bisector pass through?5 Does the perpendicular bisector of EW bisect EBW?

A

B T C

B

E T W

Using technologyGeometry

5-05Properties of triangles and quadrilaterals

Worksheet 5-03

Investigating the geometrical constructions

STAGE5.3



RhombusStep 1: Construct rhombus ABCD with all sides equal.Step 2: Construct the diagonal AC.Step 3: Construct the perpendicular bisector of AC.

6 Does the perpendicular bisector pass through B and D?

Step 4: Measure ACD, and ACB.

7 Does AC bisect DCB?8 What properties of a rhombus have you demonstrated with this construction of a

rhombus?9 Construct a parallelogram and, by drawing the diagonals and using the Measure

tool, demonstrate the properties of a parallelogram.

D C

BA

1 XYT is isosceles, so TX = TY. TW is drawn so that XW = WY.a Prove TXW TYW. b Hence prove that the angles opposite the equal sides of XYT

are equal (that is, show X = Y).

2 LMNP is a rhombus (all sides are equal).a Prove LMN LPN.b By drawing the diagonal PM, prove

PMN PML.c Hence prove that the angles of a

rhombus are bisected by the diagonals.

3 ABCD is a parallelogram, so AB || DC and AD || BC.a Prove ABD CDB.b Hence prove that the opposite sides of

a parallelogram are equal.

X W Y

T

L

P

M

N

B A

DC

Exercise 5-04

Example 8



4 ABC is an equilateral triangle (AC = AB = BC). CX is drawn so that AX = BX. Show that each angle is 60.a Prove:

i AXC BXCii A = B

b Now draw the interval BY, so that CY = AY (as shown). Prove:i AYB CYB

ii A = Cc Hence calculate the size of an angle in an equilateral triangle.

5 WXYV is a parallelogram, so WX || VY and WV || XY.a Prove WXY YVW.b By drawing the diagonal XV, prove VWX XYV.c Hence prove that the opposite angles of a

parallelogram are equal.

6 DEFG is a parallelogram, so DE || FG, DG || EF, DE = FG, and EF = GD (as proved in Question 3).a Prove DEX FGX.b Hence prove that the diagonals of a

parallelogram bisect each other.

7 XYW is an isosceles triangle. The perpendicular line (the altitude) from the vertex X to side WY, bisects that side. Prove this result by following the steps below:a Show XWT XYT. b Hence show WT = YT.

8 ABC is isosceles, so AB = AC. If AX bisects BAC, then AX BC. Prove this result for an isosceles triangle by answering the following:a Prove AXB AXC.b Hence show AXB = AXC.c Hence prove AX BC

(show AXB = AXC = 90).

A BX

C

A B

C

Y

V Y

W X

D E

G F

X

X

TW Y

A X

B

C

STAGE5.3



9 In TPR, P = R and TX PR.a Prove PXT RXT.b Hence prove that the sides opposite the equal angles are equal

(that is, prove TP = TR).

10 CDEF is a rhombus.a Prove CBF EBF.b Explain why the diagonal CE is bisected.c Explain how the diagonal DF can be shown to be bisected.d i Explain why CBF = EBF.

ii Hence prove FD CE.e What property of a rhombus has been proved?

11 LMNP is a kite, with LM = LP and NM = NP.a Use congruent triangles to prove:

i LMN = LPNii MLP and MNP are bisected by the diagonal LN.

b Prove that LMT LPT and, hence, show that the diagonal MP is bisected at right angles by the diagonal LN.

T

XP R

F E

B

DC

L

N

M T P

Working mathematicallyCommunicating and applying strategies: The converse (true or false)?The converse is the reverse of a geometrical rule or theorem written in a back-to-front way. The converse of a statement or rule is not necessarily true. Consider the following example:If a quadrilateral is a rectangle, then both pairs of opposite sides are equal.The converse is If both pairs of opposite sides of a quadrilateral are equal, then the quadrilateral is a rectangle. This is false. (The quadrilateral could be a parallelogram.)For the geometrical result, If lines are parallel, then the alternate angles are equal, the converse is If alternate angles are equal, then the lines are parallel.The converse is also true and is often used to test if two lines are parallel (see Start up, Question 4 on page 183).1 a For each of the following, write the converse statement and decide whether the

converse is also true:i If two angles are co-interior they have a sum of 180.

ii If two sides of a triangle are equal, then the angles opposite those sides are equal.iii In a right-angled triangle, the square on the hypotenuse is equal to the sum of the

squares on the other two sides.iv If the opposite angles of a quadrilateral are equal, then it is a parallelogram.v If a quadrilateral is a square, then the diagonals bisect each other at right angles.

vi If the quadrilateral is a rectangle, then the diagonals bisect each other.b Compare your answers with those of other students.

2 The diagonals of a kite are perpendicular and one bisects the other.Write the converse to this theorem and prove it to be true.



Tests for quadrilateralsThe properties of special quadrilaterals (such as the parallelogram, rectangle, and so on) can be used as minimum conditions to prove or test whether a given quadrilateral is a parallelogram, rectangle, square or rhombus.

Example 9

ABCD is a quadrilateral on which AD = BC and AB = CD. BD is a diagonal. Prove that, if the opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram.

SolutionIn triangles ABD and CDB,AD = CB (given)AB = CD (given)BD is common to both triangles. ABD CDB (SSS) ABD = CDB (matching angles of congruent triangles) AB || CD (alternate angles ABD and CDB are equal)Also, ADB = CBD (matching angles of congruent triangles) AD || CB (alternate angles ADB and CBD are equal) ABCD is a parallelogram (opposite sides are parallel)

D C

BA

Worksheet 5-02


1 a ABCD is a quadrilateral in which A = C and B = D. Prove that, if the opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram.

b LMNP is a quadrilateral in which LM = NP and LM || NP. PM is a diagonal. Prove that, if a pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.

c FGHL is a quadrilateral and the diagonals LG and FH bisect each other. Prove that, if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

D C

A B

P N

L M

L H

F

M

G


STAGE5.3



d PQRT is a quadrilateral with all sides equal. PR is a diagonal. Prove that, if the sides of a quadrilateral are equal, then it is a rhombus.

e CDEF is a quadrilateral and the diagonals CE and FD bisect each other at right angles. Prove that, if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

f WXYV is a quadrilateral. The diagonals WY and VX are equal and bisect each other. Prove that, if the diagonals of a quadrilateral are equal and bisect each other, then it is a rectangle.

g ABCD is a quadrilateral in which A = B = C = D = 90. Prove that, if the four angles of a quadrilateral are right angles, then it is a rectangle.

h TWME is a quadrilateral with all sides equal and M = 90. Prove that, if the sides of a quadrilateral are equal and one angle is a right angle, then it is a square.

i GHKL is a quadrilateral. G = H = K = L = 90 and GH = GL. Prove that, if the angles of a quadrilateral are right angles and a pair of adjacent sides are equal, then it is a square.

j MNPT is a quadrilateral. The diagonals MP and TN are equal and bisect each other at right angles. Prove that, if the diagonals of a quadrilateral bisect each other at right angles and are equal, then it is a square.

T R

P Q

F E

C

H

D

V Y

W

T

X

D C

A B

E M

T W

L K

G H

T P

M

X

N



2 a ABCD is a parallelogram and BX = DY. Prove: i ABX CDY

ii AXCY is a parallelogram.

b AECD is a rhombus. AE = EB. Prove: i CBE DAE

ii BCDE is a parallelogram.

c ABCD is a parallelogram. AP = AS = CQ = CR. Prove: i RQ = PS and PQ = RS

ii PQRS is a parallelogram.

d AC and DB are diameters of concentric circles, centre O. Prove ABCD is a parallelogram.

e PR and SQ are diameters of concentric circles, centre O. TU SQ. Prove PQRS is a rhombus.

f DEFG is a rectangle. W, X, Y and Z are the midpoints of the sides. Prove:

i WXYZ is a parallelogramii WXYZ is a rhombus.

B

A Y D

X C

A E B

CD

A

S Q

D R C

P B

A

EB

C

OD

F

T

UR

O

P

QS

D W

Y

E

X

G

Z

F

STAGE5.3



Summary of tests for quadrilaterals

A quadrilateral is:a a parallelogram if

both pairs of opposite angles are equal, or both pairs of opposite sides are equal, or both pairs of opposite sides are parallel, or one pair of opposite sides are equal and parallel, or the diagonals bisect each other.

b a rectangle if all angles are 90, or diagonals are equal and bisect each other.

c a rhombus if all sides are equal, or diagonals bisect each other at right angles.

d a square if all sides are equal and one angle is 90, or all angles are 90 and two adjacent sides are equal, or diagonals are equal and bisect each other at right angles.

Working mathematicallyReecting: Islamic designsThe Islamic rules for art were collected in the Hadith, a ninth-century text. Islamic art derives its unique style from combining the art of the Byzantines, the Copts, the Romans and the Sassanids.Islam believes in the balance and harmony of all things in existence. One of the vital beliefs is that the totality of things, all good and evil, proceed from the Lord of all being.There are strict rules against depicting humans or animals, which might result in idol worship, so an art form developed that was based on geometric designs and calligraphy, which are often interwoven.Geometric patterns appear in architecture and interiors to organise space and to beautify the environment. All patterns reect the pure beauty of numbers, considered to be of divine origin. By their very nature, geometric patterns show variation and order as expressions of unity, which is an attribute of God.Islamic geometric designs are constructed with skillful use of ruler and compasses (skills that you now have). Interactive geometry enables you to be exact with your constructions.

Geometry 5-06

Islamic design



ConstructionThis Islamic design starts with the drawing of a square.Step 1: Use axes and a grid to draw a 6 cm

square.Step 2: Construct a point at the centre of

each side and construct the diagonals of the square.

Step 3: Construct a circle with centre at the intersection of the two diagonals and with the side midpoints on the circumference.

Step 4: Continue to construct the gure shown on the right.

1 Are there other patterns you could have drawn using the construction lines?2 Can you tessellate the pattern you have drawn?

Converting decimals and percentages to fractionsTo convert a decimal to fraction form, rst count the number of decimal places in the decimal. If there is one decimal place, then write the decimal part over 10 and simplify. If there are two decimal places, then write the decimal part over 100 and simplify.1 Examine these examples.

a

(two decimal places, two 0s in the denominator)

b

(one decimal place, one 0 in the denominator)

c

d

2 Now convert these decimals to fraction form:a 0.75 b 0.28 c 0.3 d 0.14 e 0.06f 0.85 g 0.32 h 0.49 i 0.56 j 0.9k 0.72 l 0.65 m 0.2 n 0.24 o 0.53

0.35 = 35100---------

= 720------

7

20

0.8 = 810------

= 45---

4

5

0.64 = 64100---------

= 1625------

16

25

0.22 = 22100---------

= 1150------

11

50

Skillbank 5B

SkillTest 5-02

Converting decimals and

percentages to fractions



Properties of similar figuresSimilar gures are the same shape, but not necessarily the same size.The symbol for is similar to is |||.Enlarging or reducing a given gure will always produce a similar gure and the amount by which a gure is enlarged or reduced is called the scale factor or similarity factor.

3 To convert a percentage to a fraction, we write it over 100 and simplify. Examine these examples.

a b

c d

4 Now convert these percentages to fractions:a 76% b 10% c 80% d 45% e 88%f 56% g 75% h 31% i 68% j 5%k 60% l 54% m 6% n 49% o 82%

26% = 26100---------

= 1350------

13

5040% = 40

100---------

= 25---

2

5

8% = 8100---------

= 225------

2

2595% = 95

100---------

= 1920------

19

20

Skillsheet 5-06

Finding sides in similar triangles

Scale factor = image lengthoriginal length-----------------------------------

If two gures are similar, then: the matching angles are equal the matching sides are in the same ratio.



Example 10

If ABC ||| DEF, nd the value of k.

Solution

= (matching sides in the same ratio)

k = 40

= 50

40 mm

20 mm25 mm

k mm

A B

C

D E

F

k40------

2520------

2520------

1 Are the gures similar in each of the following pairs?a b

c d

e f

2 Find the value of the pronumerals in the following pairs of similar gures:a b c

9 cm

12 cm

6 cm

8 cm

70

70

6 cm

4 cm

5 cm

3 cm

10 cm8 cm

5 cm

4 cm6.4 cm

8 cm

10 cm

4 cm

3 cm

6 cm

5 cm

2 cm

10 cm

4 cm

9 1527

yk

10 18

12 74

6

m

Exercise 5-06

Example 10

Geometry 5-07

Similar triangles



d e f

3 Find the value of the pronumeral in each of the following pairs of similar triangles:a b

c d

e f

g h

4 Find the similar gures and determine the scale factor:a

b

x6

15

9

8 3

15

h 22

8

14

p

12

30

45

d30

30

9

11

6

y

3535

k

8

14

18

25

10

35a

m24

16

2030

26

26 y

7 15

22e

w

70

70

22

16 12

10 cm

8 cm60

A6 cm

7.5 cm

60B

6.25 cm

5 cm60

C

18 cm

20 cm

60 D

16 m

m

32 mm

A

10 m

m

20 mm

B

13 m

m

25 mm

C

48 mm

24 m

m

D



Tests for similar trianglesThere are four conditions or tests for proving that two triangles are similar.

5 ABC is similar to ADE. Find m, correct to one decimal place.

6 XYW is similar to XZV. Find p, correct to one decimal place.

7 Colin, who is 1.83 m tall, casts a shadow 1.2 m long. At the same time a light tower casts a shadow 14 m long. What is the height of the tower?

8 State whether each of the following is true (T) or false (F).a The angles of similar gures are equal.b All circles are similar.c The sides of similar gures are in the same ratio.d If two triangles have two pairs of matching angles equal, they are similar.e Since the angles of two rectangles are equal, they must be similar.f All squares are similar.g All equilateral triangles are similar.h Any two isosceles triangles are always similar.

9

8

6

A

BC

DE

m

5 8

10

V

W

X Y Z

p

1.2 m 14 m

1.83 m

Tests for similar triangles1 Side, Side, Side (SSS)

If the three sides of one triangle are proportional to the three sides of another triangle, then the two triangles are similar.

C5

2

104

84

F

A B D E

Worksheet 5-07

Congruent and similar triangle

proofs

STAGE5.3



2 Side, Angle, Side (SAS)If two sides of one triangle are proportional to two sides of another triangle, and the included angles are equal, then the two triangles are similar.

3 Angle, Angle (AA)If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

4 Right angle, Hypotenuse, Side (RHS)If the hypotenuse and a second side of a right-angled triangle are proportional to the hypotenuse and a second side of another right-angled triangle, then the two triangles are similar.

C5

30

50

3

F

AB

D

E

C F

AB D

E

156 5

2

Example 11

Prove that the triangles in each of the following pairs are similar:a b

9

12

6

84

6

E

C

F

G

T

D

50

9

6

508

12

RM

N

W

P B



Solutiona In CDE and FTG:

= =

= = (ratios of matching sides)

= =

= =

CDE ||| FTG (matching sides are in the same ratio or SSS)b In MWN and BPR:

= =

= =

=

M = B = 50 (included angles are equal) MWN ||| BPR (two pairs of matching sides are in the same ratio and the included

angles are equal or SAS)

Prove ABE ||| ACD.SolutionIn ABE and ACD:A is commonAEB = ADC (corresponding angles, BE || CD) ABE ||| ACD (two pairs of matching angles equal or AA)

a Prove MRP ||| XRW.b Find the value of k.

Solutiona In MRP and XRW:

P = W (alternate angles, PM || XW)M = X (alternate angles, PM || XW) MRP ||| XRW (two pairs of matching angles are equal or AA)

b = (matching sides in similar triangles) k = 12

= 4.8

EDGT--------

64---

32---

CEFG--------

96---

32---

CDFT--------

128

------

32---

EDGT--------

CEFG--------

CDFT--------

MWBP

----------

68---

34---

MNBR---------

912------

34---

MWBP

----------

MNBR---------

Example 12

A B

E

D

C

Example 13

P

M

Wk

XR

12

20

8

k12------

820------

820------

STAGE5.3



1 Which similarity test (SSS, AA, SAS, RHS) can be used to prove that each of the following pairs of triangles are similar? Where possible state the similarity factor (scale factor) between the triangles.a b c

d e f

g h i

2 a Prove XYW ||| XLM. b Prove ABC ||| EDC. c Prove PLM ||| PTW.

d Prove TYX ||| KGX. e Prove GMK ||| LHK. f Prove WYZ ||| ZYX.

18

60

40

27

8

6

9

12

10

6

12

8

4.8

9.630

16.875

16

9 42

7563

42

3030

2.4

9.66

12

3

7.5

18

2510.8

15

X

L

Y

W

M

6

7.5

10

8E

D

B

C

A

P L

M

T

W

18

28.8

7.5

12

Y

T XK

G

M

L

K

H

G

W YX

Z


Example 12



STAGE5.33 a i Prove GKF ||| HKM. b i Prove CDE ||| FGE.

ii Find the value of m. ii Find the value of k.

c Prove BCF ||| DCE and, hence, nd d Prove AXM ||| AYN and, hence, ndthe value of x. the length of AX.

e i Prove that TFR ||| TPD. f ABCD is a parallelogram.ii If FR = 9 cm, DP = 5 cm and i Prove EFC ||| EAB.

TD = 8 cm, nd the length of TR. ii Prove EAB ||| AFD.iii Prove EFC ||| AFD.iv If EF = 20 cm, FC = 8 cm and

AF = 16 cm, nd the length of AB.

4 In PWT, W = 90 and WN PT.a Prove PWN ||| WTN.b If PN = 4 cm, NT = 9 cm, nd the lengths of WN, WP and WT.

5 a Prove that EFG ||| HFE.b If FG = 16 cm and EF = 20 cm, nd the length of FH.

F

KH

m

M

G

12

6 8

D

C

kE

G

F16

1810

F E

DBx

C10

68

A X Y

M

N

9

8

15

T P

DF

RD F

E

C

A B

P N

W

T

F G H

E

Example 13



Proving geometrical resultsGeneral geometrical results can be proved by writing a geometrical argument, where reasons are given at each step of the argument. Many of the following proofs involve congruent or similar triangles.

Example 14

AB is the diameter of a circle, with centre O. P is any point on the circumference. Prove that APB, the angle in a semi-circle, is a right angle.

SolutionLet OAP = x

OPA = x (OAP is isosceles, equal radii) BOP = 2x (exterior angle of OAP)

OPB + OBP = (180 2x) (angle sum of OPB)But OPB = OBP (OPB is isosceles, equal radii)

OPB = (180 2x) 2= (90 x)

APB = OPA + OPB= x + (90 x)= 90

The angle in a semi-circle is a right angle.

ABCD is a parallelogram. HG is any interval joining parallel sides AB and DC and passing through the midpoint, T, of diagonal BD.Prove that the interval through the midpoint of a diagonal of a parallelogram divides opposite sides equally. (That is, prove DG = BH.)SolutionCongruent triangles are used to prove this result.In DGT and BHT:

DT = BT (T is the midpoint of BD)DGH = BHT (alternate angles, DC || AB)GDT = HBT (alternate angles, DC || AB)

DGT BHT (AAS) DG = BH (matching sides of congruent triangles)

Prove that the interval joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.

A

P

OB

A

P

Ox B

Example 15

T

D

A H B

G C

D E

B C

AExample 16

Worksheet5-08

Geometrical proofs order

activity

Worksheet5-09

Complete the proofs



Pythagoras theoremPythagoras theorem can be stated as follows:

SolutionSimilar triangles are used to prove this result.

In ADE and ABC:

=

= (since D and E are the midpoints of AB and AC respectively)

=

A is common.

ADE ||| ABC (two pairs of matching sides are in the same ratio and the included angles are equal or SAS)

ADE = ABC (matching angles of similar triangles) DE || BC (corresponding angles proved equal)Also = (matching sides of similar triangles)

But =

=

DE = BC

The interval joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.

ADAB--------

12---

AEAC--------

12---

ADAB--------

AEAC--------

DEBC--------

ADAB--------

ADAB--------

12---

DEBC--------

12---

12---

The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

Working mathematicallyApplying strategies and reasoning: Proving PythagorasThere are many proofs of Pythagoras theorem and they are based on: algebra geometry dissection of areas

STAGE5.3



The converse of Pythagoras theoremThe converse of Pythagoras theorem is as follows:

Proof of the converse of Pythagoras theoremGiven: ABC where AC2 = AB2 + BC2

Aim: To prove ABC = 90

Construction: Draw PQR where QR = BC,PQ = AB, and PQR = 90.

Proof: In PQR,PR2 = PQ2 + QR2 (Pythagoras theorem)

= AB2 + BC2 (since PQ = AB, QR = BC)= AC2 (given)

PR2 = AC2 PR = AC

So, in triangles ABC and PQR,PQ = AB (by construction)QR = BC (by construction)

and PR = AC (proved) PQR ABC (SSS) PQR = ABC (corresponding angles of congruent triangles) ABC = 90 (since PQR = 90)

1 Proof by dissection

Explain how this dissection proves Pythagoras theorem.

2 Find two other proofs of Pythagoras theorem by dissection of areas.

A

bc

aC B

a

a

b

b

A

C B

Geometry 5-08

Pythagorean dissection

If the square on one side of a triangle equals the sum of the squares on the other two sides, then the angle between these other two sides is a right angle.

C

A

B R

P

Q



Example 17

1 Determine whether ABC, shown, is right-angled at C.

Solution202 + 992 = 400 + 9801

= 10 201= 1012

The angle between AC and BC is a right angle. C = 90

2 PQTS is a square with SQ = 10 cm. Find:a the length of PQb the area of PQTS

Solutiona Let PQ = x cm

PS = x cm x2 + x2 = 102 (by Pythagoras theorem)

2x2 = 100x2 = 50x =

=

PQ = cmb Area = PQ PS

=

= 50 cm2or since PQTS is a rhombus (why?)

Area =

=

= 50 cm2

20 cm 99 cm

101 cm

C

B A

10 cm

P

S T

Q

505 25 2

5 2( ) 5 2( )

12--- SQ PT12--- 10 10

(Using dynamic geometry software, rst conrm each result by construction and measurement before proving the result.)1 Prove that the exterior angle of a triangle is equal to the sum of the

interior opposite angles (that is, prove CBD = CAB + ACB).

AB

C

D

Exercise 5-08

Example 14

Geometry

STAGE5.3



2 The line joining the centres of two intersecting circles is the perpendicular bisector of the common chord. Prove this result by working through the following:a Prove that AXB AYB and, hence, prove that

XAC = YAC.b Prove that XAC YAC.c Hence, show that XC = YC and AB XY.

3 The line through the midpoint of a side of a triangle parallel to another side bisects the third side. (This result is the converse of the result in Example 16.) Follow the steps below to prove this result.a In the diagram, Q is the midpoint of PM and QR || MN. Show

that PQR ||| PMN.b Hence, show that PR = RN.

4 In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Prove Pythagoras theorem using the following steps.a Consider ABC with ABC = 90 and CD AB. Show

ADC ||| ACB and, hence, that AC2 = AB AD (using the ratio of matching sides).

b Show BDC ||| BCA and, hence, that BC2 = AB DB.c Hence, show that AC2 + BC2 = AB2.

5 DEF is a triangle and DG EF. Prove, using Pythagoras theorem, that:DE2 + GF2 = DF2 + EG2.

6 Determine whether these are right-angled triangles:a b c

Y

C

X

A B

P

R

NM

Q

A D

C

B

D

E G F

153

135

71

H

J

I X Y

Z

5_

8

3_

8

1_

2 2F

D

E

2

2

Example 15

Example 16



7 Use Pythagoras theorem to calculate the following. (Express your answers in surd form where necessary.)a Calculate the length of the diagonal of a rectangle with sides 28 cm and 45 cm.b A square has diagonals of length 24 m. Calculate the side length.c Find the altitudes of an equilateral triangle with sides 10 cm.d A rhombus has sides of 52 m and one diagonal 40 m. Find the length of the other diagonal.

8 If the diagonals of a quadrilateral are perpendicular then the sums of the squares on opposite sides are equal. Prove this result by the following steps.a Use Pythagoras theorem to write expressions for AB2,

BC2, CD2 and AD2.b Hence, show AB2 + CD2 = BC2 + AD2.

9 a For the diagram on the right, prove ABC ||| EDC.b Find the value of m.

DA

B

X

C

3

10

54

m

A B

C

D E

1 a Use Pythagoras theorem to prove that the area of the semi-circle on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semi-circles on the other two sides.

b Prove that, in any right-angled triangle, the area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides.

2 The median is a line joining a vertex of a triangle to the midpoint of the opposite side. Prove that the medians of a triangle are concurrent (that is, they meet at one intersection point). A

C

B

Power plus

STAGE5.3



Topic overview Copy and complete the table below.

The best part of this chapter was

The worst part was

New work

I need help with

Language of mathsAA AAS acute-angled angle sumargument axis of symmetry bisect concurrentcongruence test congruent () converse convexdeductive diagonal equilateral exterior anglehence hypotenuse interior angle isosceleskite matching midpoint obtuse-angledopposite order parallel parallelogramperpendicular polygon Pythagoras theorem quadrilateralratio reason rectangle regular polygonrhombus RHS right-angled rotational symmetrySAS scale factor scalene similar (|||)similarity factor similarity test square SSStest trapezium triangle vertices

1 What is the formal denition of a trapezium?2 What word in the list above has the same meaning as therefore?3 What is the most general name for a shape with four equal sides?4 What does a test for a quadrilateral do? Give an example of one.5 Copy and complete: Matching sides in similar gures are in the same _______.6 Are all squares similar?7 What does AA mean?8 Write the converse of Pythagoras theorem.

Worksheet 5-10

Geometry crossword

Worksheet 5-11

Geometry summary poster

?



Copy and complete this overview of the topic into your workbook. Add extra words and use pictures to make the overview useful to you. Have your overview checked by other students or by your teacher to make sure nothing is missing or incorrect.

DEDUCTIVEGEOMETRY

Convex polygons

Propertiesof triangles

and quadrilaterals

Congruence Similarity

angle sum = 180 or 360 or 540 or regular polygons

Scale factor

SSS SAS

AAS RHS

Tests for congruent triangles

Geometrical proofs using congruence/similarity

Tests for similar triangles: SSSSASAARHS

c a

b

c2 = a2 + b2



1 Find the value of the pronumeral in each of the following. (Give reasons for your answers.)a b c

2 a In PRT, PR = PT. PR is produced to M so that RM = RT. Prove PTR = 2 MTR.

b BCD is equilateral, DBA is isosceles with BD = AB. Prove AD CD.

3 a Show that the exterior angle of a regular dodecagon (12 sides) is 30.b The size of an interior angle in a regular polygon is 175. How many sides has the

polygon?

4 Which congruence test (SSS, SAS, AAS or RHS) can be used to prove that the triangles in each of these pairs are congruent?

a b c

5 Which of the following pairs of triangles are congruent? Give reasons.

a b c

Chapter 5 Review

Ex 5-01

110

c

C

B

A

D

FE

114

4w

w

36

y

P

T

R M

Ex 5-01

A B

D

C

Ex 5-02

Ex 5-03

5

58

878

78

60

60

8 cm 8 cm

Ex 5-03

Topic testChapter 5



6 a KMNP is a square and XMN is equilateral. Prove:

i KMX PNXii KPX is isosceles.

b In the diagram, DEFG is a quadrilateral.DG = EF and DF = EG.

i Prove that DEF EDG.ii Prove that DEY is isosceles.

iii Hence, prove that FGY is isosceles.iv Show that DE || GF.

7 ABCD is a parallelogram. BC = BY = DX.a Explain why DAX = BCY.b Show that AD = DX.c Prove that DAX BCY.d Hence, prove that BXDY is a parallelogram.

8 If the diagonals of a quadrilatral bisect each angle at a vertex, then the quadrilateral is a rhombus. Prove this result by following these steps.a Show + + + = 180.b Prove = 180 ( + + ) using ABD and, hence, prove

that = .c Hence, show = and that AB = BC.d Now prove ABCD is a rhombus.

9 Find the value of the pronumerals in the following if the plane shapes in each pair are similar.a b

c d

P

X

N

K M

Ex 5-03

G F

ED

Y

A X B

D Y CEx 5-04

A B

D C

Ex 5-05

Ex 5-06

7 cm

10 cm k cm

9 cm

6 mm

9 mm

d mm

4 mm

y

10

6

7

m

11 6

3



10 If ABC is similar to AED nd y (correct to one decimal place).

11 ABE is similar to ACD. Find the value of d.

12 AC bisects BAD. Prove ACD ||| ABC.

13 ABCD is a square. a Prove WBA ||| CYD.b Prove WA DY = AB CD and, hence, that AD2 = WA DY.

14 a JKLM is a rhombus. Show JL2 + KM2 = 4JK2.

b PQMN is a square and AM = BQ. Prove that NPC is isosceles.

y

12

515

A

B

C

DE

Ex 5-06

d cm

7 cm 9 cm

5 cm

A

B

C

DE Ex 5-06

D C

BA

8

20

3.2

Ex 5-07

W B X

C

D

A

Y Ex 5-07

M L

J K

Ex 5-08

P Q

N M

A

BC


Student textImprint pageContentsPrefaceHow to use this bookHow to use the CD-ROMAcknowledgements Syllabus reference gridCh 1 - Working with numberFractions Approximation Accuracy of results when rounding Recurring decimals Expressing fractions in decimal form Expressing decimals in fraction form Solving percentage problems Scientific notation Ratios Scale drawings Rates Topic overview Chapter review

Ch 2 - Surface area and volumeSurface area of prisms and cylindersSurface area of composite solids made of prisms and cylinders Surface area of a pyramid Surface area of a cone Surface area of a sphere Surface area of composite solids Volume of right prisms and cylinders Volume of pyramids, cones and spheres Volume of a sphere Summary of surface area and volume formulas Volumes of composite solidsAreas of similar figures Surface area and volume of similar solidsTopic overview Chapter review

Ch 3 - Surds and indicesRational and irrational numbers Simplifying surds Multiplying and dividing surds Adding and subtracting surds Binomial products and surds Rationalising the denominator Index notation Summary of index laws and propertiesTopic overview Chapter review

Mixed revision 1 Ch 4 - Equations and inequalitiesSolving linear equations Solving word problems Working with formulas Changing the subject of the formula Solving linear inequalities Simultaneous equations Solving problems using simultaneous equationsSolving quadratic equations of the form x2 = cGeneral form of a quadratic equation: ax2 + bx + c = 0Solving quadratic equations by completing the squareSolving quadratic equations by using the quadratic formulaSolving quadratic equations: which method?Solving problems involving quadratic equationsVariable substitutionSimultaneous linear and quadratic equationsTopic overview Chapter review

Ch 5 - Deductive geometryDeduction in geometry Properties of triangles and quadrilaterals Properties of convex polygons Proving congruent triangles Proving properties of triangles and quadrilaterals Tests for quadrilaterals Properties of similar figures Proving geometrical results Pythagoras theorem Topic overview Chapter review

Ch 6 - Saving and borrowingInterest Simple interest Compound interest Depreciation Methods of purchasing goods Loan repayments Credit cards Topic overview Chapter review

Mixed revision 2 Ch 7 - Coordinate geometryFormulas for the midpoint, distance and gradient Points lying on a line Parallel and perpendicular lines Graphing linear equations The gradientintercept form of a linear equation Graphing equations of the form y=mx + bThe general form of a linear equation The pointgradient form of a linear equation Equations of parallel and perpendicular linesCoordinate geometry problems Graphing regions on the number plane 309Graphing non-linear equations Topic overview Chapter review

Ch 8 - TrigonometryRight-angled triangle trigonometry Angles of elevation and depression Bearings Trigonometric relations Trigonometric ratios of obtuse angles The sine rule The cosine rule The area of a triangle Applications of the sine and cosine rules Topic overview Chapter review

Ch 9 - Analysing dataAnalysing data Features of a display of data Comparing the mean, median and modeCumulative frequency tables and graphs Grouped data Measures of spread Standard deviation Comparing sets of data Comparing the range, interquartile range and standard deviation 40Topic overview Chapter review

Mixed revision 3 Ch 10 - ProbabilityProbability Equally likely outcomes Probabilities of simple events Range of probability Complementary events Relative frequency Theoretical probability Independent and dependent events Two-stage events Compound events Topic overview Chapter review

Ch 11 - GraphsDescribing change Distancetime graphs Graphs and other rates of change Interpreting graphs Graphs of equations The axis of symmetry and the vertex of a parabola Summary The cubic curve The hyperbola The exponential curve The circle Identifying graphs Topic overview Chapter review

Mixed revision 4 General revision Ch 12 - Circle geometryThe language of circles Chord properties of circles Angle properties of circles Tangents to a circle Tangent and secant properties of circlesProofs using circle theorems

Ch 13 - Curve sketching and polynomialsThe cubic curve y=a(x -r)(x - s)(x - t)The graph of y = axn The curve y = axn + k The curve y = a(x r)n The equations of circles Point of intersection of a line with a curve Polynomials Adding, subtracting and multiplying polynomials

Ch 14 - Functions and logarithmsFunctions Inverse functions Logarithms Logarithm laws The graphs of y = ax and y = logax Exponential and logarithmic equations

AnswersIndex

Glossary AB CDE FG HIK L M NO PQ RST UV X Y

All files menuExplanation menuLearning technologyPractice menuRevision menuUsing technology menuLicence agreement

Cabri:

GSP:

geometry:

glossary:

new century maths 10: stage 5.2/5.3 - chapter 5

Documents