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21 APR 1970
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僑榮
70英文中學會考試題預習專欄
數學科
文書院主編
(廿五)
(b) ABCD is a quadrilateral in which A = __ B = 90°
j
AB 6 BC100AD18". Prove that BD hissote
SOLUTION
ADC.
A
const.
19770英文中學會考試題預習專欄
堅道英文書院主編
(a) Draw AD LBC
in ABD / ADB = 90°
3- 45°
BAD 180°
(90° + 45°)
45° (/ sum of A)
BD AT siden opp. L
By Pythagora#,
AD
Hence, area of AABC = 3(AD)(30)
Important Theoreme
MATHEMATICS (25)
Aran theeraus on-triangle:
1. A - † (bane z height)
2. Beren's Formulas Azna
1 * † (a+bro)
u on the name Bars (or aqual bages) and
between the same parallels are equal in area,
- rasi theorems on quadrilateral
1. Area of reot. – Length x breadth"
2. área of square. » square on ene sideā
3. Aron of // gran a base x height
4.
(180. 2008. and (50) as the Base batt
(or equal bases) and between the sano parallel are equal in area.
åres of trapezium - Average of two barsa z
6. área of rhombus
Pythagoras 'theodezy
height
médian x height.
of the product or two diagonala
In a rì.A, the area of the square on hypotenuse equals the sum of areas of aquazen, on two adjacRAS sides.
Az pla 1: Find the area of a triangle whose sides
are. (p2 - 42)
inches
SOLUTION: Let a,
o be the sides of the A
Jush that a
200
1.8. the A is right-angled
bo = 1(281)(p2-
pu (p2 - g2)
HOTE: THẢO Hwables can also be solved by using the
erga termula,
where ca
-2. (a+b+c)
* P *Fq • \p{p+q)
(P-Q) STO.
Bennelo Zc AÐ in an altitude of AABOT P, Q are
points on AD produced such that DP:
RA DO MAG. Prove that 34 - Cr
PROOF IS AB> 1G, then from the fig. as showr
ADP BDC.
ABD, BDQ, PDC, GEA Are
By Pythagoras thegroni From 21.▲ 804 : BD BA From rt.A BQD: BQ" -
Q2 = BD2 +
(1).
2 D
CDA: CD2 A
GP2 - CD2 + DP2
(2)
(1) & (2) DP AB & AC = Dal
CP
BQ
adding the case
From It. XADO AD
DO
By Pythagoras
+ √22) (6)
►10.6 (uq.in)
(35) + (6-2)2
60-60√
18.57
√18.57 4.31 (in.)
(b) LA - LB - 90° //BC
"Take a point I on AD. such that XD
TD BO
XBCD is a //gram.
Prom 15, ▲ ABX、
IX-18- 10
By Pythagoras, BX"
x = 10
BD
diag.
Hence, in //gam XECD, BX - BC 10 (in.)
XBCD is a Thombus
BD bisects ADC
Exemple 5: As enown,
RS // AB
Prove that
(a) ASPA+ ASPQ - ASPD (b) APR AASQ AABD (o)APKR KSCQ- ABKD
PROOF (a) Join KD, PR
(b) Join KA
100
and PQ //AD
ASPD. AKPD+AKPS + AKSD
As on the same base and between the Saame // are equal in area / KPD
ДЕРЕ - ДАРЕ - ДАРЕ
AXSD- AKSQ
APS la in common
.
▲KPD + AKSD + AKFS - AAPS + AkSQ
AKPS
ASPA+ ASPQ - ASFD
ASQ - AKSA Z\XSQ + \\QKA
AKSA-AESP - † of //gram K3BP
ARSQ = 1 of //gran KSCQ- Aoki - AQKR - 4 of //gran QDRK AASQ + APR
♣ of (KBBP + XSCQ + QIRK + AFKR )
of ABCD
(0) Join KC
the proof is then).
Example 31 As shown in figure, ABP is a st. line and
Mais the mid-point of AP., Frove that KK bisects the area of
PROGF) Join PK
PMK, AMK are on
ABC
Дарс - Двок + Дрск + Аск ABDC-of // gram ABOD Дрок Драк + Деск
• 1 of (DQKR + QCSK) ACBK - ACSK + ASBK
or (CSKQ + SBPX)
英文科
(廿五)
meaning HER (B) moaning-lear man
nean man (D) man of means (1) moann man
snoo said that she would come
wouldn't she, (B) won'4 {(0) did she didn't sho
15. He dares na to fight,^.
10. Let
daren't ha (B) dosan's dare doesn't" ka
isn't he
go before it is dark,
(A) will you (B) won't you (C) shall wa {D} aren't wa
17. Let me gojenia?
(A) will you (B) wouldn't you (0) aren't YOL (D) don't you
18. You dare not go,
(A) dare you (R) don't you (C) do you
D) are you
19. As wenal, she
har father.
(A) dara not
look at the stern face.
(D) hopes no dares not (0) will not
20. He returned home
long distance.
can hardly
breathing heavily. He
(A) ran (B) should have run (C) was running (D) had run (B) must have run
21. Not even
when he found the house on rize,
(A) he felt a bit sorry (3) ■ bit sorry, he felt (C) Terry did he feel (D) he did feel sorry a bi (B) did he feel a bit sorry
Ko has never seen Joan again, arran
so were his mother (B) so and not his mot neither had his mother (D) nor had his mother either had his mother
21. I can never understand him and he can nevez
understand me ....
A】 tos (3) nino, (C) as well (D) neither
the door, he remembered leaving the key in the house. (A) Having looked (3) Looked (C) Looking. (D) Being lanked: (B) To look
.... he had been sick, he would have won. (A) If not that (B) Unless (C) But for (D) Bub that: (1) Anyway
26. Refrigerators now .... a very important part in
our daily life.
(A) take (3) do (c)
((( (D) play (F) have-
27. You ... every book in my library.
(A) welcome (B) are welcomed
(C) will be welcomed (D) welcome v
(B) are welcome to
28. I am quite sorry ........ your unbrella for no many
months.
(4) (3) having used to: (C) for borrowing
to have borrowed (1) borrewing
The musiuens vont pappened last night MAR the carelessness of the driver.
oving to (B) osused to (0) dus to ((7) for the sake of (B) forsed by An accident happened here last nighw CarGLOBURGOm of the driver, so
ewing to (3) caused to (C) dus to for the sake of (E) formed by
Coses him, name. Defore sinse you two came out from the same university an
(A) sught to know (B) should koar
(C) would have heard - (D) ought to have heard.
must hear.
32. I get used to English food but I can never
accustomed to.... late.
go to bed (B) #leep (0) have slept
(5) ** ** 100). (E) *leeping
43. Kovaz .......... punotual and this makes us very
with her.
Oak she... (3) she 'oan (C) will she` be dinho, (3) is her
14. I akauld have worked much harder, in that conse:2.
(A) have not failed (3) should not fall
(C) should not have failed
8
must not have failed (E) could not have railek
the shot, he threw himself down innodiatel Haring heard ( (B)By hearing
After begring (D) On hearing (X) For hearing
equal bases (PM
KA)
and of the same altitude.
ARMP AKMA
Since AKMP - AKMB + AKEP
KBP,
KBC are on the same base (KB) and between the same // (BK//PC)
ДКВР - ДКВС
Hence, KMP AIIB + AKBC KHBC
AKMP - AKA proved
AKBIA KHEC
K bisects the area of AABC.
Example 4
(a) In AB, 4B = 5" EC
Calculelo the area of AEC and length of AC
± of ABCD - ABDK + 1【DQKR + QOSK) SBPX)
or ABCD = 2,ⱭBDK + DQKR + 2.QCSK + SBPK
ABCD = (DQKR + QCSE + SBPE) + APKR APKR 2.BDK +QCSK
APKR
QCSK
BDK
Example 63 ABCD is a parallelogram, any line parallel
to BA cuts BC, AG, AD at X, Y,, 2 resp. Prove that AAXY - ADYZ.
uiveni ABCD in a //gram
XIZ // AP
To prove: AAXY - ADYZ
Proofs Join CZ
In A CAZ, CXA
They are on the same base (CX) and between
Of which, ACXY is common to both
By subtraction,
In As ZD. ZIC
ACZY
They are on the same base (ZY) and between the same parallels (DC // ZX)
CZY - / DYZ
DOZY - DAXY and Aczy - ADYZ
AAKY - ADIZ
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日六十月三年戌庚歷夏 WAH KIU YAT PO