Congruency and Similarity

1. 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
Chapter 9

2. Chapter 9
Prove that the following two triangles are congruent.
A
B C
2.3 cm
4.2 cm
E
F
G
2.3 cm
4.2 cm

3. Chapter 9
Solution :
EC
GB
FA



(given)
cm4.2
cm3.2
FEAC
GEBC
FGAB



(SSS)FGEABC 

4. Chapter 9
Copy and complete the proof to show that the following
two triangles are congruent.
A
B C12.5 cm
10 cm
35
Y
X
Z
12.5 cm 10 cm35

5. Chapter 9
Solution :
ZC
XB
YA






35ˆˆ
cm10
cm5.21
XZYBCA
YZAC
XZBC
(SAS)YXZABC 

6. 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.
A B
D C
15 cm
O

7. Chapter 9
Solution :
DB
OO
CA



(given)
s)opp.(vert.ˆˆ
(given)
ODOB
DOCBOA
OCOA



(SAS)CODAOB 
A B
D C
15 cm
O
(i)

8. Chapter 9
Solution :
Since , then all the corresponding
sides are equal.
CODAOB (ii)
cm15 ABCD

9. Chapter 9
Show that the following two triangles are congruent.
A
B
C
8 mm
110
48
Q 8 mm
110
22

10. Chapter 9
Solution :
 4822110180ˆQRP
QC
PB
RA



In ΔPQR,
mm8
48ˆˆ
110ˆˆ



RQAC
QRPCAB
QPRCBA
(AAS)RPQABC 

11. Chapter 9
Show that the following two triangles are similar.
C B
A
5 cm
13 cm
FE
D
5 cm
12 cm

12. Chapter 9
Solution :
cm13
512 22
22


 DEEFDF
FC
EB
DA


By Pythagoras’ Theorem,
cm5
cm13
90ˆˆ



DEAB
DFAC
FEDCBA
(RHS)DEFABC 

13. Chapter 9
Show that the following two triangles are similar.
C
B
A
65
Y Z
X
50

14. Chapter 9
Solution :
 65ˆˆ CBABCA


 65
2
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

15. Chapter 9
Show that the following two triangles are similar.
C
B
A 2 cm
3 cm3.5 cm 3 cm
F
E
D
5.25 cm
4.5 cm

16. Chapter 9
Solution :
FC
EB
DA



is similar to (3 pairs of corr. sides equal).ABC DEF 
3
2
25.5
5.3
3
2
5.4
3
3
2



DF
AC
EF
BC
DE
AB

17. Chapter 9
Show that ABC and AEF are similar.
C
B
A
E
F

18. Chapter 9
Solution :
FC
EB
AA



is similar to
(2 ratios of corr. sides equal and included equal).
ABC AEF 
AF
AC
AE
AB
AF
AC
AE
AB
FAECAB




2
1
2
1
angle)(commonˆˆ C
B
A
E
F

19. 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.
CB
A
D
F
E
10 cm
8 cm
6 cm
G

20. Chapter 9
Solution :
BC
CF
AE



(SAS)ACBEFC 
(given)
ˆˆ (corr s)
(given)
EF AC
EFC ACB
CF BC

 

CB
A
D
F
E
10 cm
8 cm
6 cm
G
(i)

21. Chapter 9
Solution :
s)opp(vert.ˆˆ 



GDECDA
GC
DD
EA
is similar to (2 pairs of corr. s equal).ADC EDG  
congruent,areandSince ACBEFC 
CB
A
D
F
E
10 cm
8 cm
6 cm
G
(ii)
DEGDAC
CEFBAC
ˆˆi.e.
,ˆˆthen



22. Chapter 9
Solution :
2
10
6


DG
EG
AC
DC
DG
then,similar toisSince EDGADC 
10
6
2
30 cm
DG  

CB
A
D
F
E
10 cm
8 cm
6 cm
G
(iii) Since EF = AC = 10 cm,
then EG = 10 – 8 cm = 2 cm.

23. Chapter 9
Prove that OE is the angle bisector of
O
A
B
C
D
E
x
y
.ˆBOA

24. Chapter 9
Solution :
EE
OO
DC



(SSS)DOECOE 
side)(commonOEOE
DECE
ODOC



O
A
B
C
D
E
x
y
ˆ, i.e. is the angle bisector of .x y OE AOB 

25. Chapter 9
Prove that OE is the angle bisector of
O
A
B
C
D
E
x
y
.ˆBOA

26. Chapter 9
Solution :
EE
OO
DC



(SSS)DOECOE 
side)(commonOEOE
DECE
ODOC



O
A
B
C
D
E
x
y
ˆ, i.e. is the angle bisector of .x y OE AOB 

27. 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
l1= 5 cm l2= 3.5 cm
l1= 8 cm
l2= 18 cm

28. Chapter 10
Solution :
2
2
2
1
2
1
2
5
5.3
12














A
l
l
A
A
(a)
100
49

15
100
49
2  A
2
cm35.7
2
1
2
2
1
2
1
18
8
45














A
l
l
A
A
(b)
81
16

45
81
16
1  A
2
cm
9
8
8

29. 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.
C
Q
8 cm

30. Chapter 10
Solution :
12
36
8
ofArea
ofArea
2
2














AQ
ABC
APQ
AC
AQ
Since BC is parallel to PQ,
ΔABC and ΔAPQ are similar.
C
Q
8 cm
2
36
8 12
AQ 
 
 
192
13.86 (to 4 s.f.)
AQ 

13.86 8
5.86 cm (to 3 s.f.)
CQ AQ AC 
 

2 36
64
12
192
AQ  

( 13.86 is rejected since 0.)AQ 

31. 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
C
A
D

32. Chapter 10
Solution :
CF
BF
hCF
hBF
AFC
ABF






2
1
2
1
ofArea
ofArea
(i) ΔAFC and ΔABF have a common height corresponding to the
bases CF and FB respectively.
C
A
D
5
3
25
ofArea

ABF
2
3
Area of 25
5
15 cm
ABF  


33. Chapter 10
Solution :
2
2
10
3
25
ofArea
ofArea
ofArea

















AED
AF
AE
AFC
AED
(ii) Since DE is parallel to CF, ΔAED and ΔAFC are similar.
C
A
D
100
9
25
ofArea

AED
2
9
Area of 25
100
2.25 cm
AED  


34. Chapter 10
The figure shows two similar cylinders. Find the volume
V2 of the smaller cylinder.
d1= 12 cm
V1= 81 cm3
d2= 4 cm
V2= ?

35. Chapter 10
Solution :
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
81
27
3 cm
V  


36. 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.

37. Chapter 10
Solution :
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
27
81
17
4
27
m
g
   
 

The mass of the smaller figurine is .g
27
17
4
3
1 45
27 81
m    
 

38. 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.
10 cm

39. Chapter 10
Solution :
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
3
1
2
10 5
2
10
5
6.32 cm (to 3 s.f.)
h
h

 

The depth of water is 6.32 cm.
(i)

40. Chapter 10
Solution (Cont’d) :
10
325.6
2
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
2
6.325
10
0.400 (to 3 s.f.)
A r
A 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)