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英中會考數學(三)答案
育教、僑華
英中會考數學(二)答案
( 籟 ) - 堅道英文書院撰答
14.(a
Givens AB in the tankent of
O AFC
Užvent AD
ED
BE EQ
AELEB
CFED is a at." line
to prove: AFBD is a //gram. Proof:
Lax, = LI (ext / cyclic
A AFE, BDE
ruad,
L (Ls in alt-
segment)
L52
Lx, (La în the
To prove : DEFG is a cyclic quad
Proofi
Since D, 0, 5 are mid-pts, of 13, BC, AC
same sapment.)
rasps otively
DG
BC
DE
AC
GF
AB
LI - LI1 (OPD. Ls of // gram/
堅道英文書院撰答
Suggested Answer
of
.K. Certificate of Education Examination (English)
Section A
Mathematics III
(1) The polygon has 12 sides.
(2) Area of AABE - 9 sq.m.
3) area of inscribed oirola
area of circumscribed circle
(4(a) rectangle, squar
(b) square
(c) Thọabus
(5) DC the longes
"AB is the shortest
6) The locus of P is a straight line passing through
F parallel to BC
(7) only (d) is wrong
AB - BE (given)
ADF BED (vart. opp. / s
• 13 (proved)
AAFE ABDE {A.A.Ş.
* EM
DE
AFBD is e
orr. aidea of conguent As)
gran (diag. bisect each other)
_x. (corr./ 8 UG // EG
L3
LX (alt. Lo DG // EC)
AGN, FON
NP (intercept Th.).
Given
B
ABCD 18 a
//gran
LONY
GNA - 90° (alt. / e DG // BC)
E in any pt. sing AABD
(9
CN common
To provas ABEC -
ABED
rroof: AAEB
ADEC-
ABCD
ABEC
ABOD (ДАЕВ
ADEC)
AAED
HABOD - AET
17.(a)
AABB + ABED - SABCO - AED
ABEC-AAEBAREL
GAFIN (S.A.S.).
(corr. /s of congruent (s).
dince / EFG + × • 180° (adj. is on a st.
EFG
180*
line)
DEFG is a cyclic quadrilater
I want ABCDE is a regular
pantagon
To prove: AB+ B- A
Proof
Each int. of the pentagon
1080
800
72
FG
Given ABFE, BCGF, CDHG are 3 oonė
of each side 'equals to
To próvas (1)
36
AF
FH
(25)
Proofi
L
OF PR
(sidea opf, oqual (8)
AB+ B AE + BP.
2a (Pyth. Th.)
s(2a)
EP + BP
BE
• AG
(11) tanla
FG FH
cantarir
- tangtann
(C)
17.(b) Let a, b be lengths of two lines, in which
La longer than b..
date methou::wing or cusana rorm
@pplicablm).
Givens In
ABC
A
90°
To prove
Froofi
AN AN
- AB+ AC BC).
BM: BP
ON
CP
Since AMON 18 a rectangle
AN AC CH 40 -AN-AB- BK - AB
• AB + AC ~ (OP + BM)
(AB+ AC - BG)
ja + b) 2
* b)
Lab)
(a - b)
+4ab
23
(20)
(10)
(11)
Diameter
Set square:
ADE, ABC are similaz
鍋子
BD
is the given pt.
APB is the longest chord
which must pass through the
centre, and CPD in the shortest chord which mua † perpendicular to AB.
OP
132-
12 O
a
from measurement, lengths of the two lines are
232 co
and a 10.9 ca
Constructions -
1 Draw BQ- BO + CA
2. Take any pt. Non BQ erect a perpendicular
MN-equals stoTABASA
3. From M.construct a line MP //-3Q
At B construct an angle equals to the given
and meet MP at A
Draw the perpendicular bisector of AQ and let
it out BQ at C
ABC is the required
Section B
(12)
EN
ER 2
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