The document discusses the properties of congruent figures, including: - Corresponding sides are proportional and corresponding angles are equal (congruence) - Triangles are congruent if three sides are equal (SSS), three angles are equal (AAA), two sides and the included angle are equal (SAS), or one side, two angles (ASA, AAS, SAA) - The document provides examples of using proportional reasoning to solve problems involving similar figures and triangles. It includes 10 practice problems with solutions.

1. CONGRUENCE
By: Afida Zahara Adzkiya (IX. 4)

2. • All the corresponding sides are proportional, and
• All the corresponding angles are equal in measure.
C
Z Y
Question : Are the trapezoids congruent?
Answer : 1) AB = WX, BC = XY, CD = YZ, DA = ZW, and
2) A = W, B = X, C =  Y, D = Z.
Based on description above we can concluded that, both of
trapezoids are congruent.
D
A B
W X

3. A picture or scale model has the same form as the
original form. Corresponding parts of the pictures or scale
model will have the same comparison with the original build.
Since the corresponding sides equal then the comparison can
be made as follows.
Length of the model = Width of the model = Height of the model
Length of the original Width of the original Height of the original

4. • The Properties of Two Triangles
1) The three Corresponding Sides are Equal (SSS property / side, side,
side)
C
A B
• Explanation:
AB = ED
BC = DF
AC = FE
F
D E

5. 2) The Three Corresponding Angles are Equal (AAA Property /
angle, angle, angle)
H I
Explanation:
H = K
I = L
J = M
J
M
K L

6. 3) The Two Sides and The Included Angles are Equal (SAS /
side, angle, side)
O
Explanation:
P
OP = PQ (given)
OPR = QPR = 90
PR = RP (coincident)
Q
R

7. 4) One Side, Two Angles Property (angle, angle, side), (angle,
side, angle), (side, angle, angle).
V
S T
Explanation:
U
UST = SUV (alternate interior angle)
SU = US (coincident)
VSU = TUS (alternate interior angle)

8. 1)
or
a
b
c
d
e
f
a / a+b = c / c+d = e / f
a / b = c / d

9. 2)
a
b
c
e
d
e = (a . d) + (b . c)
a + b

10. 3)
D
C
A B
AD² = BD . CD
AB² = BD. BC
AC² = CD. CB
AB. AC = BC. AD

11. Exercise!
1)
If AC // DE, with AC = 12 cm, DE = 6 cm,
BE = 4 cm, so the length of CE is …
C
E
A D B
Answer:
EB + CE / EB = CA / ED
4 + CE / 4 = 12 / 6
4 + CE / 4 = 2 / 1
4 + CE = 8
CE = 8 – 4
CE = 4 cm So the length of CE is 4 cm

12. 2) Height and the width of a tower in the photograph is 24 cm
and 6 cm. If the actual width of tower is 2 m, so the actual
height is?
Answer:
6 cm
24 cm
2 m
x
x / 24 = 200 / 6
x / 24 = 100 / 3
3 x = 100 . 24
3 x = 2400
x = 800 cm
x = 8 m

13. 3)
c
d
e
a b
If bac = edc, with
ae = 12 cm, ce = 6 cm
and cd = 8 cm,
so the length of db is?
Answer:
cd / ca = ce / cb
8 / 16 = 6 / cb
1 / 2 = 6 / cb
cb = 6 . 2
cb = 12 cm
db = cb – cd
db = 12 – 8
db = 4 cm

14. 4)
F
D C
A E
B
9 cm
15
cm
Asked: Area of
ABCD…?
D F 15
cm
9 cm
A E
C 9 cm
B
?
F E
Answer:
CF / DA = CB / DF
CF / 9 = 9 / 15
CF / 9 = 3 / 5
5 CF = 9 . 3
5 CF = 27
CF = 5,4 cm
Area ABCD = DC. DA
= (15 + 5.4) x 9
= 20.4 x 9
= 183.6 cm²

15. 5)
D C
F E
A B
Known:
DF = 3
FA = 9
DC = 12
AB = 18
Asked: The length of FE?
Answer:
FE = (DF . AB) + (FA . DC)
DF + FA
FE = (3 . 18) + (9 . 12)
3 + 9
FE = 54 + 108
12
FE = 162
12
= 13. 5 cm

16. 6)
A N
K
I
L U
12 cm
Asked: The length of AI
?
Answer:
AU² = AL² + LU²
= 12² + 12²
= 144 + 144
= 288
AU = √288
= √144 x 2
= 12 √2 cm
AI = AU – IU
AI = 12 √2 – 12
AI = (12 √2 – 2) cm

17. 7)
C
A D B
At the figure above known
the length of AD = 9, CD =
12, the length of CB is?
I.
CD² = AD . DB
12² = 9 . DB
144 = 9 . DB
DB = 144 / 9
DB = 16
II.
CB² = BD . BA
= 16 . (16 + 9)
= 16 . 25
= 400
CB = √400
= 20

18. 8)
D (10 x + 2) cm C
12 cm
6 cm
E
A B
(18 x + 6) cm
In the picture above,
ΔAEB congruent with
ΔDEC. The value o f x
is?
AB / DC =
EB / ED
18 x + 6 / 10 x + 2 =
12 / 6
18 x + 6 / 10 x + 2 = 2
/ 1
18 x + 6 = 20 x + 4
18 x – 20 x = 4 – 6
-2 x = -2
x = 1
So, the value of x is 1.

19. 9)
9 6
a
21
b 12
Determine the value
of a and b !
a / 6 = 21 / 9
a / 6 = 7 / 3
3 a = 7. 6
3 a = 42
a = 14
b / 21 = 12 / 14
b / 21 = 6 / 7
7 b = 6 . 21
7 b = 126
b = 18

20. 10) A photograph measuring 24 cm and 30 cm placed in a carton. At the top, left, and
right images it remaining 3 cm. If the images and carton congruent. Find the area of
carton that is not covered by the photograph!
3 cm 3 cm
24
cm
3 cm
30
cm
30
cm
24
cm
30
cm
x
x / 30 = 30 / 24
x / 30 = 5 / 4
4 x = 5 . 30
4 x = 150
x = 37.5 cm
Area carton not
covered:
= A carton – A
photograph
= (30 x 37.5) – (24 x 30)
= 1025 – 720
= 305

21. THANK YOU 