term 3 unit 1 congruency and similarity

Chapter 9

Learning Objectives: understand and identify congruency and similarity in polygons by matching

corresponding sides (emphasising the point to point correspondence between figures)

state conditions and prove for congruency in triangles: SSS, SAS, AAS, RHS state conditions and prove for similarity in triangles and other polygons: AA,

3 ratios are equal, 2 ratios+1 incl angle are equal understand and apply relationship between ratios of areas and corresponding

sides of similar figures, i.e. understand and apply relationship between ratios of volumes and

corresponding sides of similar solids, i.e. understand and apply on two triangles sharing common base/height, i.e.

where l represent the different base (but common height) or height (but common base)

apply concepts of congruency and similarity in problem solving Understand map scales and know how to convert between actual and scaled

lengths and areas Understand and describe line and rotational symmetries in 2D figures

Term 3 Unit 1- Congruency and Similarity

Chapter 9

Prove that the following two triangles are congruent.

Worked Example 1

A

B C

2.3 cm

4.2 cm

E

F

G

2.3 cm

4.2 cm

Chapter 9

Worked Example 1Solution :

EC

GB

FA

(given)

cm 4.2

cm 3.2

FEAC

GEBC

FGAB

(SSS) FGEABC

Chapter 9

Copy and complete the proof to show that the following two triangles are congruent.

Worked Example 2

A

B C12.5 cm

10 cm

35

YX

Z

12.5 cm 10 cm35

Chapter 9


ZC

XB

YA

35ˆˆ

cm 10

cm 5.21

XZYBCA

YZAC

XZBC

(SAS) YXZABC

Chapter 9

In the diagram, AOC and BOD are straight lines, OA = OC, OB = OD and AB = 15 cm.

(i) Prove that ΔAOB is congruent to ΔCOD.(ii) Find the length of CD.

Worked Example 3

A B

D C

15 cm

O

Chapter 9


DB

OO

CA

(given)

s) opp. (vert. ˆˆ

(given)

ODOB

DOCBOA

OCOA

(SAS) CODAOB

A B

D C

15 cm

O

(i)

Chapter 9


Since , then all the corresponding sides are equal.

CODAOB (ii)

cm 15 ABCD

Chapter 9

Show that the following two triangles are congruent.

Worked Example 4

A

B

C8 mm

11048

R

P

Q 8 mm

110

22

Chapter 9


4822110180ˆQRP

QC

PB

RA

In ΔPQR,

mm 8

48ˆˆ

110ˆˆ

RQAC

QRPCAB

QPRCBA

(AAS) RPQABC

Chapter 9

Show that the following two triangles are similar.

Worked Example 5

C B

A

5 cm13 cm

FE

D

5 cm

12 cm

Chapter 9


cm 13

512 22

22

DEEFDFFC

EB

DA

By Pythagoras’ Theorem,

cm 5

cm 13

90ˆˆ

DEAB

DFAC

FEDCBA

(RHS) DEFABC

Chapter 9


Worked Example 6

C

BA

65

Y Z

X

50

Chapter 9


65ˆˆ CBABCA

652

50180ˆˆ YZXZYX

65ˆˆ

65ˆˆ

YZXBCA

ZYXCBA

ZC

YB

XA

is similar to (2 pairs of corr. s equal).ABC XYZ

C

B

A 65

Y Z

X

50

Chapter 9


Worked Example 7

C

BA 2 cm

3 cm3.5 cm 3 cm

F

E

D

5.25 cm

4.5 cm

Chapter 9


FC

EB

DA

is similar to (3 pairs of corr. sides equal).ABC DEF

3

2

25.5

5.33

2

5.4

33

2

DF

ACEF

BCDE

AB

Chapter 9

Show that ABC and AEF are similar.

Worked Example 8

C

BA

E

F

Chapter 9


FC

EB

AA

is similar to

(2 ratios of corr. sides equal and included equal).

ABC AEF AF

AC

AE

ABAF

ACAE

AB

FAECAB

2

12

1

angle)(common ˆˆ C

BA

E

F

Chapter 9

In the diagram, BC is parallel to EF, EF = AC = 10 cm, FC = BC, DC = 6 cm and GF = 8 cm.

(i) Show that ΔEFC is congruent to ΔACB.

(ii) Show that ΔADC is similar to ΔEDG.

(iii) Find the length of DG.

Worked Example 9

CB

A

D

FE

10 cm

8 cm

6 cm

G

Chapter 9


BC

CF

AE

(SAS) ACBEFC

(given)

ˆˆ (corr s)

(given)

EF AC

EFC ACB

CF BC

CB

A

D

FE

10 cm

8 cm

6 cm

G

(i)

Chapter 9


s) opp (vert. ˆˆ

GDECDA

GC

DD

EA

is similar to (2 pairs of corr. s equal). ADC EDG

congruent, are and Since ACBEFC CB

A

D

FE

10 cm

8 cm

6 cm

G

(ii)

DEGDAC

CEFBAC

ˆˆ i.e.

,ˆˆthen

Chapter 9


2

10

6

DGEG

AC

DC

DG

then, similar to is Since EDGADC

10 62

30 cm

DG

CB

A

D

FE

10 cm

8 cm

6 cm

G

(iii) Since EF = AC = 10 cm, then EG = 10 – 8 cm = 2 cm.

Chapter 9

Prove that OE is the angle bisector of

Worked Example 10

O

A

B

C

D

Exy

.ˆBOA

Chapter 9


EE

OO

DC

(SSS) DOECOE

side)(common

OEOE

DECE

ODOC

O

A

B

C

D

Exy

ˆ, i.e. is the angle bisector of .x y OE AOB

Chapter 9

Prove that OE is the angle bisector of

Worked Example 10

O

A

B

C

D

Exy

.ˆBOA

Chapter 9


EE

OO

DC

(SSS) DOECOE

side)(common

OEOE

DECE

ODOC

O

A

B

C

D

Exy

ˆ, i.e. is the angle bisector of .x y OE AOB

Chapter 10

Find the unknown area of each of the following pairs of similar figures.(a) A1 = 15 cm2 A2 = ?

(b) A1 = ? A2 = 45 cm2

Worked Example 11

l1= 5 cm l2= 3.5 cm

l1= 8 cm

l2= 18 cm

Chapter 10


2

2

2

1

2

1

2

5

5.3

12

A

l

l

A

A(a)

100

49

15100

492 A

2cm 35.7

2

1

2

2

1

2

1

18

8

45

A

l

l

A

A(b)

81

16

4581

161 A

2cm9

88

Chapter 10

In the figure, BC is parallel to PQ, AC = 8 cm and the areas of ΔABC and ΔAPQ are 12 cm2 and 36 cm2 respectively. Find the length of CQ.

Worked Example 12

C

B

A

PQ

8 cm

Chapter 10


12

36

8

of Area

of Area

2

2

AQ

ABC

APQ

AC

AQ

Since BC is parallel to PQ, ΔABC and ΔAPQ are similar.

CB

A

PQ

8 cm

236

8 12AQ

192

13.86 (to 4 s.f.)

AQ

13.86 8

5.86 cm (to 3 s.f.)

CQ AQ AC

2 36 6412192

AQ

( 13.86 is rejected since 0.)AQ

Chapter 10

In the figure, DE is parallel to CF, CF = 5 cm, FB = 3 cm, AE = 3 cm and EF = 7 cm. Given that the area ΔAFC is 25 cm2, find the area of (i) ΔABF (ii) ΔAED

Worked Example 13

C

B

A

FD

3 cmE

7 cm

5 cm

3 cm

h cm

Chapter 10


CF

BF

hCF

hBF

AFC

ABF

2121

of Area

of Area

(i) ΔAFC and ΔABF have a common height corresponding to the bases CF and FB respectively.

C

B

A

FD

3 cmE

7 cm

5 cm

3 cm

h cm

5

3

25

of Area

ABF

2

3Area of 25515 cm

ABF

Chapter 10


2

2

10

3

25

of Area

of Area

of Area

AED

AF

AE

AFC

AED

(ii)

Since DE is parallel to CF, ΔAED and ΔAFC are similar.

C

B

A

FD

3 cmE

7 cm

5 cm

3 cm

h cm

100

9

25

of Area

AED

2

9Area of 251002.25 cm

AED

Chapter 10

The figure shows two similar cylinders. Find the volume V2 of the smaller cylinder.

Worked Example 14

d1= 12 cm

V1= 81 cm3

d2= 4 cm

V2= ?

Chapter 10


d1= 12 cm

V1= 81 cm3

d2= 4 cm

V2= ?

3

2

3

1

2

1

2

12

4

81

V

d

d

V

V

2

3

1 8127

3 cm

V

Chapter 10

Two similar clay figurines have heights of 45 cm and 81 cm respectively. Given that the mass of the bigger figurine is 27 g, find the mass of the smaller figurine.

Worked Example 15

Chapter 10


3

2

1

2

1

2

1

h

h

V

V

m

m

Let m1, V1 and h1 be the mass, volume and height of the smaller figurine respectively, and m2, V2 and h2 be the mass, volume and height of the bigger figurine respectively.

3

1

45 2781

17427

m

g

The mass of the smaller figurine is . g27

174

3

1 4527 81m

Chapter 10

The figure shows an inverted conical container of height 10 cm. It contains a volume of water which is equal to two-fifth of its full capacity. Find(i) the depth of the water,(ii) the ratio of the area of the top surface of the water to the area of the top surface of the container.

Worked Example 16

10 cm

Chapter 10


3

2

1

2

1

h

h

V

V

Let V1 and h1 be the volume and height of the smaller cone respectively, and V2 and h2 be the volume and height of the bigger cone respectively.

5

2

10

105

2

3

1

3

1

h

h

1 3

31

210 5

2 105

6.32 cm (to 3 s.f.)

h

h

The depth of water is 6.32 cm.

(i)

Chapter 10

Worked Example 16Solution (Cont’d) :

10

325.62

1

2

1

h

h

r

r

The top surfaces of the water and that of the container are circles. Let r1 and r2 be the radii of the smaller circle and the larger circle respectively. Using similar triangles,

Let A1 and A2 be the areas of the smaller circle and the larger circle respectively

2

1 1

2 2

26.325

10

0.400 (to 3 s.f.)

A rA r

The ratio of the area of the top surface of the water to the area of the top surface of the container is 2 : 5.

(ii)

term 3 unit 1 congruency and similarity

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