9.

1 - Students should be able to use

adjacent angles on straight line adj.s on st.line Ex.9A Q.3
angles at a point s at a pt. Ex.9A Q.5
for reasoning and calculations.
9.2 - Students should be able to use
vertically opposite angles vert.opp. s Ex.9A Q.6
9.3 - With a pair of parallel lines and a transversal (), students should be
able to use
corresponding angles corr. s, __//__ Ex.9B Q.1
alternate angles alt. s, __//__ Ex.9B Q.2
interior angles int. s, __//__ Ex.9B Q.3
9.4 - Students should be able to use
angle sum of triangle sum of
exterior angle of triangle ext. of Ex.9C Q.2
base angles, isosceles triangle base s, isos. Ex.9C Q.3
- Students should be able to solve some specific problems. Ex.9C Q.9, 10
9.5 - Students should be able to use
sum of ext. s of polygon Ex.9D Q.1-7
sum of polygon Ex.9E Q.1-6
- Students should be able to solve problems involving interior and exterior
angles.
Supp - Revision should be given to students to prove 2 lines are parallel.
alternate angles equal alt. s eq.
corresponding angles equal corr. s eq.
interior angles supplementary int. s supp.

9.6 (tessellation) - Students should be able to find the regular polygons for tessellation.
- Students should be able to recall the techniques (translation , rotation
, reflection ) to construct tessellation figures.
9.7 ** This part will not be taught and will leave as vacation homework in
Easter.

(a)
(b)

(Platonic solid) 5 4 8 20
6
12
 Handout 5-1

a+b = 180o adjacent angles on straight line adj.s on st.line

a+b+c = 360o angles at a point s at a pt.

a=b vertically opposite angles vert.opp. s

b=d AB//CD corresponding angles, AB//CD corr. s, AB//CD

a=c AB//CD alternate angles, AB//CD alt. s, AB//CD
int. s, AB//CD
o
b+c = 180 AB//CD interior angles, AB//CD
corresponding angles equal corr. s equal

alternate angles equal alt. s equal

AB // CD
interior angles supplementary int. s supp.
SSS
SAS
ASA / AAS
RHS
3 sides proportional 3 sides prop.
2 sides in ratio, included angles equal 2 sides in ratio,
inc. s
equiangular AAA

a+b+c = 180o angle sum of triangle sum of

a+b = c exterior angle of triangle ext. of

angle sum of polygon sum of polygon

= (n-2)180o

sum of exterior angles of polygon sum of ext. s of

= 360o polygon

a=b base angles of isosceles triangle base s, isos.

a=b sides opposite equal angles sides opp.equal s

ABC

AB = AC base angles equal base s equal

ABC is an
isos. with
AB = AC
Construction exercise of Polygons

1. Following the steps shown in the textbook P.12.20 12.25, construct

(1) equilateral triangle, (2) square, (3) regular pentagon,
(4) regular hexagon, and (5) regular octagon
of side length 2 cm.

2. AB is the base of a triangle ABC, and the length of BD is the sum of the lengths of BC and CA.

Given that ABC = 60o, construct the triangle ABC. (HKMO08, Construct Q2)

3. Use the following line segment AB. Construct a triangle ABC such that AC : BC = 3 : 2 and
ACB = 60 .o

A B (HKMO09, Construct Q3)

Chapter 12 Polygons Quiz 12-1
F.2_____ Name:___________________________________________________( ) Marks:_____/ 6
1. According to the figure, which of the following must 3. In the figure, x = 5. In the figure, AY and CY are the angle bisectrs of BAX
be true? and DCX respectively. If AXC = 100o, then
AYC =

A. a+b=c+d A. 50o
B. a+d=b+c B. 60o A. 40o
C. a + b + c + d = 360o C. 70o B. 50o
D. a + b + c + d = 540o D. 90o C. 60o
E. 2a + 2b - c - d = 720o D. 80o

2. In the figure, ABCD is a trapezium with AB // DC. 4. According to the figure, which of the following must 6. In the figure, x =
AH bisects BAD and DH bisects ADC. Which be true?
of the following must be true? A. 50o
B. 55o
C. 60o
D. 65o
E. 70o

A. a+b=c
c AHD = 90o B. a + b = c + 90o
d ADC = BCD C. a + c = b + 540o
e BAD + BCD = 180o D. a + b + c = 720o

A. conly
B. donly
C. eonly
D. cand eonly
E. dand eonly
Chapter 12 Polygons Quiz 12-2
F.2_____ Name:___________________________________________________( ) Marks:_____/ 16
1. In the figure, x = 4. In the figure, a + b + c + d + e + f = 6. Find the value of x. (3 marks)

A. 270o
B. 360o
C. 450o
A. 50o D. 540o
B. 60o E. 720o
C. 70o 100o + 120o + xo = 360o
D. 80o (sum of ext. s of polygon)
x = 140
5. Find the smallest interior angle of the pentagon
ABCDE if A : B : C : D : E= 2 : 3 : 4 : 5 : 6.
2. If each interior angle of a regular n-sided polygon is (4 marks)
144o, then n = 7. Find the value of x in the following figure.
(5 marks)
A. 10 Let the size of A be 2ko
B. 12 2k + 3k + 4k + 5k + 6k = (5 2)180
C. 14 ( sum of polygon)
D. 16 20k = 540
k = 27

Size of smallest angle (A) = 54o Join CE.

3. In the figure, ABCDE is a regular pentagon and
ABFG is a square. Find x. In CDE,
DCE + DEC + 155o = 180o
A. 18o
( sum of )
B. 27o
DCE + DEC = 25 o
C. 30o
D. 36o
In quadrilateral ABCE,
E. 45o
120o + xo + (55o + DCE) + (DEC + 35o)
= (4 2)180o
( sum of polygon)
x + 120 + 55 + 35 + 25 = 360
x = 125
Chapter 12 Polygons Quiz 12-supp
F.2_____ Name:___________________________________________________( ) Marks:_____/
1. The sizes of ABD and BDE are 110o and 120o 2. Find the sum of all the marked angles in the 3. If QPS is a right angle, PQR = 36o, QRS = 3xo
respectively. Find the sizes of CBD and BCD. following figure. and RSP = xo. Find the value of x.
P
A
B

R
E D C
Q S
(6 marks) (4 marks)

(4 marks)

CBD + ABD = 180o (adj. s on st.line) In AEI, Extend the line QR and meet PS at T.
CBD = 70o IAE + AEI + EIA = 180o ( sum of ) P

In BCD, In CGK,
BDE = CBD + BCD (ext. of ) CGK + GKC + KCG = 180o
120o = 70o + BCD ( sum of )
BCD = 50o T
IAE + KCG + AEI + CGK + EIA + GKA
= 360o R

Q S

In RST,
3xo = xo + RTS (ext. of )
RTS = 2xo

In PQT,
2xo = 90o + 36o (ext. of )
x = 63
Chapter 9 Plane Geometry Quiz 09-supp
F.2_____ Name:___________________________________________________( ) Marks:_____/
1. Find the size of marked angles. (5 marks) 2. Find the size of the marked angle. (3 marks) 3. Note: No reasons are required in this question.
The sum of the interior angles of a convex polygon is
ten times as great as the sum of its exterior angles.
Find the number of sides. (3 marks)

a = 47o (alt. s, AB // EC) 110o + x + 130o + 120o + x + 160o = (6 2)180o Let the polygon have n sides.
( sum of polygon) (n 2)180 = 360 10
(ext. of )
o
125o = a + b x = 100 n 2 = 20
b = 78o n = 22

The polygon has 22 sides.

4. Note: No reasons are required in this question.

The size of each interior angle of an equiangular
convex polygon is greater than each exterior angle by
132o. Find the number of sides. (4 marks)

Let the size of an exterior angle of the equiangular

polygon be xo.
(180 x) x = 132
x = 24

360
Number of sides = 24
= 15
5. In the figure, AC // ED, BD = BE, BCD = 110o and 6. In the figure, L1, L2, L3 and L4 are straight lines 7. Answer-only-Question: No reasons are required in
BD is the angle bisector ( ) of CDE cutting PQ at four different points. Determine, with this question.
(ie. CDB = BDE). Find x, y and z. (8 marks) reasons, whether Write down the sum of the marked angles in the
(a) L1 is parallel to L4, following figure. (2 marks)
(b) L2 is parallel to L4,
(c) L3 is parallel to L4. (5 marks)

BCD + CDE = 180o (int. s, AC // ED) (a) L1 is parallel to L4 (int. s supp.) The sum of all marked angles = 360o
CDE = 70o
(b) L2 is parallel to L4 (alt. s eq.)
Given CDB = BDE
x = 35o (c) L3 is not parallel to L4

BED = x (base s, isos. )

z = BED
= 35o (alt. s, AC // ED)

CBD = x (alt. s, AC // ED)

CBD + y + z = 180o (adj. s on st. line)

y = 110o