貞三第張三第
日十二月正年午芮歷复
WAH KIU YAT PO
中中證試題預習專欄
華僑教育|
OEF
(附註)如ABC三莫在一直线上, 則结果如右图日园半穰芮AC AB** BC #12 AB
數學科
(七)
梁覽
(E) ABAC 為兩推音直线,包括冠器
第五次預習問題解答
求作一园典一已知园相切目典两相交直线相切 已知AB AC為两相百直线。因為已知园
解析過A作ADI.AB.並截
(解析)設如上图中园為求得之园,則要必在2BAC之分角线
ADIOPE PELAB 并联OP 則PO=PE+0园種(外相切時 又延長PE至F,使EF=0國半授B即 Po=PF
又過大作 HK | AB 則奌之軌跡為興定要0及定直线 HK等 距離真不軌跡但此執亦為以口為鯊吳(Focus) HK為準讜 (Dureetrix)=抛物线非普通?歐惠曾德(Enolia)我何時能解 之問題
不過P真必在AD上放在DA其延待上取通宜
ONI HK QMI PO, HOㄗ於M.則因PO=PF
由是M要求将之後ㄗ吴亦可以求出
(作法)作人CAB之分角线 AD
2.作AB之平行线HK,使與AB 距離:
等於0回半徑,并诚DA之延長线:
18 ARE AN 1 HR
QN=
SAAN 131⁄2 & HOM MRM" 218 AM 2AM! 6.15 0 AE AMB AM 69 #11 13 AD PAP
我遇P或PAKAB的垂线支AB柃E及E孟实 HK&于F及F
9. 中国及下图均美丽求
= HA: HT.
RAM: PO HA: HP
AMA AN (作法):
AHI PO)
PF =PO(两比例式有三相當項相等第四相當顯如等) PE = P园半社 EF=0园半狸(作法 平行线間距離1
7. PA至ABAC等距離(分角结上实至两边等距離)
8. P国切於 AB 及 AC (~线奖园心距離等於半程,鸟菜园协)
(討論)讀者試依第四次習題第一題结果討論之,
QE.F.
同理過A作AKIAC AA
KY典AB 於Y,則又為適合,
3.AD=AKARAY
葛西求軌跡部分(読者可留電等護三角形底辺と佐一 予西腰で距離嶌知第一定島等
3
*高的練習題)
32 P/F PELAB PELAC XZ4 EP É DI HIF' §] PF - PF (: nt. a PXF = rt a PXF") ? Å PE + PF #RE+PF'
「故分角线亦為軌跡之ㄧ部分 作法1.你AB的平行线 DX使典旧 的距離AD=2,並設DXFACEX.
2 作 AC的平行线KY使典AC的距
5. 结经XY及半射线 XV, YZ ±P =
PEAB PFLAC, RLX44XHLAB BE AP §JA
1. AD=AK = 2 (48)
B: AX=AY(对应边)
« J^AXY=1 AY-XH
(因-DA8-2CAB=-FAC-<BAC...
@APX=—AX DE =— AY PE
@ APY = AY PE 《三角形面積等於底梨高之半) 5:DAIYFAAPL+△APY(全量等於諸分量之和)
6. LAY TH=1⁄2AY PF+AY PE· PE+PF=x8=L(EIA)
7. £ ? § # × * * Y2 £ £ of PE ?F=&_ (^188*8)
故軌跡上任意奌適合條件
· AP KP IP 并通X作 XHIAY时困
J& PE+PF = {= xH(軌跡條件 平行结間距離) 2吉AYPE+吉AY PF=EAY KH(等量合理:
AY PE+AX PE=2AY XH (#25)
By AAPY A APX = SAXY
42 QAPI‡6APX+ ¿PAY = AAXY (3$ ¥1+ft
OPXY * 面積為零(等景公理)
J設中為適合條件之要&P要在<BAC外遇XXHLAT
1. PE+PF = 2=xH(軌跡條件,平行线間距離 蘿相等)
RONG
日九月二年六六九一夏公年五十五阐民赛中
道週會健
暴行 《琳》
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一、玉二氏洗裏,異 两氏年來分別在
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宴梁國材 王劍鋒
劍食
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英中会考試題預習專欄
化學科 (七)
CHEMISTRY (6)
a) Copper carbonate. Cu *
(5) Lead carbonate, P
(e) Amonium carbonate. (Why)2 091
(a) Sodium carbonatḥ. Naz 20
王錦釗!
Q.2 (414herical change that is influenced by light is the combination
of hydrogen and chlorine to form nydrogen chloride Chlorine and hydrogen mixed in approximately equal volumes will combinë stowly and quietly in ordinary daylight to for nydrogen chlorian
If the mixturexposed to oright swilight, the mixture mik
Ib) Another example is the maction of chlorine, will methane In
the presence of ltraviolet light or diffused sunlight, a mixture of chloro-gerivatives is formed)
CHL
之
But the mixture la exposed to bright sun light, the mixture Mall Sxplode «forming hydrochloride and carbon.
(a) It is used in the manufacture of sulphuric acid:
b) It is used to vilcanise rubber for the making of motor car tyres.
fo) It is used to make calcium bisulphite, which is used to pisach"
"wood pulp in the paper industry,
(d) it is used in the manufacture of maconne aitu BAJKERMAN (a) (1) oxygen: a) nitrogen (11) carbon monoxide.
(b) (ii) carbon dioxane, (ii) sulphur dioxide, (111) nitrogen dioxide ca) Copper can be obtained from black copper oride: ay passing a
gaseous reducing agent such as nydrogen, over the red not oxide.
The coppers' sulohace is dissolved in water. Some powdered zinc L群 added to the aqueous solution, and the mixture is thoroughly. shaken. The zinc will dispiace metallic copper, which is red in colour, from the solution:
USD; Zh⚫ Cu ZHC1 `* {0} • H2O • Clg
2 litres of
-HO
22.4 litres
·AL NIT. P.
Volume of onlorine, at H.T.P. that can be obtained from 2 Litres of N-HC3- * 22.4 litres
兹並附典。园内切的P图如上图开示 AH等於口国半方 為<CAB的分角线PO典AM同向平行文AM=AH.
(2)以三已知莫為园心作三互切园
3. EF=XH (1AX PA 1638) A:PF=PF(等量相减!
5 84 PX 81.4 TFXxt. OPEX.
Le
DF N-HCI W 11,2. litres
of Hal 1.12 11üres
(b) Ca CO
2HC) (40+12+16 × 3). .100 g..
CaCl2
22.4 litres * N/TP.
<PXF=<px#"(对应】
PLAXD Z111 xv
(一角。有一分角线)
综合上述(a)(b)两奌適合條件卖在飘球上
AD
(附註)),若两建线PEPF
配啄鸟四條“半射线”
Volume of CO2 at NT.P. that can be obtained from 100 gm CaCO3
22.4 litres..
2.24 litres.
10 gm CaCO
(a) The gas given off is either oxygen or nitrous oxide. If the
gas is nitrous oxide, then would be ammonium nitrate, which. on heating would leave no residue. The gas given off, therefore is oxygen and, x is either potassium chlorate, potassium or sodium 2 rates an
Equationsı» | 2K010y| 28C3-30.
2KND
* 2KNO2 02
(270) A B C = $
(求作):以A,B,C三奌為园心
「解析)設如右图:ABC 香旰求园
PF為切英則 ADB BEC CFA的為 直线且AD=AF BD= BE CF = CE
.可用联立方程式以求出值(代數解洁)
(47) + AB SC CA
Að BC CAZUL L L 18]$ D E F 3.分别以ABC為园心 AD86
4 A, B, C
bas
(证明) ADE AF-B眙= BE LE=CF
AB = AD1BD = ADI BE (金量等
XMXSYNYT.
讀者可自行研討之
本題如此解答,仍有錯漏否?
「第一次預習試題
2.
(美街之 H.GF. 及tc.m.問題)
茶數除 34100則剌餅63.1
除307100 則剩餘 93.中某數
(2)有ABC三船,同時同地同方向出發,環繞周圍為21哩之小复 航行A時速6哩B時速80哩(時速10哩間三船出發後第一次相 會要幾小時?又各行熊週
(3)一條大路,左李每隔24尺種梧桐一株大旁每隔这尺櫝 楊柳一株关梧桐典楼柳相对的有78株求大路的長又梧 *相與楊柳各有幾株
14)有四位數署中個他們的史
(5)三連續數的乘積花35404
(6)两默的HCF為情
19,有一数,其实 3420,3600 Anc. F. 為36而其术Cm总给 1來此數(明因子連束穫表之
(8)以一整數案 308308 所得之路-完全平方問以數最小差? (9)有三间两位數其HCFA85 萬560顫
(b) When dilute hydrochloric acio is added to the residue obtained
from the thermaldecomposition of potassium chlorate, uners: would be no chemical reaïion, as nyurochloric acid has no seriam on potassium chloride, Dilutė nydrochloric acid reacts with nitrites to farm nitrous acio which decomposes rapidly in one cold, evolving nitric oxide and jeaving dilute nitric oxide. The nitric oxipe given off turns into brown alprogen alozjue on coking into contact with air.7.
SHNO2
2N02
2ND
This Indicates that a le either polissing of 1001um nitrate
Potassium balta give N
@itiać, estour to the Dunsani LAIDE,
Therefore in potassium nitrate, KNO',
Questions Fur next week.
1. Describe in detail how you would prepare Litre of normal sodiim
carbonate solution if you were provided with some pure, but not
moisture: free, sodium carbonate and livre measuring flask:
would you do if only sodium bicarponate was avoidable?
2. A solution of sodium hydroxide contains 5. Waoн oer liure of 14 neutralise 27.5 ml. of solution of sulomuric acid.. Calculatet
(a) the normality of the sodium nydroxide solution.
to the concentration of the acid solution in grame of sulphuric acid,
per litre.
(5) the weight of sodium carbonate, say Cowich donte se dissolved
(a) the volume of carbon dioxide collected as 17°C and 750 am pressure,
whsen would be evolved by complete solution of inta wesent of
Tự) the weight of sodium sulphate-arystals Naŋ 50, 110 H2 0 mLSO
- could be obtained..
19.1