Download pdf - CONGRUENCE AND SIMILARITY - Amazon S3 · CONGRUENCE AND SIMILARITY ... Two triangles are congruent if Two Angles and a Corresponding Sides are equal . ... 1.2 CONGRUENCE AND TRANSFORMATION

CONGRUENCE AND SIMILARITY

The figures that have the same size and the same shape, i.e. one

shape fits exactly onto other is called Congruent figures .

CONGRUENT TRIANGLES:

1. Two triangles are congruent if they have the same size and the same shape.

2. If two triangles are congruent, then

a) Their corresponding angles are equal and

b) Their corresponding sides are equal.

3. If Δ ABC = Δ PQR, then

A P

B C Q R

1.1CONGRUENT FIGURES

A = P AB = PQ

B = Q BC = QR

C = R AC = PR

TIPS FOR STUDENTS: 1. The symbol ‘ ≡ ’ means ‘ is congruent to ’ 2. The matching diagram shows how the corresponding vertices match . Δ ABC ≡ Δ PQR, The corresponding vertices must be named in the correct order. e.g. Δ BCA ≡ Δ QRP and Δ ACB ≡ Δ PRQ

TEST OF CONGRUENCY BETWEEN TWO TRIANGLES:

TEST

DESCRIPTION

SSS

Two triangles are congruent if all Three Corresponding Sides are equal .

A P B C Q R This is known as the SSS rule. ( side, side, side )

If AB = PQ

BC = QR

AC = PR

Δ ABC = Δ PQR .

TEST

DESCRIPTION

SAS

Two triangles are congruent if Two Corresponding Sides and the Included angle are equal .

A P B C Q R This is known as the SAS rule. ( side, angle, side )

TEST

DESCRIPTION

AAS

Two triangles are congruent if Two Angles and a Corresponding Sides are equal . A P

B C Q R This is known as the AAS rule . ( angle, angle, side )

If AB = PQ

AC = PR and

A = P then

Δ ABC = Δ PQR .

If A = P and

B = Q then

BC = QR

Δ ABC = Δ PQR .

TEST

DESCRIPTION

ASA

Two triangles are congruent if Two angles and the Included Sides are equal .

A P B C Q R This is known as the ASA rule. ( angle, side, angle )

If A = P and

B = Q then

AB = PQ

Δ ABC = Δ PQR .

EXAMPLE1:

In the Δ ABE ≡ Δ CDE . BAE = 900, AED = 600, BC = 3 cm and DE = 18 cm.

FIND A

a) The length of AE ,

C

b) Δ CDE . B E

3 600

18

D

TEST DESCRIPTION

RHS

Two triangles are congruent if both triangles have a Right Angle, equal Hypotenuse and another Side which is equal. A P B C Q R This is known as the RHS rule. ( right angle, hypotenuse, side )

If C = R

AB = PQ and

AC = PR then

Δ ABC = Δ PQR .

SOLUTION :

A

a) Δ ABE = Δ CDE ( Given )

BE = DE = 18 cm C 600

B E

3 15 300

CE = 18 - 3 cm = 15 cm

AE = CE = 15 cm 18

D

b) DCE = BAE = 900

CED = 600 ÷ 2

= 300

CED = 1800 - 900 – 300 ( sum of Δ )

EXAMPLE2:

In the quadrilateral ABCD, BE = CE and AE = DE .

Corresponding sides are equal .



AEB = CED sinceCorresponding sides are equal .

a) Prove that triangle AEB is congruent to B C

triangle DEC .

b) Name a triangle that is congruent to

triangle ABD .

c) Name a triangle that is congruent to

triangle ABC . A D

SOLUTION:

a) BE = CE

Given

AE = DE

AEB = DEC ( vert. opp. s )

Δ AEB = Δ DEC ( SAS rule )

b) Δ DEC

E

BD = CA

BDA = CAD ( Base s of isos . Δ AEB )

AD ( common sides )

Δ AEB = Δ DEC ( SAS rule )

c) Δ DCB

1. Under a transformation, an object is formed onto its image .

2. A figure and its image are congruent under a translation, a rotation and a reflection .

Translation Rotation Reflection

Two figures are similar if

AC = DB

ACB = DBC ( Base s of isos . Δ EBC )

BC ( common sides )

Δ ABC = Δ DCB ( SAS rule )

1.2 CONGRUENCE AND TRANSFORMATION

1.3 SIMILAR FIGURES

a)The corresponding angles are equal and

b)The corresponding sides are in the same ratio .

Two similar figures have the same shape but not necessarily the same size.

TIP FOR STUDENTS:

When two figures are congruent, they are also similar . However the converse is not true.

SIMILAR TRIANGLES:

1. Two triangles are similar if


b) Their corresponding sides are in the same ratio.

2. If Δ ABC is similar to Δ PQR, then

A = P

B = Q and ��

�� =

��

�� =

��

��

C = R

P

A

B C Q R

TEST FOR SIMILARITY BETWEEN TWO TRIANGLES:

One of the following conditions is sufficient for two triangles to be similar.

1. Two triangles are similar if two of their corresponding angles are equal.

A P

B C

Q R

If A = P and

B = Q then

Δ ABC is similar to Δ PQR .

2. Two triangles are similar if all three corresponding sides are proportional in

the same ratio .

P

A

kd kf

d f

B e C Q ke R

3.Two triangles are similar if two of their corresponding sides are proportional and

the included angle is equal.

If ��

�� =

��

�� =

��

�� = k

Where k is a constant, then


If ��

�� =

��

�� = k, and

B = Q, then


P

A

kd

d

B e C Q ke R

EXAMPLE:1

In the diagram, AB is parallel to DE, AB = 7 cm, BC = 4 cm, CD = 8 cm and CE = 16 cm .

a) Show that triangle ABC is similar to triangle EDC.

b) CALCULATE A 7 B

4

i) The length of AC, C

8 16

ii) The length of DE.

D E

SOLUTION:

a) ABC = EDC (alt. s, AB // DE )

BAC = DEC (alt. s, AB // DE )

Two triangles are similar if two of

their corresponding angles are equal.

.

Δ ABC is similar to Δ EDC.

b) i)Since Δ ABC is similar to Δ EDC .

��

�� =

��

��

��

�� =

�

�

AC = 8

4 X 16 2

= 8 cm

ii) ��

�� =

��

��

��

� =

�

�

DE = 4

82 X 7

= 14 cm

EXAMPLE2:

The diagram shows triangle ACD. BE is parallel to CD, AE = 9 cm, DE = 3 cm, BC = 2 cm

and CD = 6 cm .

Corresponding sides of similar

triangles are equal.


triangles are proportional.

CALCULATE A

a) BE , 9

b) AB .

E

B 3

D

C 6

SOLUTION:

a) Δ ABE is similar to Δ ACD .

��

�� =

��

��

��

� =

�

��

BE = �

�� X 6

= 4 �

� cm

ABE = ACD (corr. s BE // CD )

AEB = ADC (corr. s BE // CD )

Δ ABE is similar to Δ ACD



b) ��

�� =

��

��

��

�� =

�

��

4 AB = 3 ( AB + 2 )

4 AB = 3 AB + 6

AB = 6 cm

EXAMPLE3:

In the diagram triangle ABC is a right – angle. D lies on AB and CD is perpendicular

to AB = 5 cm, AC = 4 cm and BC = 3cm .

a) Show that triangle ABC is similar C

to triangle CBD .

4 3

b) Find The length of BD.

A D B

5



Cross - multiply

SOLUTION:

a) C C

4 3 3

A 5 B D B

b) Since Δ ABC is similar to Δ CBD.

��

�� =

��

��

��

� =

�

�

BD = �

� X 3

= 1.8 cm

1. An enlargement is a transformation that charges the size of a figure (enlarged /

reduced) without changing its shape .


triangles are equal.

1.4 SIMILARITY AND ENLARGEMENT

2. An enlargement is determined by the centre of enlargement and a scale factor.

3. The scale factor of an enlargement is the common ratio between pairs of

corresponding sides of the image and the original figure .

Scale Factor = ��

��

4.In the diagram, Δ ABC is mapped onto A' B' C' by an enlargement with centre O and

scale factor 2 .

A’

A

B

B’

O

C C’

��

�� =

��

�� =

��

�� = 2 or

��

�� =

��

�� =

��

�� = 2

5. In the diagram, Δ ABC is mapped onto AB' C' by an enlargement with centre A and

scale factor �

� . C

C’

A B’ B

6. In general, if Δ A' B' C' is an enlargement of Δ ABC with centre O, then

A'

A

O C C '

B

B '

SCALE FACTOR

= ��

�� ( or

��

�� =

��

�� ) or

= ��

�� ( or

��

�� or

��

�� )

7. A figure and its image under an enlargement are similar.

TIPS FOR STUDENTS:

1. If the scale factor is greater than 1, then the image is enlarged. 2. If the scale factor is between 0 and 1, then the image is reduced. 3. If the scale factor is 1, then the image is congruent to the original figure.

EXAMPLE:

In the diagram, Δ ABC is mapped onto AB' C' by an enlatgement with centre A and scale

factor k . AB = 4 cm, BB' = 6 cm, BC = 3.5 cm, and CC' = 9 cm .

FIND C '

9

a) The value of k ,

C

b) The length of B ' C ' ,

A

c) The length of AC . 4

B

6

B '

SOLUTION:

3 . 5

a)Scale Factor, k = ��

��

= � � �

�

= ��

�

= 2 . 5

b) ��

�� = Scale Factor

��

�.� = 2.5

B'C' = 2.5 X 3.5

= 8 . 75 cm

c) ��

�� = Scale Factor

��

�� = 2.5

AC + 9 = 2.5 AC

1.5 AC = 9

AC = �

� .�

= 6 cm

If two figures are similar, then the ratio of their areas is equal to the square of the

ratio of the lengths of any pair of corresponding sides .

Multiply both sides by AC.

1.5 AREAS OF SIMILAR FIGURES

If A1 and A2 denoted the areas of the two similar figures and l1 and l2 denoted their

corresponding lengths, then

EXAMPLE:

In the diagram, BC is parallel to DE, AC = 5 cm and CE = 3 cm, find the value of

a) ��

�� , E

3

b) ��

�� , C

5

c) ��

�� .

A B D

SOLUTION:

Δ ABC is similar to Δ ADE

2

1

A

A

=

2

2

1

l

l

ABC = ADE (corr. s BC // DE)

ACB = AED (corr. s BC // DE)

a) ��

�� =

��

��

= �

� � �

= �

�

b) ��

�� =

2

AE

AC

= 2

8

5

=64

25

c) ��

�� =

��

��

= ��

��

= ��

��

EXAMPLE:

In the diagram, QR is parallel to ST and

��

�� =

��

��

a) Find the value of ��

�� , P

b) Given that the area of triangle PQR is 12 cm2.

Find the area of the quadrilateral QRTS. Q R

c) What is the special name given to S T



2

1

A

A

=

2

2

1

l

l

quadrilateral QRTS ?

SOLUTION:

Δ PQR is similar to Δ PST

a) ��

�� =

��

��

2

PS

PQ =

��

��

��

�� =

49

16

=

�

�

b) ��

�� =

��

��

��

�� =

��

��

Area of Δ PST = ��

�� X 12

= 36.75 cm2

Area of Quadrilateral QRTS = Area of Δ PST - Area of Δ PQR

= 36.75 - 12

= 24.75 cm2

c) Quadrilateral QRTS is a trapezium.

PQR = PST ( corr. s QR // ST )

PRQ = PTS ( corr. s QR // ST )

Take positive square root on both sides.

A trapezium is a quadrilateral which

has a pair of parallel opposite sides.

EXAMPLE:

In the diagram, ABCD and EFBGH are straight lines . AE, GC and HD are parallel, BG = 6

cm . BF = 2 cm, GC = 8 cm and HD = 13 �

� cm .

a) Calculate the length of GH.

b) Write down the numerical value of

i) ��

�� ii)

��

��

A E

F

B 2

6

G C

8

H 3

113 D

SOLUTION:

Δ BGC is similar to Δ BHD

��

�� =

��

��

Corresponding sides of similar triangle

are proportional.

��

� =

83

113

X 6

GH + 6 = 10

GH = 4 cm

b) i) ��

�� =

2

HD

GC

=

2

3

113

8

= 2

5

3

64

25

= 25

9

ii) F

2

B

10

H D

2

1

A

A

=

2

2

1

l

l

h

��

�� =

XHBXh

XBFXh

2

12

1

= ��

��

= �

��

= �

�

EXAMPLE:

a) The surface areas of two similar figures are in the ratio 16 : 81 . Given that the

length of the smaller figure is 18 cm, find the length of the corresponding side of

the larger figure .

b) Two statues are geometrically similar and one 3 �

� times as tall as the other.

i) Given that the length of the smaller statue is 12 cm, find the length of foot of

the larger statue .

ii) The surface area of the larger statue is 98 cm2. Calculate the surface area of

the smaller statue .

SOLUTION:

a)

2arg

eallerFigurLengthOfSm

erFigureLengthOfL =

��

��

Area of Δ

= �

� X Base X Height

= �

� X b X h

b

h

2

1

A

A

=

2

2

1

l

l

��

�� =

16

81

Length of Larger Figure = �

� X 18

= 40.5 cm

b) i) ��

�� =

�

�

��

�� =

�

�

Length of Foot of Larger Statue = �

� X 12

= 42 cm

ii) ��

�� =

2

7

2

��

�� =

2

7

2

Surface Area of Smaller Statue = 2

7

2

X 98

= 8 cm2


The ratio of the volumes of two similar solids is equal to the cube of the ratio of any

two corresponding length of the two solids .

If V1 and V2 denoted the volume of the two similar solids and l1 and l2 denoted their

corresponding lengths, then

EXAMPLE:

The two cones are geometrically similar. The height of the smaller cone is 9 cm

while the height of the larger cone is 12 cm.

a) Find the ratio of the curved surface area

of the smaller cone to that of the Larger

Cone.

b) Given that the volume of the smaller cone is 189 cm2,

find the volume of the larger cone .

SOLUTION:

1.6VOLUME OF SIMILAR SOLIDES:

2

1

V

V

=

3

2

1

l

l

9

12

a) ��

�� =

2

4

3

12

9

= �

��

b) ��

�� =

3

9

12

��

�� =

3

9

12

Volume of Larger Cone = 3

9

12

X 189

= 448 cm3

EXAMPLE:

The surface areas of two similar spheres are in the ratio 4:25.

a) The radius of the larger sphere is 10 cm. Find the radius of the smaller sphere.

b) If the mass of the smaller sphere is 1.6 kg, find the mass of the larger sphere.

SOLUTION:

a) ��

�� =

�

�� ( Given )

��

�� =

�

��

2

1

A

A

=

2

2

1

l

l

2

1

V

V

=

3

2

1

l

l

2

1

A

A

=

2

2

1

l

l

2

arg

erSphereRadiusOfL

eallerSpherRadiusOFSm =

25

4

��

�� =

�

�

Radius Of Smaller Sphere = �

� X 10

= 4 cm

b) ��

�� =

��

��

��

�.� =

3

2

5

Mass of Larger Sphere = 3

2

5

X 189

= 25 kg

TIPS FOR STUDENTS:

The mass of the sphere is proportional to its volume.

EXAMPLE:

A shop sells bottles of pasta sauce in two sizes as shown below. The two bottles are


2

1

V

V

=

3

2

1

l

l

geometrically similar. The base area of the larger bottle is 75 cm2 and the base area

of the smaller bottle is 48 cm2

a) CALCULATE

i) The height of the smaller bottle if the height of the larger bottle is 15 cm .

ii) The ratio of the volume of pasts in the smaller bottle to the volume of pasta in

the larger bottle .

b) The smaller bottle costs $ 9.60 . Calculate the cost of the larger bottle, given that a

discount of 20 % is given for buying the larger bottle .

a) i) ��

�� =

25

16

75

48

2

arg

erBottleHeightOfL

eallerBottlHeightOfSm =

��

��

��

�� =

25

16

= �

�

Height of Smaller Bottle = �

� X 15

= 12 cm

ii) ��

�� =

3

5

4

= ��

��

The Required Ratio is 64 : 125

b) ��

�� =

��

��

��

$ �.�� =

��

��

Cost of Larger Bottle = ��

�� X $ 9.60

= $ 18 .75


2

1

A

A

=

2

2

1

l

l

2

1

V

V

=

3

2

1

l

l

Cost of Larger Bottle after Discount

= ��

�� X $ 18.75

= $ 15

TIPS FOR STUDENTS:

The cost of the bottle of pasta is proportional to its volume.

PROBLEM1:

In the diagram, PTQ = RSQ and QT = QS. Prove that triangle PQT is Congruent to

triangle RQS.

Q

S T

U

P R

SOLUTION:

Q

S T

U

P R

PTQ = RSQ (Given)

PQT = RQS (Q is Common)

QT = QS (Given )

Δ PQT = Δ RQS (ASA Rule )

PROBLEM2:

In the diagram, AB = 13.5 cm, AE = 22.5 cm and FE = 9 cm.

a) Write down a triangle that is congruent to triangle ABE, giving reasons for your answer .

b) Find the length of BC.

A

B F

D

C E

SOLUTION:

A

13.5 (13.5) 22.5

B F

9

D

C E

a) AF = 22.5 - 9

= 13.5 cm

= AB

ABE = AFC = 900 ( Given )

BAE = FAC (A is Common )

Δ ABE = Δ AFC ( ASA Rule )

b) AC = AE

= 22.5 - 13.5 cm

= 9 cm

PROBLEM3:

Are triangles ABC and PQR similar ? Explain your answer.

a)

C P

R

350 750 700

A B 750

Q

b)

C R

3.6 4.2

6.6 6.3

A 4.8 B

P 7.2 Q

c) P

A 5.6

B 3.5 C Q 7 R

SOLUTION:

a) YES .Δ ABC is similar to Δ PQR.

C P

700

R

350 750 700

A B 750

Q

250

500

ABC = 800 - 350 - 750 ( sum of Δ)

= 700

Since B = Q = 750 and C = R = 700

Δ ABC is similar and Δ PQR.

b) NO. Δ ABC is not similar to Δ PQR.

��

�� =

�.�

�.� = 1.5

��

�� =

�.�

�.� = 1.5

��

�� =

�.�

�.� = 1.83

Since ��

�� ≠

��

�� , Δ ABC is not similar to Δ PQR.

c) NO . Δ ABC is not similar to Δ PQR.

B ≠ Q

PROBLEM4:

Find the value of p if triangles ABC and XYZ are similar.

a)

C X

3.6 300

300 800

A 6.8 B 700 Z

Q P

b)

A Z Y

4.6 3.8

P 6.9

B 3.4 C

X

c)

Z 13 Y

A

4.6 7.2 P

B 8 C X

SOLUTION:

a) Since Δ ABC is similar to Δ XYZ ,

��

�� =

��

��

�

�.� =

�.�

�.�

P = �.�

�.� X 3.6

= 4.5 cm

b) Since Δ ABC is similar to Δ XYZ ,

��

�� =

��

��

�

�.� =

�.�

�.�

P = �.�

�.� X 3.8

= 5.7 cm

c) Since Δ ABC is similar to Δ XYZ ,

��

�� =

��

��

�

�.� =

��

�

P = ��

� X 7.2

= 11.7 cm







PROBLEM5:

a) b)

P D

4 A y

Q R 1.4

3 x E 4.5

8 1.2

B 3.5

S x T C

c) d)

E P

6

C S

9 4

T

A 3.6 B 2.2 D y 5

Q R

e) f)

A T

x 5 13 R

B C 7.5

4

D E P 6 Q x S

1.8

5

g) h)

A y B E

2.8 4 5

C D C 11

x

x 8

A 7.5 B y D

E

i) j)

A 4 F 3 B S R

10.8 x

T

8 8

16.2 16.8

P 18.6 Q

C 7 D

SOLUTION:

a) Δ PQR is similar to Δ PST.

5.4

5.5

x

E

4.9

�

� =

��

�

x = ��

� X 3

= 9 cm

b) Δ ABE is similar to Δ CDE.

�

�.� =

�.�

�.�

x = ��

�.� X 4.5

= 1.8 cm

c) Δ ABC is similar to Δ ADE.

�

�.� =

�.� � �.�

�.�

�

�.� =

�.�

�.�

x = �.�

�.� X 41.8

= 2.9 cm

d) Δ PQT is similar to Δ RST.

�

� =

�

�

x = �

� X 9

= 7.5 cm



�

� =

�

�

x = �

� X 4

= 74.8 cm

e) Δ ABC is similar to Δ ADE.

�

� � � =

�

��

13 x = 5 ( x + 4 )

13 x = 5 x + 20

8 x = 20

x = 2.5 cm

f) Δ PQR is similar to Δ PST .

� � �

� =

�.�

�

� � �

� = 1.5

x + 6 = 1.5 X 6

x + 6 = 9

x = 3 cm

g) Δ ECD is similar to Δ EAB .

�

� � �� =

�

� � �

�

� � �� =

�

��

Cross multiply

.

Cross multiply

.

3 x = 2 ( x + 2.8 )

3 x = 5 x + 5.6

x = 5.6 cm

�

�.� =

��

�

y = ��

� X 5.4

= 8.1 cm

h) Δ AEF is similar to Δ DEC .

�

� � � =

�.�

��

�

� � � =

�

�

2 x = x + 5

x = 5 cm

� � �.�

�.� =

��

�.�

y + 7.5 = 2 X 7.5

y + 7.5 = 15

y = 7.5 cm

i) Δ AEF is similar to Δ DEC .

�

�.� =

�

�

x = �

� X 4.9 cm

= 2.8 cm

Cross multiply

.

j)

S R

10.8 x

T

16.2 16.8

P 18.6 Q

PTQ = STR ( Vert. opp. s )

QPR = QSR ( s in the same segment )

QPT = RST

Δ PTQ is similar to Δ STR.

�

��.� =

��.�

��.�

x = ��.�

��.� X 18.6 cm

= 12.4 cm

PROBLEM6:

In the diagram, BAC = DEC, AB = 8 cm, BC = 12 cm, CD = 18 cm and CE = 15 cm.

a) Give reason why triangle ABC and EDC are similar.

b) Calculate AC and DE.

E

15

A

8 C

12 18

B D

SOLUTION:

a) ACB = ECD ( vert . opp . s )

BAC = DEC ( Given)

Δ ABC is similar to ΔEDC.

b) ��

�� =

��

��

AC = ��

�� X 15

= 10 cm

��

� =

��

��

DE = ��

�� X 8

= 12 cm



PROBLEM7:

In the diagram, Δ ABC is similar to ΔPQR. Given that AB = 6 cm, BC = 8 cm, PQ = ( 21 –

x ) cm and QR = ( 21 + x ) cm, from an equation in x and solve it .

P

A 21 – x

6

B 8 C Q 21 + x R

SOLUTION:

Since Δ ABC is similar to Δ PQR.

��

�� =

��

��

��

� =

�� − �

�

8 (21 - x ) = 6(21 + x )

168 – 8 x = 126 + 6 x

42 = 14 x

3 = x

x = 3

PROBLEM8:

In the diagram, PWQ, PXR, WYR, XYZ and QZR are straight lines . WX is parallel to QR

Cross – Multiply

and PQ is parallel to XZ . QZ = 3 cm, ZR = 5 cm and WQ = 4 cm.

a) Name two triangles, which are similar to triangle RWQ .

b) Use similar triangles to calculate

i) YZ ,

ii) PW.

P

W X

4

Q 3 Z 5 R

SOLUTION: P

W X

4

Q 3 Z 5 R

8

Y

a) Δ RYZ is similar to Δ RWQ .

Δ RYZ is similar to Δ WYX .

Δ RYZ and Δ WYX are similar to Δ RWQ.

b) i) Δ RYZ is similar to Δ RWQ .

��

= �

�

YZ = �

� X 4

= 2.5 cm

ii) WX = QZ = 3 cm since WQZX is a parallelogram.

Δ PWX is similar to Δ PQR .

��

�� =

�

�

8 PW = 3 ( PW + 4 )

8 PW = 3 PW + 12

5 PW = 12

PW = 2.4 cm

RYZ = RWQ ( Corr . s, YZ // WQ )

RZY = RQW ( Corr . s, YZ // WQ )

WXY = RZY (Alt . s, WX // ZR )

WYX = RYZ (Ver. opp s )

Cross – Multiply

PROBLEM9:

In the diagram, PXQ, PYW, PZR, XYZ and QWR are straight lines. XZ is parallel to QR

and PQ is parallel to ZW. XY = 3 cm, YZ = 4 cm and WZ = 8 cm.

a) Show that triangle XYP is similar to triangle ZYW.

b) Calculate PX.

c) Show that triangle RQP is similar to triangle RWZ.

d) Calculate RW.

P

X 3 Y 4 Z

8

Q W R

SOLUTION:

P

(6)

X 3 Y 4 Z

8 8

Q 7 W R

a) XYP = ZYW ( vert . opp . s )

PXY = WZY (Alt. s PX // ZE )

Δ XYP is similar to Δ ZYW. (Shown)

b) ��

= �

�

PX = �

� X 8

= 6 cm

c) RQP = RWZ ( Corr . s, PQ // ZW )

RPQ = RZW ( Corr . s, PQ // ZW )

Δ RQP is similar to Δ RWZ . ( Shown )

d) QW = XZ

= 3 + 4

Opposite Sides of the parallelogram

XQWZ are equal.

= 7 cm

XQ = ZW = 8 cm

PQ = PX + XQ

= 6 + 8

= 14 cm

Since Δ RQP is similar to Δ RWZ .

��

�� =

�

��

7 RW = 4 ( RW + 7 )

7 RW = 4 RW + 28

3 RW = 28

RW = ��

� cm

PROBLEM10:

In the diagram, quadrilateral PQRS is mapped onto quadrilateral P'Q'R'S' by an

enlargement with centre O and scale factor k . OQ' = 6 cm, Q'Q = 9 cm ,QR = 8 cm and

P' S' = 3 cm.

FIND

a) Find the value of k,

b) The length of Q'R',

c) The length of PS.

Opposite Sides of the parallelogram

XQWZ are equal.

Cross – Multiply

S

P

S'

P'

O Q' Q

R'

R

SOLUTION:

S

P

3 S'

P'

O 6 Q' 9 Q

R' 8

R

a)Scale factor, k = ��

��

= �

� � �

= �

��

= �

�

b) ��

�� = Scale factor

��

�� =

�

�

�� = �

� X 8

= 3.2 cm

c) ��


�

�� =

�

�

��

� =

�

� X 3

PS = �

� X 3

= 7.5 cm

PROBLEM11:

In the diagram, pentagon PQRS is mapped onto pentagon P' Q' R' S' by an

enlargement with centre O and scale factor 3 .given that PQ = 60 cm and O S' = 450

cm .

Take reciprocals on

both sides.

P'

FIND T'

a) The length of P' Q',

b) The length of OS . Q' P

Q T

O

S

R S'

R'

SOLUTION:

a) ��


��

�� = 3

�� = 3 X 60

= 180 cm

b) ��


��

�� = 3

��

� =

�

�

OS = �

� X 450

= 150 cm

Take reciprocals on

both sides .

PROBLEM12:

In the diagram, figure ABCDE is mapped onto figure AB' C' D' E' by an

enlargement with centre A . ADE is quadrant of a circle, centre D . BC = 4.75 cm, B'C' =

9.5 and D' E' = 21cm .

a) Find the scale factor of the enlargement.

b) Take � to be , ��

� find the length of the arc AE .

c) Given that the area of figure ABCD is 110 cm2, find the area of figure AB' C' D' E'.

E'

E 21

C'

C D'

D

4.75

A B B'

SOLUTION:

E'

E 21

10.5 C'

C D'

D 9.5

4.75

A B B'

a) Scale factor = ��

��

= �.�

�.��

= 2

b) ��


��

�� = 2

��

�� =

�

�

DE = �

� X 21

= 10.5 cm

Length of arc

= �

� X 2 X

��

� X 10.5

= 16.5 cm

Take reciprocals on

both sides.

Circumference of circle = 2 �r

c)

TIP FOR STUDENTS: A figure and its image under an enlargement are similar .

Figure ABCDE is similar to figure AB' C' D' E'.

��

�� =

2

75.4

5.9

��

�� =

2

1

2

Area of AB'C'D'E' = 4 X 110

= 440 cm2

PROBLEM13:

In the diagram, ABCD is quadrilateral. The diagram AC and BD intersect at E . AD =

15 cm, AE = 10 cm, BC = 7.5 cm and ABE =DCE . AD is parallel to BC and triangle

AED is an isosceles triangle.

a) Show that triangle ABE is congruent to triangle DCE.

2

1

A

A

=

2

2

1

l

l

b) Name a triangle that is similar to Δ AED.

c) Show that triangle ACD is isosceles triangle.

d) Write down the numerical value of

i) ��

�� ,

ii) ��

�� .

A 15 D

10

E

B 7.5 C

SOLUTION:

A 15 D

10

E

(5)

B 7.5 C

a) ABE = DCE ( Given )

AE = DE ( Sides of isos . s )

ABE = DCE ( Vert . opp . s )

Δ ABE = Δ DCE ( AAS rule )

b) Δ ABE is similar to Δ DCE

(Δ BEC is also similar to Δ AED since Δ AED is an isosceles triangle .

BEC is also an isosceles triangle .)

c) ��

�� =

��

��

��

�� =

�.�

��

CE = �.�

�� X 10

= 5 cm

AC = AE + CE

= 10 + 5

= 15 cm

= AD

AED = CEB ( Vert . opp . s )

EAD = ECB ( Alt . s AD // BC )



Δ ACD is an isosceles triangle.

PROBLEM14:

PSU is a triangle. QR is parallel to SU and SU and PS is parallel to RT. Give that ��

�� =

�

� and the area of triangle PQR is 8 cm2 .

CALCULATE

a) The value of ��

�� ,

b) The area of trapezium QSUR,

c) The area of parallelogram QSTR.

P

Q R

X

S T U

SOLUTION:

P

Q R

2 parts

S 2 parts T 3 parts U

a) Since Δ PQR is similar to Δ PSU,

��

�� =

��

��

= �

� � �

= �

�

b) ��

�� =

2

QR

SU

��

� =

2

2

5

ST = QR = 2 parts since QSTR is a

parallelogram .

2

1

A

A

=

2

2

1

l

l

Area of Δ PSU = 2

2

5

X 8

= 50 cm2

Area of trapezium QSUR

= Area of Δ PSU - Area of Δ PQR

= 50 - 8

= 42 cm2

c) ��

�� =

2

2

3

��

� =

�

�

Area of Δ RTU = �

�

X 8

= 18 cm2

Area of parallelogram QSTR

= Area of trapezium QSUR - Area of Δ RTU

= 42 - 18

= 24 cm2

PROBLEM15:

The diagram shows two geometrically similar solid cylinders. the height of Cylinder

A is 6 cm and the height of Cylinder B is 24 cm.

a) If the base area of Cylinder B is 80 cm2, calculate the volume of Cylinder A.

2

1

A

A

=

2

2

1

l

l

b) If the mass of Cylinder A is 0.18 kg, find the density of the material used to make

solid, giving your answer in g/cm3 .

24

6

Cylinder A Cylinder B

SOLUTION:

a) ��

�� =

2

24

6

��

�� =

2

4

1

2

1

A

A

=

2

2

1

l

l

Base Area of Cylinder A = 2

4

1

X 80

= 5 cm2

Volume of Cylinder A

= Base Area X Height

= 5 X 6

= 30 cm2

b) Density = ��

��

= ��

��

= 6 g / cm3

SUMMMARY AND KEY POINTS

1.) The figures that have the same size and the same shape, i.e. one shape fits

exactly onto other is called Congruent figures .

2.)Two triangles are congruent if they have the same size and the same shape.

3.) If two triangles are congruent, then


b) Their corresponding sides are equal.

4.) The symbol ‘ ≡ ’ means ‘ is congruent to’

0.18 kg

= 0.18 X 1000g

= 180g.

5.) TEST OF CONGRUENCY BETWEEN TWO TRIANGLES:

a.) Two triangles are congruent if all Three Corresponding Sides are equal

This is known as the SSS rule. (side, side, side)

b.) Two triangles are congruent if Two Corresponding Sides and the Included

angle are equal.

This is known as the SAS rule. (side, angle, side)

c.) Two triangles are congruent if Two Angles and a Corresponding Sides are equal .

This is known as the AAS rule . ( angle, angle, side )

d.) Two triangles are congruent if Two angles and the Included Sides are equal

This is known as the ASA rule. (angle, side, angle)

e.) Two triangles are congruent if both triangles have a Right Angle, equal

hypotenuse and another Side which is equal.

This is known as the RHS rule.( right angle, hypotenuse, side )

6.) CONGRUENCE AND TRANSFORMATION:

a.) Under a transformation, an object is formed onto its image.

b.) A figure and its image are congruent under a translation, a rotation and

a reflection .

Translation Rotation Reflection

7.)Two figures are similar if

a)The corresponding angles are equal and

b)The corresponding sides are in the same ratio .

KEY POINTS:

Two similar figures have the same shape but not necessarily the same size.

When two figures are congruent, they are also similar. However the converse

is not true.

8.) Two triangles are similar if


b) Their corresponding sides are in the same ratio.

9.) If Δ ABC is similar to Δ PQR, then

P

A

A

B C Q R

A = P

B = Q and ��

�� =

��

�� =

��

��

C = R

10.) TEST FOR SIMILARITY BETWEEN TWO TRIANGLES:

One of the following conditions is sufficient for two triangles to be similar.

a.) Two triangles are similar if two of their corresponding angles are equal.

A P

B C

Q R

b.) Two triangles are similar if all three corresponding sides are proportional in

the same ratio.

If A = P and

B = Q then


If ��

�� =

��

�� =

��

�� = k

Where k is a constant, then


P

A

kd kf

d f

B e C Q ke R

c.) Two triangles are similar if two of their corresponding sides are proportional

and the included angle is equal.

11.) a.) An enlargement is a transformation that charges the size of a figure

(enlarged/reduced) without changing its shape .

b.) An enlargement is determined by the centre of enlargement and a scale factor.

c.) The scale factor of an enlargement is the common ratio between pairs of

corresponding sides of the image and the original figure .

Scale Factor = ��

��

d.) A figure and its image under an enlargement are similar.

If ��

�� =

��

�� = k, and

B = Q, then


KEY POINTS:

1. If the scale factor is greater than 1, then the image is enlarged.

2. If the scale factor is between 0 and 1, then the image is reduced.

3. If the scale factor is 1, then the image is congruent to the original figure.

12.) If two figures are similar, then the ratio of their areas is equal to the square

of the ratio of the lengths of any pair of corresponding sides .

where A1 and A2 are the areas of the two similar figures and l1 and l2 are their

corresponding lengths.

13.)a.) The ratio of the volumes of two similar solids is equal to the cube of the ratio

of any two corresponding length of the two solids .

where V1 and V2 are the volumes of the two similar figures and l1 and l2 are their

corresponding lengths.

b.)The mass of the sphere is proportional to its volume.

2

1

A

A

=

2

2

1

l

l

2

1

V

V

=

3

2

1

l

l