二星張四第
日万十月二十年 WAH KIU YAT PO
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10. ATTHE TQ==QR=RS=SP,
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「四期星 日二十月一年〇七九一周公年九十五國民華中育儒
<PAN=CORS === C R S P = < SPQ=90° (@ 127)
11. PQRS為正方形(各电相等,备角相等)
(B)希 AB=8吋, AP-34求正方形 PQRS ABCD定比
(14) ABCD == AB2—§ 64*t.
78-8-3-54
APBA - 4×5×3 = 1 + † (~aQCR.=ARDS=---)
• PORB=L4-4×4=3437
ABCD
64
17 (答)
32
(ZA) PORS= PQ*=PB*+ BQ*= 5+3°— 34 (#T) (4)E與正方形ABCD,求作其內接正方形 PQRS使
PARS - ABCD.
余慶
學會考試題預習專欄
數學科 (十二)
·喬仲强•
(已)正方形ABCD.
第十一次預習題解答
AY元比值奖選之直我無關,又着两国内切時其结果树?
(已知)O,P两回外切於A,XAY為過A卓元任意直线
(EAK:AY 典子選之直线無關.
(1)西园外切於ASA作任意直线各交两园拾X,Y,試EAX
OPALEA A
【两团相切,中心线通切
2. 將OP两端各延長之
D. X. CX, DY
(直线可任意延层,两桌决定一直线)
3. <x<Y=t(闼国周角尊直角,凡直角相等)
4. LKACULYAD (丁頂角相等)
5. AXACO AYAD (AS有两角等,則而相似形
b.Ax:AY=AC:AD=两园直往比(相似形之对应边)
因两國直径比為定比,故AX:AY與選直线無關。
(二)!A作两园的公切线:AT.
(两
* 切過切桌可作公切傳)
品遇A再作任意直线 K'AY,
XXRYY
(两桌决定一直线)
3. 21-2 (TA44)
44=4Y, 42=4X,
wh)
5.2.XY(等量代换)
6. *£*£ <X'=LY' (#3£4)
7. AAKXOAAYY'(两角被此相等) 8.AX:AY=AX':AY(相似形对比)
B: AKAYA選之直线無關。
Q.E.D.
(答)酒园内相切AX:AY 假
辅等茶西园半径(实直径之比,
(2)0.P滿熱音款A.BA奌一直
CADEROB**c☀D BE CA=2AD,
(EX) 0,798TA, 6.
(求作)遇A引一直线CAD各
與P目相交於CD
使 CARAD.
(分析)談CD区作出如有
BW, KOEL CA. PFL AD,
則EA CA, AF=EAD.
(弧弦平分理)
∴EA=2AF
又遇A作CD之海线Q
OPQOEI QAIPF|
(CD)
∴ Q=2QP(理)
(求作)正方形PARS 内接卡 ABCD M
PORS-ABCD.
(分析)設AB=a,AP=x,則
* ABCD
7Q 123=7Q*= x2+(a−x}" 依題意
x2 + (a−x) = za x2+a-zax+x=ZL
D
R
1770英文中學會考試題預習專欄
堅道英文書院主編
生物科
BIOLOGY (12)
saswer to the questions of last vack
(+)
I. Chooms the best answers that answer the questioÞAR
A
(R-X)
Ane.
Bp x(a¬x)=₤α 板本題為知线段和(a)及两线段積(58) 文作問題。
(作法)1.以AD直狸作半团,
2. ADIABAD.
3.1E1EFLAD & AD
园 F:
4.以AB為直徑作洋國
5. DAEGA~AG=AF..
6.邏G作AB平行B
阿琴HHH')
不過H作AB
AS
8. 在BC, CD, DA上各截
BQ=CRUD$ =AP.
9.TH TO, AR, RS, SP. 71 PORS že⭑.
0袋7至9方法,得P'Q'R'S‘亦為求,
R
(討論)。本題恒有两解,2.若改為主,則只有一海军
3.笼番饭鸡小柃文分數,則本題無解。
(别解)OABCD及PERS之中心五
PARS PO
op
ABZ OA
∴OP(OA)XOA
即OPMOA QA之比例中項
(作法)). AC, BD使:0奐。
志在0°C上5=400.
3.以E為直徑作园交於F
4以口為心,OF為半話作國與 A
正方形怱兒於PPQQRR S. S各点.
5賴PQ,OR,RS,SP及PA'Q'R', R'S'SP,則正方形 PQRS
RF'O'R'S HAMA.
(5)設PQR急ÇASC的AB, BC, CA边上的奌,蕉
APAB, BQ=18C, CR=ŹCAZMIN.
△ABC元坑
(24x) AP=1AB, BQ=10C.
CRCA..
(京) AFQR: AABC 比值
(A) 1. AR=4AC (~ CR=4CA)
2.
P
(14) 1. OP. 品在P上取Q奌使OQ= OP
3. AQ 佐A作AQ之法线 CAD 即為將碗
(証明)可参看分析,自行証明衣
本題亦可用下法解之
(分析)設作区因院PE
為ㄗ园半之2倍,則
通A总之任意直线,在QP\
二國內文弦CA, AD恒有
CA==AD 係
故D,Q二之交與即骂C無
(作法)1.PA延長之至
QAQ 2PA.
3.联CA並延長之交PD, 則CAD為求之直缕
(三)如作與P园内切之Q因,亦可得C'A==ZAD'直线
但方向則相反,(图中未顯出) D
(3)(a)假如右图: ABCD為正方形,
潘AP=Q=CR=DS,試龇
PARS亦正方形
(E) ABCD為正方多
AP-BQ=CR=DS
(MIE) PQRS É EIF
(*) 1. A&-AD (ÉSHALT)
2. AP=DS-BQ (Eke)
3. AB-AP=AD-DS 即 PB= SA (等量相減)
4. <A=<日==党人(正方形各角為直角)
5. MLAATS SOMABQP (S. A. S.)
6. <1><2, $P=PQ (全等形对应部分) 7. <1+23=90° (B<A=90°, rt. A MILĪŔŽKA) 8. 42+23=90′ (4‡*).
43PQ=180°(<2+3}=90°(APB為一直
APZ APIAS ZABIZAC A
AAEC
3
H恕
ABKAC
C
∴AAPR GAABC.
APQR
(可在一角架,西積地等於其夹边相乘積之比),
3. BIETJE APPO==AABC, ACRQ= /^ABC ($122, KAAPRABPOT4CPQC△ABC (等量相比) ABC(全等於諸分量和
即
440
(B) £=%BACO
(944) IL AS=80+24= 4,
XIR APS BAMORE*, A 78= a+x
APOR-S-ACC, A 44?R=ABPQ=ACRQ=ŹSABC,
▲灣
* x(a-x)=fa
(4835) NUM, ACE142418.
51 444 A0 20 WAFÉ
AFAE
ABG及G
71 G14 GPLAB SABRÉP
您在 BC, CA 上海 BQCR AP
9.联PQ,QA, PP.緡APOR
S
(有一角等,面積比等於 「爽比乘積心比
10.份7至9方法得AP'G'R,則APQR及APQR均為好求 (討論) APQR-AABC為最小之內接正三角形 ()本題能份上題第二解法求出及桌?
第十二次預習題
(1)在任意三角形ABC中,剩餘弦定律或用其他方法证明 (a) COLA CAB
a
++
abc
or complete the statemon$0.
Ans: 1. (3)
2. (4)
3. (3)
6. 7. B.
4. (3)
5.
II. (a) What is a paradite?
(b) By the help of a labalian asagram, describe
the outer feature of a named plant paradite.
How is this' parasite adapted to ate mode of
文?
(a) A parasite is an organism which lives in or on
other living organisms on which it is dependent and to which it usually does no harm. For ita survival and life change partially or completely have to rely on the environment found within the body of that living organism.
In serious forme of parasitism 17 may dumage the body of the living organise which 1e called the host while in milder forms of parasitism the damage may be merely that it
occupies room on the host's body, but more usually it deprives its host of food or actually feeds on the body tissue of the hoet, in addition to giving off poisonous metabolic pvm - products.
-Scale leaf -Haustoria -Stem of
host plant
·Stem of
Dodder
Clusters of flowers.
Sternal Feature of Dodder
-Stam of Dodder
-Haustorium
-Vascular bundle
of host
-Vascular tissue
of Dakler A portion section (T.S.) of Stems of Dodder and its host
(b) Dodder is a plant parasite round on a variety of
green plants such as clover, Nettle and GorGA. Its thread-like pink stems twine round the stan of its host. A frequent intervals haustorij penetrate the host'a tissues and enable the parasite to obtain supplies of food and water. Although the pink flowers are so small, but each one has sepals, petals, five stamens and an overy formed from the bases of two carpels. The early life of the dodder as normal, germination resulting in the production of a root and a shoot, but, as soon as the shoot comes in contact with the stem of the host plant and penetrates it with huustoria. the root dies
Hy
The dodder is adapted to its mode of lifa by possessing the following important characteristics:-
(1) There is a reduction of xylem tissue in its stem, but this is no disadvantage, since it has not the ordinary full functions to carry out.
(未完轉入第四張第三頁)
(4) a+b+c= =(b+c)cosA+(c+a)cos B+ (a+b) C. (2)誕下列恆等式:
(a) (1+tan@+sec8) (1+tan 9−sec8)= 2 sín, 8 scAA.
Taino
SOMN ALCH - Zinô ·
(3) A ABC 4, * sin A: sine : sinC=-4;5:6.
(a)求tau垒六值(用根式表示)
(B)证明 C2A.
的某人在山脚測得山素之仰角為100肉山之斜坡 上前1000尺再測得仰角150設山坡之傾角萬!求 山之高。
(5)-氯球在A地正比,測得其仰角38*30%同時在, B 測之,其方位為M.从“W.B在A正東3000尺求 鼽球之腐
(6)設如右: AB=AC, AD=DC=CB
(a)求<CAB六度數
(B)證明右图中有兩個三角形槽似
(C) AC=4吋,求肥之長
(要用根式表示) (d) 利用上述结果,
A
D
C05365.