New Century Maths 10: Stage 5.2/5.3 - Chapter 5

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

Wordbank 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.

Z deduction

05_NCM10EX2SB_TXT.fm Page 181 Monday, September 12, 2005 3:45 PM

deductive geometry S PAC E AN D G E O M E T RY 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

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 hence

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

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

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’. Z similarity

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 182 Monday, September 12, 2005 3:45 PM

Start up Worksheet 5-01 Brainstarters 5

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

63°

47° x° 82°

Skillsheet 5-01 Angles and parallel lines

d

135°

e

f

k°

42°

78° y°

127°

5a°

38°

37°

w°

152° 4a°

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

4m°

2h° 66°

127° 88°

d

e

f 81° m°

a°

w° k°

72°

104° p°

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

63°

2h° 124°

35°

71°

5h° 3h°

d

e

83°

f

a°

137° 3y° 2y°

27° 105° 3k°

182

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

163°

34°

05_NCM10EX2SB_TXT.fm Page 183 Monday, September 12, 2005 3:45 PM

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

c

132° 67°

69°

78°

48° 103°

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 specific 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.

Just for the record Euclid and Euclidian geometry Euclid was a Greek mathematician who lived in Alexandria during the third century BC. Euclid’s 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 first volume). Geometry using the methods from Euclid’s 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 briefly describe how it is different from Euclidean geometry.

Example 1 Find the values of the pronumerals in the following diagrams: a

b

E

k°

55°

D

46°

A

d° B

C

DE DUCTIVE GEOMETRY

183

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 184 Monday, September 12, 2005 3:45 PM

Solution

a k + 55 + 55 = 180 k + 110 = 180 k = 70 b ∠EDC = ∠ECD = (180 − 46) ÷ 2 ∠EDC = 67° ∠EBC = 67° d + 67 = 180 ∴ d = 113

Skillsheet 5-02 Two-dimensional shapes

(angle sum of isosceles triangle)

(angle sum of isosceles ∆ECD) (opposite angles of a parallelogram are equal) (angle sum of a straight line)

Properties of triangles and quadrilaterals Triangles Definitions of special triangles • • • • • •

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.

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

184

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 185 Monday, September 12, 2005 3:45 PM

Quadrilaterals A quadrilateral is a four-sided plane figure. A quadrilateral may be either convex or non-convex (concave). A convex quadrilateral has no interior angles greater than 180°.

Worksheet 5-02 Naming quadrilaterals Worksheet 5-03 Investigating the geometrical constructions Skillsheet 5-03 Geometrical constructions

A convex quadrilateral (all angles are less than 180°)

A non-convex quadrilateral (one interior angle is greater than 180°)

The diagonals of a convex quadrilateral lie inside the quadrilateral. P C

D N

B

L

A M

Diagonals AC and BD lie inside the convex quadrilateral ABCD.

The diagonal MP does not lie inside the non-convex quadrilateral LMNP.

In the work that follows, convex quadrilaterals will be simply referred to as ‘quadrilaterals’. Definitions of special quadrilaterals 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.

DE DUCTIVE GEOMETRY

185

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 186 Monday, September 12, 2005 3:45 PM

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

Parallelogram

Rectangle

– opposite sides parallel – opposite sides equal – opposite angles equal – diagonals bisect each other – has rotational symmetry – 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 mathematically Worksheet 5-02 Naming quadrilaterals

Reasoning: Is a square a rhombus? Looking at the definitions of the quadrilaterals above, we see that a parallelogram can also be classified 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.

186

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 187 Monday, September 12, 2005 3:45 PM

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?

quadrilaterals trapeziums parallelograms rhombuses rectangles squares

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

Example 2

STAGE

Prove that the exterior opposite angles of a parallelogram are equal.

A

B

C

5.3

Solution To prove ∠FBC = ∠FDE Proof: ∠FBC = ∠A (corresponding angles, AD || BF) and ∠FDE = ∠A (corresponding angles, AB || DF) ∴ ∠FBC = ∠FDE ∴ Exterior opposite angles of a parallelogram equal.

D E

F

Skillsheet 5-03 Geometrical constructions

Using technology Geometric constructions: Bisectors The 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 Geometer’s Sketchpad or Cabri Geometry. Construction 1 Step 1: Draw a triangle and label it ABC. Step 2: Construct the perpendicular bisectors of sides AB and BC. (In Geometer’s Sketchpad you will need to find the midpoint of the side first.) 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.

Skillsheet 5-04 Starting Geometer’s Sketchpad

B

A

O C

Skillsheet 5-05 Starting Cabri Geometry Geometry 5-01 Side bisectors

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 significant way when you drag a vertex or side of ∆ABC? DE DUCTIVE GEOMETRY

187

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 188 Monday, September 12, 2005 3:45 PM

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? Geometry 5-02 Angle bisectors

Construction 2 Step 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.

Q

O R P

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 significant 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? Extension 11 Find the radius of the in-circle of the isosceles triangle with sides 10 cm, 10 cm and 4 cm.

Exercise 5-01 Example 1

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

b

c

m°

80° 36° h° y°

d

e

f

x° a°

188

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

d°

3d°

05_NCM10EX2SB_TXT.fm Page 189 Monday, September 12, 2005 3:45 PM

g

h

i h°

y°

p°

130°

42°

70°

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

b

y°

c

w°

k°

55° 127°

53°

d

e

h°

f°

65°

f

134°

71°

a°

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

b

c y°

40°

m°

d

e

f

43°

78° w° k°

c°

m°

115° D

a° C

p°

h

78° h°

X

i

65°

w°

56° 32°

34° A

d°

w°

h°

g

120°

58°

t°

74°

B X

110° 70° p°

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.)

q° Z DE DUCTIVE GEOMETRY

w°

Y

189

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 190 Monday, September 12, 2005 3:45 PM

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°. STAGE

5.3

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

K P

x° L

M

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

N Y

X

P W

Example 2

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

T B C

F

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

E

D

H

N 93°

K

190

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

147° L

M

05_NCM10EX2SB_TXT.fm Page 191 Monday, September 12, 2005 3:45 PM

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

B

C

A

m°

k°

42°

E

D

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

V

W

A

b

E

c

y° Z x°

x° y° x°

X

C

B

40°

D

H

G

F

y° Y I

d

A

e

33°

E

H

y°

x°

y° J

x°

D

P

f T y° B

S

x° Q

54°

20°

67° C

R

108° K

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

E

D

A

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

B P

N

K

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

C

L

M

B E C D A

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

U

W

X

DE DUCTIVE GEOMETRY

Y

191

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 192 Monday, September 12, 2005 3:45 PM

STAGE

5.3

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

W

T

Y

P

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

Q B

C

D F

Worksheet 5-04 Angles in polygons

A

H

Properties of convex polygons The general name for any plane figure 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.

Worksheet 5-05 Find the missing angle

Convex polygon

Non-convex polygon

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

Regular polygons A regular polygon has all sides equal and all angles equal. The first eight regular polygons are:

3 sides

4 sides

5 sides

6 sides

Equilateral triangle

Square

Pentagon

Hexagon

7 sides

8 sides

9 sides

10 sides

Heptagon

Octagon

Nonagon

Decagon

192

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 193 Monday, September 12, 2005 3:45 PM

Just for the record Canberra: The national capital Canberra 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 Griffin, 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 Griffin directly along Anzac Parade to the Australian War Memorial. What other geometrical features can you see in Canberra’s design?

DE DUCTIVE GEOMETRY

193

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 194 Monday, September 12, 2005 3:45 PM

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.

Solution

Angle sum = (16 − 2) × 180° = 14 × 180° = 2520° The size of each angle of a regular polygon with n sides is given by the following formula. angle sum One angle = ------------------------ , where n is the number of sides n (n – 2) × 180° = ---------------------------------n

Example 4 1 Find the size of an angle of a regular octagon.

Solution

Angle sum = (8 − 2) × 180° = 6 × 180° = 1080° 1080° ∴ Each angle = -------------8 = 135°

or

(8 – 2) × 180° Each angle = ---------------------------------8 = 135°

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

a (n − 2) × 180 = 3960 3960 n − 2 = -----------180 n − 2 = 22 n = 24 ∴ Number of sides is 24.

3960 b Each angle = -----------24 = 165°

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

194

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 195 Monday, September 12, 2005 3:45 PM

Example 5 1 The exterior angle of a regular polygon is 18°. How many sides does the polygon have?

Solution Let n be the number of sides of the polygon. Sum of exterior angles = n × 18° = 360° 18n = 360 360 n = --------18 = 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°) = 40° Number of exterior angles = 360 ÷ 40 =9 ∴ The polygon has 9 sides. Method 2: (n – 2) × 180 -------------------------------- = 140 n (n – 2) × 180 n × -------------------------------- = n × 140 n1 (n − 2) × 180 = 140n So: 180n − 360 = 140n 180n = 140n + 360 180n − 140n = 140n + 360 − 140n 40n = 360 ∴n=9

1

Geometry 5-03 Exterior angles of a polygon

Using technology Exterior angle sum of a convex polygon Step 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?

DE DUCTIVE GEOMETRY

195

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 196 Monday, September 12, 2005 3:45 PM

Exercise 5-02 Example 3

1 Use the rule, Angle sum = (n − 2) × 180°, to find the angle sum of a polygon with: a 20 sides b 10 sides c 15 sides d 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 900° d 6840° e 2160° f 1260°

Example 4

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 Example 5

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?

Skillbank 5A Converting fractions and decimals to percentages SkillTest 5-01 Converting fractions and decimals to percentages

To convert a fraction or decimal to a percentage, multiply it by 100%. 1 Examine these examples: 20

a

2 2 2 --- = --- × 100% = --- × 100% = 2 × 20% = 40% 5 5 51 4

18 18 18 b ------ = ------ × 100% = ------ × 100% = 18 × 4% = 72% 25 25 251 20

3

c

24 24 24 3 ------ = ------ × 100% = ------ × 100% = --- × 100% = 3 × 20% = 60% 40 40 405 51 4

7

21 21 21 21 d ------ = ------ × 100% = ------ × 100% = ------ × 4% = 7 × 4% = 28% 75 75 753 31 2 Now convert these fractions to percentages: 7 33 27 a -----b -----c -----10 50 60 30 -----40 15 k -----50 f

196

g l

60 -----75 16 -----20

4 h --5 54 m -----60

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

22 d -----25 11 -----20 18 n -----40

i

e j o

24 -----32 28 -----80 13 -----25

05_NCM10EX2SB_TXT.fm Page 197 Monday, September 12, 2005 3:45 PM

3 Examine these examples: a 0.41 = 0.41 × 100% = 0.41 = 41% c 0.9 = 0.9 × 100% = 0.90 = 90% 4 Now convert these decimals to percentages: a 0.25 b 0.68 c 0.17 f 0.333 g 0.59 h 0.702 k 0.428 l 0.055 m 0.91

b 0.08 = 0.08 × 100% = 0.08 = 8% d 0.375 = 0.375 × 100% = 0.375 = 37.5% d 0.6 i 0.84 n 0.7825

e 0.1 j 0.7 o 0.314

Proving congruent triangles Congruent figures have exactly the same size and shape. The symbol for ‘is congruent to’ is ≡. In congruent figures, 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. 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.

Worksheet 5-06 Congruent triangles proofs

Geometry 5-04 Congruent triangles

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.

DE DUCTIVE GEOMETRY

197

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 198 Monday, September 12, 2005 3:45 PM

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 48°

30 mm 35°

48°

35° 30 mm

Solution a AAS STAGE

5.3

b SAS

Example 7 In the diagram on the right, ∠W = ∠V and TX ⊥ WV. Prove: a ∆WXT ≡ ∆VXT b X is the midpoint of WV.

T

Solution

a In ∆WXT and ∆VXT ∠W = ∠V ∠WXT = ∠VXT TX is common ∴ ∆WXT ≡ ∆VXT b WX = VX ∴ X is the midpoint of WV

W

V

X

(given) (TX ⊥ WV) (AAS) (matching sides of congruent triangles)

Exercise 5-03 Example 6

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

10 cm 4 cm

5 cm 8 cm

8 cm

4 cm 10 cm

d

e

5 cm

f

70°

7 cm

40° 9 cm 40° 9 cm

198

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

70°

15 cm

15 cm

7 cm

05_NCM10EX2SB_TXT.fm Page 199 Monday, September 12, 2005 3:45 PM

g

h

60°

i

70° 10 cm

55°

10 cm

8 cm 8 cm 70°

55°

60°

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

b 40° 31 mm

d mm

40°

50° 50°

58°

58°

25° k°

12 cm w°

c

d a°

9 cm

p cm 33°

25° y cm

e

f 53 mm

35 mm

109° 35 mm

9 cm

k cm

30 mm 53 mm

60°

60°

p° 9 cm

30 mm

11 cm

For Questions 3 and 4 following, identify matching vertices before writing the proof. 3 a AB = CB and EB = DB. Prove ∆ABE ≡ ∆CBD. A

D

b QT ⊥ WT, PW ⊥ WT and QW = PT. Prove ∆QTW ≡ ∆PWT. P

Example 7

W

B E

C

T

Q

DE DUCTIVE GEOMETRY

199

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 200 Monday, September 12, 2005 3:45 PM

STAGE

5.3

c CH || EG, DH || FG and CH = EG. Prove ∆CDH ≡ ∆EFG. H

d YX bisects ∠UXW and UX = WX. Prove ∆UXY ≡ ∆WXY. U

G

X C

D

E

Y

F W

e ABCD is a square and AY = CX. Prove ∆ABY ≡ ∆CBX. X

D

f

LM = NP and LP = NM. Prove ∆LMP ≡ ∆NPM.

C

N

P

Y A

B

g O is the centre of the circles. Prove ∆AOB ≡ ∆COD.

L

M

h ∆FGH is an isosceles triangle, so FH = FG. HN ⊥ FG and GM ⊥ FH. Prove ∆FHN ≡ ∆FGM. G

H

B C O

N

M

A D

F

4 a ∆PQR is isosceles and QA = RB. Prove: i ∆PQA ≡ ∆PRB ii ∆PAB is isosceles.

b TP = XP and AP = CP. Prove: i ∆TAP ≡ ∆XCP ii TA || XC. T

P

A

P

C Q

200

A

B

R

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

X

05_NCM10EX2SB_TXT.fm Page 201 Monday, September 12, 2005 3:45 PM

c AB || CD and AD || CB. Prove: i ∆ABD ≡ ∆CDB ii AD = CB and AB = CD. D

d A and B are the centres of two circles which intersect at C and D. Prove: i ∆ADB ≡ ∆ACB ii AB bisects ∠DAC.

C

D A

A

B

B

C

e ∠HEF = ∠GFE and EH = FG. Prove: i ∆HEF ≡ ∆GFE ii ∠EHF = ∠FGE. H

f

O is the centre of the circle and LT = MN. Prove: i ∆LOT ≡ ∆MON ii ∠LOT ≡ ∠MON. N

G K

T

E

F

O

M

L

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 = ∠DCA iii ∆BCY ≡ ∆DCY iv BY = DY. B

O A C

D

E

Y

C

D

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

DE DUCTIVE GEOMETRY

201

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 202 Monday, September 12, 2005 3:45 PM

Using the word ‘hence’ When the word ‘hence’ is used, it means use or follow on from the previous result. For example, if a question states: a Prove ∆ABC ≡ ∆DBC b Hence prove ∠A = ∠D, then part b follows from what was proved in part a and the final statement in the proof would be: ∠A = ∠D (matching angles of congruent triangles) STAGE

5.3

Example 8 ABCD is a rectangle. a Prove that ∆ABD ≡ ∆BAC. b Hence show that the diagonals of a rectangle are equal.

Solution

a In ∆ABD and ∆BAC: AD = BC AB is common. ∠DAB = ∠CBA = 90° ∴ ∆ABD ≡ ∆BAC

D

C

A

B

(opposite sides of a rectangle) (angles in a rectangle) (SAS)

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

Using technology Geometry 5-05 Properties of triangles and quadrilaterals Worksheet 5-03 Investigating the geometrical constructions

Properties of triangles and quadrilaterals Isosceles triangle 1 Step 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. B 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 2 Step 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? E

202

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

A

T

C

B

T

W

05_NCM10EX2SB_TXT.fm Page 203 Monday, September 12, 2005 3:45 PM

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

D

C

A

6 Does the perpendicular bisector pass through B and D?

B

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.

Exercise 5-04 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).

T

X

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.

P

Y

W

Example 8

N

L

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.

M

C

B

D

A

DE DUCTIVE GEOMETRY

203

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 204 Monday, September 12, 2005 3:45 PM

STAGE

5.3

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 ≡ ∆BXC ii ∠A = ∠B

C

A

b Now draw the interval BY, so that CY = AY (as shown). Prove: i ∆AYB ≡ ∆CYB ii ∠A = ∠C c Hence calculate the size of an angle in an equilateral triangle.

X

B

C

Y

A

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.

B

V

Y

W

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.

X

G

F

X

D

E

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.

X

W

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. A b Hence show ∠AXB = ∠AXC. c Hence prove AX ⊥ BC (show ∠AXB = ∠AXC = 90°).

204

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

T

Y B

X

C

05_NCM10EX2SB_TXT.fm Page 205 Monday, September 12, 2005 3:45 PM

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).

T

P

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?

X

F

R E

B

C

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

D L M

T

P

N

Working mathematically Communicating 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. DE DUCTIVE GEOMETRY

205

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 206 Monday, September 12, 2005 3:45 PM

STAGE

5.3

Worksheet 5-02 Naming quadrilaterals

Tests for quadrilaterals The 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.

C

D

B

A

Solution In 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)

Exercise 5-05 Example 9

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.

D

C

B

A

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.

P

N

M

L

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.

L M

F

206

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

H

G

05_NCM10EX2SB_TXT.fm Page 207 Monday, September 12, 2005 3:45 PM

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.

T

R

P

Q

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.

E

F

H D

C

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.

V

Y T

W

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.

X D

C

A

B

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

j

E

M

T

W

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

K

G

H

MNPT is a quadrilateral. The diagonals MP and TN are equal and bisect T 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.

P

X M

DE DUCTIVE GEOMETRY

N

207

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 208 Monday, September 12, 2005 3:45 PM

STAGE

5.3

2 a ABCD is a parallelogram and BX = DY. Prove: i ∆ABX ≡ ∆CDY ii AXCY is a parallelogram.

B

X

C

A

b AECD is a rhombus. AE = EB. Prove: i ∆CBE ≡ ∆DAE ii BCDE is a parallelogram.

Y

A

D

E

B

D

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

A

C

P

B

S

Q

D

C

R

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

A E B O D F C

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

T P O

S

Q

R U

f

DEFG is a rectangle. W, X, Y and Z are the midpoints of the sides. Prove: i WXYZ is a parallelogram ii WXYZ is a rhombus.

D

W

Z

G

208

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

E

X

Y

F

05_NCM10EX2SB_TXT.fm Page 209 Monday, September 12, 2005 3:45 PM

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.

Geometry 5-06 Islamic design

Working mathematically Reflecting: Islamic designs The 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 reflect 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. DE DUCTIVE GEOMETRY

209

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 210 Monday, September 12, 2005 3:45 PM

Construction This 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 figure 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?

Skillbank 5B SkillTest 5-02 Converting decimals and percentages to fractions

Converting decimals and percentages to fractions To convert a decimal to fraction form, first 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. 7

35 7 a 0.35 = --------- = -----100 20 20 (two decimal places, two 0s in the denominator) 4

8 4 b 0.8 = ------ = --10 5 5 (one decimal place, one 0 in the denominator) 16

c

64 16 0.64 = --------- = -----100 25 25 11

22 11 d 0.22 = --------- = -----50 50 100 2 Now convert these decimals to fraction form: a 0.75 b 0.28 c 0.3 f 0.85 g 0.32 h 0.49 k 0.72 l 0.65 m 0.2

210

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

d 0.14 i 0.56 n 0.24

e 0.06 j 0.9 o 0.53

05_NCM10EX2SB_TXT.fm Page 211 Monday, September 12, 2005 3:45 PM

3 To convert a percentage to a fraction, we write it over 100 and simplify. Examine these examples. 13

26 13 a 26% = --------- = -----100 50 50 2

c

8 2 8% = --------- = -----100 25 25

4 Now convert these percentages to fractions: a 76% b 10% c 80% f 56% g 75% h 31% k 60% l 54% m 6%

2

40 2 b 40% = --------- = --100 5 5 19

95 19 d 95% = --------- = -----100 20 20 d 45% i 68% n 49%

e 88% j 5% o 82%

Properties of similar figures

Skillsheet 5-06 Finding sides in similar triangles

Similar figures are the same shape, but not necessarily the same size. The symbol for ‘is similar to’ is |||. Enlarging or reducing a given figure will always produce a similar figure and the amount by which a figure is enlarged or reduced is called the scale factor or similarity factor.

image length Scale factor = ----------------------------------original length

If two figures are similar, then: • the matching angles are equal • the matching sides are in the same ratio. DE DUCTIVE GEOMETRY

211

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 212 Monday, September 12, 2005 3:45 PM

Example 10 If ∆ABC ||| ∆DEF, find the value of k. C

F

20 mm 25 mm A

B

40 mm

D

E

k mm

Solution k 25 ------ = -----40 20

(matching sides in the same ratio)

25 ∴ k = ------ × 40 20 = 50

Exercise 5-06 Geometry 5-07 Similar triangles

1 Are the figures similar in each of the following pairs? a

b

9 cm

4 cm

70°

5 cm 6 cm

8 cm

3 cm

12 cm 70° 6 cm

c

d 8 cm

10 cm

8 cm

5 cm

4 cm 6.4 cm 6 cm

e

f

3 cm

2 cm

4 cm

5 cm

10 cm

10 cm 4 cm

Example 10

2 Find the value of the pronumerals in the following pairs of similar figures: a

b

c

12

7 4

9

27

15

10

18 k

y

212

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

m 6

05_NCM10EX2SB_TXT.fm Page 213 Monday, September 12, 2005 3:45 PM

d

e

f

15

9

22

h

15 6

8

x

14

3

p

8

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

b

12

y 30°

45

6

9

30

30° 11

d

c

d

35

a

35°

8 35° 14

10 25

k

18

e

f 30 24

20

m

26

y

26

16

g

h

70° w

22

22 e

16

12 70°

7

15

4 Find the similar figures and determine the scale factor: a 18 cm 60° 6 cm B

10 cm A

20 mm

C 25 mm

24 mm

32 mm

B

20 cm

13 mm

A

D

60°

7.5 cm

10 mm

b

16 mm

60° 8 cm

6.25 cm 60° C 5 cm

D 48 mm DE DUCTIVE GEOMETRY

213

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 214 Monday, September 12, 2005 3:45 PM

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

A 8 9 B

C 6 E

m

D

V 10

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

W p X

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?

5

Y

8

1.83 m 1.2 m

14 m

8 State whether each of the following is true (T) or false (F). a The angles of similar figures are equal. b All circles are similar. c The sides of similar figures 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.

STAGE

5.3

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

Tests for similar triangles 1 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.

Worksheet 5-07 Congruent and similar triangle proofs

F

C

10

5

4

2 A

214

4

B D

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

8

E

Z

05_NCM10EX2SB_TXT.fm Page 215 Monday, September 12, 2005 3:45 PM

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. F 50

C 5 B

3

A

E 30

D

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. C

F

E

A B

D

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.

5

6

15 2

Example 11 Prove that the triangles in each of the following pairs are similar: a

N

b

E 6

9

9 D 12

C

R

50° M T

8

6

W

12 4

F 6

G

50° P

8

DE DUCTIVE GEOMETRY

B

215

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 216 Monday, September 12, 2005 3:45 PM

STAGE

5.3

Solution

a In ∆CDE and ∆FTG: ⎫ ED -------- = 6--- = 3--⎪ 4 2 GT ⎪ ⎪ CE -------- = 9--- = 3--⎬ FG 6 2 ⎪ ⎪ CD 12 3 -------- = ------ = --⎪ 8 2 FT ⎭

(ratios of matching sides)

ED CE CD ∴ -------- = -------- = -------GT FG FT ∴ ∆CDE ||| ∆FTG

(matching sides are in the same ratio or SSS)

b In ∆MWN and ∆BPR: MW ---------- = 6--- = 3--8 4 BP MN 9 3 --------- = ----- = --12 4 BR MW MN ∴ ---------- = --------BP BR ∠M = ∠B = 50° ∴ ∆MWN ||| ∆BPR

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

Example 12 Prove ∆ABE ||| ∆ACD.

D E

Solution

In ∆ABE and ∆ACD: ∠A is common ∠AEB = ∠ADC ∴ ∆ABE ||| ∠ACD

(corresponding angles, BE || CD) (two pairs of matching angles equal or AA)

A

B

C

Example 13 a Prove ∆MRP ||| ∆XRW.

P

12

b Find the value of k.

R k

Solution

a 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

k -----12

=

8 -----20

∴ k = 12 × = 4.8

216

(matching sides in similar triangles) 8 -----20

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

20 M

X

8 W

05_NCM10EX2SB_TXT.fm Page 217 Monday, September 12, 2005 3:45 PM

Exercise 5-07 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 12 60

27

Example 11

9 8

40

6

18

d

e

4.8

f

9

42°

8 9.6

10

16

12

75°

30

42° 63°

6

16.875

g

h

i

3

18

30° 30° 9.6

10.8

7.5

12

25

6

15 2.4

2 a Prove ∆XYW ||| ∆XLM.

b Prove ∆ABC ||| ∆EDC.

c Prove ∆PLM ||| ∆PTW.

A

W

X

D 6

10 C

M

L

M

7.5

8

E

W

B

Y

P

d Prove ∆TYX ||| ∆KGX. G

e Prove ∆GMK ||| ∆LHK.

18

f

L

T

Prove ∆WYZ ||| ∆ZYX. Z

G

12 T

Example 12

H

X 7.5 K

K W

X

Y

L

28.8 M Y

DE DUCTIVE GEOMETRY

217

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 218 Monday, September 12, 2005 3:45 PM

Example 13

STAGE

5.3

3 a

i Prove ∆GKF ||| ∆HKM. ii Find the value of m.

i Prove ∆CDE ||| ∆FGE. ii Find the value of k.

b

D

F

F

16 M

k

18

E

12

10

m G

6

H

C

K

8

c Prove ∆BCF ||| ∆DCE and, hence, find the value of x.

d Prove ∆AXM ||| ∆AYN and, hence, find the length of AX.

E

F

G

N M

C

10 x

8 D

B

e

15

9

6

X 8 Y

A

i Prove that ∆TFR ||| ∆TPD. ii If FR = 9 cm, DP = 5 cm and TD = 8 cm, find the length of TR.

f

ABCD is a parallelogram. i Prove ∆EFC ||| ∆EAB. ii Prove ∆EAB ||| ∆AFD. iii Prove ∆EFC ||| ∆AFD. iv If EF = 20 cm, FC = 8 cm and AF = 16 cm, find the length of AB.

F

E

D D T

P

F

C

R

A

B

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

W

P

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

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

T

E

F

218

N

G

H

05_NCM10EX2SB_TXT.fm Page 219 Monday, September 12, 2005 3:45 PM

Proving geometrical results General 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.

Worksheet 5-08 Geometrical proofs order activity Worksheet 5-09 Complete the proofs

P

B

A

O

Solution Let ∠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.

P

A

B

x° O

Example 15 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.)

D

C

G T

A

Solution

H

B

Congruent 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) A

Example 16 Prove that the interval joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.

D

B DE DUCTIVE GEOMETRY

E

C

219

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 220 Monday, September 12, 2005 3:45 PM

STAGE

5.3

Solution Similar triangles are used to prove this result. In ∆ADE and ∆ABC: AD -------- = AB

1 --2

AE -------- = AC

1 --2

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

AD AE ∴ -------- = -------AB AC ∠A is common. ∴ ∆ADE ||| ∆ABC ∴ ∠ADE = ∠ABC ∴ DE || BC DE AD Also -------- = -------BC AB AD But -------- = AB

1 --2

DE ∴ -------- = BC

1 --2

∴ DE =

1 --2

(two pairs of matching sides are in the same ratio and the included angles are equal or SAS) (matching angles of similar triangles) (corresponding angles proved equal) (matching sides of similar triangles)

× BC

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

Pythagoras’ theorem Pythagoras’ theorem can be stated as follows:

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

Working mathematically Applying strategies and reasoning: Proving Pythagoras There are many proofs of Pythagoras’ theorem and they are based on: • algebra • geometry • dissection of areas

220

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 221 Monday, September 12, 2005 3:45 PM

Geometry 5-08 Pythagorean dissection

1 Proof by dissection

a

A

A

b

c

b C

a

a

b B

C

B

Explain how this dissection proves Pythagoras’ theorem. 2 Find two other proofs of Pythagoras’ theorem by dissection of areas.

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

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.

Proof of the converse of Pythagoras’ theorem Given:

∆ABC where AC 2 = AB2 + BC 2

Aim:

To prove ∠ABC = 90°

Construction:

Draw ∆PQR where QR = BC, PQ = AB, and ∠PQR = 90°.

Proof:

A

C

B

P

R

Q

In ∆PQR, PR2 = PQ 2 + QR2 (Pythagoras’ theorem) = AB 2 + BC 2 (since PQ = AB, QR = BC) = AC 2 (given) 2 2 ∴ PR = AC ∴ 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°)

DE DUCTIVE GEOMETRY

221

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 222 Monday, September 12, 2005 3:45 PM

STAGE

5.3

Example 17 C

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

99 cm

20 cm

Solution

B

202 + 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 PQ b the area of PQTS

P

Q 10 cm

S

Solution a

A

101 cm

T

Let PQ = x cm ∴ PS = x cm ∴ x2 + x2 = 102 (by Pythagoras’ theorem) ∴ 2x2 = 100 x2 = 50 x = 50 =5 2 ∴ PQ = 5 2 cm

Area = PQ × PS = (5 2) × (5 2) = 50 cm2 or since PQTS is a rhombus (why?) b

Area = =

1 --2 1 --2

× SQ × PT × 10 × 10

= 50 cm2

Exercise 5-08 Geometry Example 14

(Using dynamic geometry software, first confirm 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).

C

A

222

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

B

D

05_NCM10EX2SB_TXT.fm Page 223 Monday, September 12, 2005 3:45 PM

Example 15

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.

X C

A

B

Y Example 16

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.

P

Q R

M N

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 = 90º and CD ⊥ AB. Show ∆ADC ||| ∆ACB and, hence, that AC 2 = AB × AD (using the ratio of matching sides). b Show ∆BDC ||| ∆BCA and, hence, that BC 2 = AB × DB. c Hence, show that AC 2 + BC 2 = AB 2.

C

A

D

5 DEF is a triangle and DG ⊥ EF. Prove, using Pythagoras’ theorem, that: DE 2 + GF 2 = DF 2 + EG 2.

D

E

6 Determine whether these are right-angled triangles: 3_ a b 135 I

H

G

F

c

8

X

B

D

Y 2

153

71 1_ 2

5_ 8

F

2

J 2 Z E

DE DUCTIVE GEOMETRY

223

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 224 Monday, September 12, 2005 3:45 PM

STAGE

5.3

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, BC 2, CD2 and AD2. b Hence, show AB2 + CD2 = BC 2 + AD2.

A D X

C B

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

3

D 4

E 5

C

m

A

10

B

Power plus 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. C 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

B

224

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 225 Monday, September 12, 2005 3:45 PM

Language of maths AA argument congruence test deductive hence kite opposite perpendicular ratio rhombus SAS similarity factor test

AAS axis of symmetry congruent (≡) diagonal hypotenuse matching order polygon reason RHS scale factor similarity test trapezium

acute-angled bisect converse equilateral interior angle midpoint parallel Pythagoras’ theorem rectangle right-angled scalene square triangle

angle sum concurrent convex exterior angle isosceles obtuse-angled parallelogram quadrilateral regular polygon rotational symmetry similar (|||) SSS vertices

Worksheet 5-10 Geometry crossword

1 What is the formal definition 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 figures are in the same _______. 6 Are all squares similar? 7 What does AA mean? 8 Write the converse of Pythagoras’ theorem.

Topic overview

Worksheet 5-11 Geometry summary poster

• Copy and complete the table below. The best part of this chapter was… The worst part was…

New work…

?

I need help with…

DE DUCTIVE GEOMETRY

225

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 226 Monday, September 12, 2005 3:45 PM

• 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.

Properties of triangles and quadrilaterals

Convex polygons • angle sum = 180° or 360° or 540° or … • regular polygons …

DEDUCTIVE GEOMETRY

Similarity

Congruence

Scale

SSS

SAS

Tests for congruent triangles AAS

RHS

b

Tests for similar triangles: SSS SAS AA RHS

Geometrical proofs using congruence/similarity c

226

factor

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

a

c 2 = a2 + b2

05_NCM10EX2SB_TXT.fm Page 227 Monday, September 12, 2005 3:45 PM

Chapter 5

Topic test Chapter 5

Review

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

a

E F c°

b

c 36°

4w° B

110°

Ex 5-01

D

y°

114°

A

w°

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

Ex 5-01

T

P

R

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

M D

A

B

C

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?

Ex 5-02

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

Ex 5-03

a

b 8 5 5

78° 60° 8 cm 78°

8

c

8 cm 60°

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

b

Ex 5-03

c

DE DUCTIVE GEOMETRY

227

CHAPTER 5

05_NCM10EX2SB_TXT.fm Page 228 Monday, September 12, 2005 3:45 PM

Ex 5-03

6 a KMNP is a square and ∆XMN is equilateral. Prove: i ∆KMX ≡ ∆PNX ii ∆KPX is isosceles.

P

X

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. Ex 5-04

Ex 5-05

Ex 5-06

N

K

M

G

F Y D

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.

E

D

A

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 α + φ + β + θ = 180°. b Prove φ = 180 − (α + β + φ) using ∆ABD and, hence, prove that θ = φ. c Hence, show α = β and that AB = BC. d Now prove ABCD is a rhombus.

X D α φ

C

Y

B C

α

θ

φ

β

A

θ

β

B

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

b

9 cm

4 mm

7 cm

d mm

6 mm 9 mm

k cm

10 cm

c

d 7

11 6

10 y

3 m

6

228

NE W CE NT UR Y M AT H S 10: S T AGE S 5.2/ 5.3

05_NCM10EX2SB_TXT.fm Page 229 Monday, September 12, 2005 3:45 PM

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

Ex 5-06

A 12 E

y D 5

B 15

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

7 cm

A

C Ex 5-06

9 cm

E

D

5 cm B

d cm

C

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

Ex 5-07

C

D 8

3.2

B

20

A

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

Y

Ex 5-07

D C

A W

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

M

J

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

B

X L

Ex 5-08

K N

M A C B

P

DE DUCTIVE GEOMETRY

Q

229

CHAPTER 5