1969-03-06 — Page 23

育教僑華 頁三第張六第日八十月正年商已潘 WAH KIU YAT PO

CITYHALL 四期星日六月三年九六九一曆公年八十五國民善中

HOT LICONCE LIEKAM

·6 MAR 1969

郭日僑華

國際英文商科夜書院

本屆畢業生授憑典禮

名骸下:(一)工 按照典禮,各孤射業生,三月州日磔行新舊會員

一、端八、懷遠城、何、何油潑、歐年,列 「秀瑤、劉琦恩、換一、莫高、低芟 :簡新華、洪研濁、李、不爆密、汊少女、戰

·硭撼、葉、會,與會有纈它 屐、陳澄意、石皮、會所召開首次感委員 (二)際貿易明,任江发誓,落質行動議 秘 吳斧照

【院,於本月一日在該校,改會所,開本年 「很會嬴文商科夜一後,現正斮百思不肏游

* 何韻、張志彬

三明、閣茂泰、陳朵法、佈開會,並簡述本經有 隆颺治工作計

*、林廣珠宀初

【藏沙、陳小題、華月娥 近况,你说,應引 「級會計班陳小, 謅語奘試審總營會S

|| 華、李佩骖、徐麗明、「及美術組國籍,以便 獅氷龍、沁料、簡 將酚歌音樂組織

諦、陳志民、黃凡、事項,各委殷然發

一抠出强、作文、何

·

姿意見,當决

英、

該校規發開州牛

,商家部:工商

-國際貿易斑、會計班

及商科特訓班,英

會勞班。

、徽、招飾等體漭術工作 、湖、擔品、游藝、梅

,少文等分任

、莫祗機、位雜英、

,

說任歎待;(二)

{香港教師會

少文計怱增汊燕舰班;

(三)由行購環太

增設樂理班

医編訂;《四,推定丘

國際 香蕸各項粜。

(接第六張第二頁)

群界..陰割交糊來學校

聖玫瑰中英文校

亞幼拖時,由於富的校監主辦,以管落米格見

- 〔北訊)艘在角通类術寶玫瑰中英文小學

員生昨參觀本報

老向褟進

一

文中學會考試題預習

一九六九車中文中學會考試預習力

數學科

(十八)

喬抑強·

生物科

(十八)

摩百琴

第十七次預習題解答

pinna

Canal

Stapes

Tncus.

Malleus

External

auditory

Canal

oval Window Tympanum

Semi-circuitour.

Ampulla

Auditory

(1)寫出相似三角形的三個基本定理(不須証明) 答)相似三角形的基本定理為

(Q)两個三角形有二角彼此相等。

(B)两三角形有一角相等,而夹边成比例。

(C)两相三角形的三边成比例

則此两相三角形為相似形

'2) ABCD 第一平行四边形,由B引一直线典对角线 AC 於F典CD支於G,又與ADż延线交柊E求証

EFXGF = BF"

已知) ABCD為一平行四边形:

( * ) E F X GF=BF"

Window

•*•8) 1 j <FBC=LE,

Eust -tachian

Tympanic

Cavity

Vestibule

Chilea -Outer Ear.

Mid-

Inner

-A Diagram Showing the Structures of Human tar

The ear which is known as the auditory organ

In man, the ear consists essentially of three parts,

the outer ear, middle ear and inner ear.

(1) The quter ear: it includes the pinne and th external auditory canal, which is a tube

Leading to the tympanic membrane.

The middle ear it is situated on the inner side

of the tympanic membram. The cavity of the mid-

ear is called tympanic cavity and, is witnin the temporal bone. The three small bones in the

midmear are known as oss icles extending from the tympanum to the oval window. These three ossicles are the malleus, incus and sta pes reapectively. The handle of the malleus is. attached to the tympanum, its head forming a ▼ Joint with the incus. The terminal end of the incus articulates with the head of the stapes and the foot of the atapes is fixed by the membrane of the oval window, which opens into the vestibule,

The tympanio cavity is closed to the

outside air by the tympanung from the floor af this cavity, the Eustachian tube which connects the mid-ear with the pharynx.

(iii) The inner ear or the membranous labyrintii ;-

it lying on the outer side of the oval window floats in a liquid, the perilymph, which is contained within a cavity

The membranjus part of the inner ear

consists of two parts, the vestibule and the cochlea.

: The vestibule is the main chamber of the ear. From it arise three membranous semicircular canals in planes mutually at: right angles to each other. At the base of each; of these canals, there is a swelling part called the ampulla, and the branches of the auditory nerve are cohnected with them.

The cochlea, is a coiled tube, whien : contains the actual hearing organ called the organ of corti. This organ is made by Bensitive cilia, basilar membrane and the

orial membrane.

(b) The sacculus utriclus and the thres

semicircular canals are considered as the part

which the ear is concerned with the function of balance of the body.

by

In section, hairs arise from the inner wall of the ampulla and project into a fluid, the endolymph, acting as receptors for any plane movement of the head. When the head is moving in one plane, the fluid in the canal of the corresponding plane thus bending the hairs when the hairs are bent a nerve impulse is: sent from the hair cells to the brain. The

brain will control muscular movement necessary to keep the body in equilibrium.

Questions for this week:

1. Make a fully labelled drawing of the skin of a mammal as seen in vertical section and give an

Dunn t the funotion of its various components.f

State: the general functions of the skin of a mammal. What functions of the skin do you consider› are similar to the functions performed by the epidermis of a leaf of the flowering plant?

1 < F C B = L FAE,

{平行线的內錯角相等

APIBC)

AFCB AFAE(两AS有上角彼此相等)

BF:EF= FC:FA(相似三角形对应边成比例)

BILTİF AFCG IS A FAS

OF: BF FC: FA

(由1至3)

GF:BF= BF:EF(代换)

. EFXGF=BF*(外項積等於內項積) Q.E.D.

(3)差一角之两边為二平行线开截復作好成梯形之 两对角线,则两对角线之交奌此角之頂奌,及样形兩 底之中奖,凡四奌同在一直线上 已知) BCIIDE,BC=P莫為 M. DE=453NZI

(E)A, M, P, N@££#.

(証明) ). 聯AP 并延長之典

BC 支於K,央DEㄨㄢˇ於H,则

因BCIDE (已知)

(一直线平行於三角形之底, 截两腰成三角形典

原来三角形相似)

PD DE CB

一番相似三角形对应成比例,

DE

AD

同理可证 △ADE M A ABC,

B.C

AB

A PPH CO A PCK,

PD

一路

AR

DH

$123

AB BK

A ADH CS & ABK,

CB

LAD

TAB

CK= BK

愛之傳遞律)

M真典K奌重合 (一线段只有一中莫) 即: AP经遇 M矣,

7. 同理可證 AP之延线经遇N奌 (由1至6)

故A, M. P, N 四桌共线...

QED.

(4)由0园外C奌至此园作=切线 CA及CB48與

OCÈDE, L&D 31 – 14 § 34 PDQ, QJOC FS <PCQ.

(已知) CA, CB酋0园切线:PQ蔫遇 奌之任意弦 (**) COFS 4PCQ.

証明) 1. OA,OB,則 LOAC=LOBC=tL

(切线垂直於切奌半径) 2, 0, A, C, B ★B. (四边形对角相補) 3. CD DO-AD:DB

(园内交弦定理) 4.6034.

AD⭑DB = PD-DQ (相交弦定理)

5. CD* D0=PDxDQ (***)

C,POQ回奌共园(相交弦逆定理)

7. AK OP. OQ 01 OP=OQ (13% 0B**) 8.在。

OP=OQ (432 4*23A #4)

Q.E.D. 即CO平分∠PCQ.j

Q园中孤子对之國周角相等)

(5)設AB為E知园之直径,C為因上任意奌,遇C作为 线典由A,B两奌作之切线支持E及F.联BE及 AF之安吳尊 D.則 CD || AE.

(已知AB為O园直往,

AE, BF, EF始罵0园切线

其切莫备為 A, B, C,

(E) CDI AE.

(证明). EA, FB为┻AB (切线垂直於切莫半径) 2.EAII FB(同线密线互相平行) 3. AADE COs FDB: (平行於△底边之线,截两腰成之三角

形與原来三角形相似)

AE:FB= AD:FD(相似形对应地成比例

AE=EC,FB=FC(切线等長

EC:FC=AD:FD(代换)

CD II:AE (截三角形两腰或比例之线,平行於其底边).

Q.E.D

(6) 設△ABC之内切园心為工外接园心至联AI 延長之交0园於

(a) X SI ID=DB

DC.

(已知)I為△ABC之内心, 0骂

其外心,AT延线交0园於D. (haw) ID=DB=DC.

(证明)1.21=4元(内心足理)

3. CDB(等周角形

3_CD=DB (3A5** *)

)

4联BI, 則 <3=4

(因工為AABC之内心)

25=41=42(XBXBZA) b. 26=22+23 (4 |

7. 16=25+24 (474)

Fito) I

TD=B(对等角的比相等)

ID=DB=da.

(a)

(别联DO 延長之交园周於E,又通工作 IFL AB

A AIF COLEDB

I△ABC之内心,IFLAB.

DE為0园直径.

(E) AAIFAEDS

(E88) 1. AFI—rt.2

(垂线成直角》:

2. LEBD=rt, 4(+8001).

3.<>=<E(对同弧之园周角等)

4 st. a AIF es rt.DEDB (-ALAS)

(8)誕華

(c)若R, 九分别為△ABC之外半径及内半径,試

2R r = AIXID.

設証如題

,

ED=2R (B ED ZOB II)

(IF=h(因IF⊥AB,三角形内心是理,

AAIFAEDB R DB=ID [上面证明]

AT: ED=IF:DB(相似形对应地成比例”

ED×IF = AZXDB(內項積等於外嘎積)

RRYN AI ID(代换)

(d) B 10, x 3Œ IO*=R2-2Rr 已知)工〇备為△ABC之内心及外心

A,R各為内半径及外半征

(求证) TO R2-2RL 証明)1. EI, EA 則

LEAD=rt.(*BBAM)

2 GEAGED AD"

LEIS EA

(畢氏定理)

3. ET ED-AD2+ AI2(代换)

=(2R)-(AI+ID)+AI*

=4R2+(AI+ 2AIxID+ID*J+AX**

=4R-2AII ID-ID"

CE1*+ID=4R-2AIXID=4R-4RL (因上証 AIXID=2RL,等量公理) 5.在AEID中,

EI+ID=2(10+OD) (WRFID)

《代换 4R2=4Rλ=2(10+R2)

2R2 2RA - R2 =R-2RS

(1) △ABC中,若

̇第十八次預習題

COS A: Cos B: cos C

(2) *% seco=

++

A: sin B sinc =25:1947.

証

5:6:7.誠正]

3 seco + rano ==M

secd-tand之值差何?

問在何種情 (3)AABC中,若口,及A為已知! 况下,此三角形為两解題? 又治C及C,為此 两相三角形之第三边試c+cz=2&coA. 4)(Q) ABCD為任意四边形,对角线之交角爲日 ****** £ AC x BD sin 8. – × Ž A B C D Ž

凹四边形時,其结果則若何?!

()四边形之对角线為7吋16吋,面積為10方时, 問其对角线之交角為若干度?

(C)凸四边形ABCD中,

AB=60BC=

CD=DA=100", ∠A=80°試求其面積。

(5)者:AD△ABC之分角线,試証

Be ain A

AD= (b+c) ains

(6)一正四角錐体中,底為每边5吋正方形高6吋

試求其 (a) 表面積,(6)一稜與底之傾角

(7) t tan 6+tand=p,

̇式吉誕

seco +

tan += 82 px