1965-01-27 — Page 23

CITY HALL

真三第六

月二十年度院:

華僑教育

W

日七月一年五六九一催公年四十五四:

第七次習題解答

二九六五年度

論題預習専

T

WAD3AABC÷44 LTI1⁄2E,44 BE&ACK F@AF=£AC 【設証如題)

物理 科

(八)

陸永熾

1963中文中會試試題預習專欄 正明)通D作DGIIBF(选择外卖,可作其中程為

數學科

(八)

喬仲强。

(側)P篤定角xOY內一定過P引一直线典20两安拎A&B 兩使 OA: OB 為定比

(0km) P1⁄2¿xoy * £ §. m 2n

(4) P 31#ABOX OY

克於A, B而使 OA:OB

:九

(4) P17 P Q # O X Ž O Y ★ Q

2. m, n, PQ 2/8) ‡ QR.

Wit min=PQ: \ QR

JEQYL QB =QR.

4.聨BP等延長之興OX相変然A,則,AB為求

(*) OAB AQPB ( QP HOA, #1)

(討論)本題+有一解

2. DA:OB=QP : Q B = P Q : Q R = m;n (+84]*)$/16)

{例二)ABCD为任意四边形,遇对角线

ACTO | BD Z PPQ

AB,AD相交於PQ.則DP分原形為

兩個等積部分

(Jk te k z 11)

(正明)1.聨OB,OD(点定一銭)

2.

{AAGD = ACOD=AACD.

A

P

2.A ADG, ELADY§, · AF-FG

☆《通ᅀ一腰中夹,平行為之线,必令他腰)

3. MET A A B C F + & BD=DC++ FG=GC

4, AF=FG=GC_(41%)

MAF=FAC(

B

Definition

Solutiont

Quantity

Force

Unit

Q. E. D.

Force is that which changes, or tanda to change, a body's state of rest or of uniform motion in a straight line.

dyne

Energi

Energy is the capacity to perform work

Wra

(2)ABC為等腰直角三角形,

• j < B = x 47′′ E B B C ¢§‚££ 80$AE] 之

典AC於D試証AEB=ムDEC

(詖正如題)

【証明)1.遇C作BC线BD延线FE

4)=<2(+<AKB=30, tk ul, ca ba g

ZABK REN

3. AB=8C (W

4. staÀ BÉrt a BCF (~â£ã -12)",

S, LAEB=LF, BE÷CF (4).

6. EC=CF (EBC +

**

7. ZDCE=LDCF =45° ('' AB=BC, <BCF~ 50°)

8. ADCE BADCF ( 5. A. S. DESAEL)

Q.E. F.

13

10.

LAA0B=ACOB={BACB (=#={+4,7a={$#@#$*%)

3. AAOD+AAOB & COD +ACOB = (BACD+AACA) (4€ 1870) }

Ep Of ABOD CBOD = 2 A B C D

4. ABOD=ABPD (PQI BD, M★¥'iz as f*

5. CBODAC BD + A BOD == A CBD+A BPD=CBPD (4)

故D分ABCD笃两徊等我部分

Q.E.D 注意)(1)“遇四边形之一項桌,引一直线分原形為两等積部分” 之作图題,用本題方法解之,可以不需变ABCD為等精之

(2)BQ可分ABCD両個等積部分否?

(3)一頂奌引两直线分為三個等積部分”仍可用本法否? 例三)四边形两组对边乘積之和,若等於两对角线之積,财心点 Bè ‡ miens. (WA Ptolemy I#£££*£)

(E) AB-CD + ADBOAC.BD

(st) A, B, C_D o§§.

(作图証明)小說 ADB大於<BDC

19 DIF DE IF LADE=2BDC (8%<[=22)

DE IL BD DC=AD DE, *1*4

AE, CE. #1 SADE SABDO

(AS有一角等夹边成比例,則為相似)A

2 AD: BD=AE: BC (QM=HL) • [AD\BC=BD AE:

21423=62443 (***) BY LADB=2EDO

3.

¿DAE÷LDBC (*****)

4.

5. AD:DB=ED:DC (RE, ALX)

6 AADB AEIC (~14, R2])

7. DB: DC = AB: EC (3)

AB-CDE8D-EQ

8. ABCD + AD · BC = BD· EC + BD ·AE- BD/AE+EC) (+8)

AB⋅CD+AD BC BD-AC (1)

10. AETEC=AC (£4*)

三美藩在AC上(岩AEC成三角州,則・AE+ECSAC上明抵

12. LDAG=<DAE = LDBC ( AEC 3-4, NO. 3).

13. - A‚B.C.DEB (as Art✯e

ED

√例四)AB為半国之直;逾A,B作两弦AC,BD使相交於E奌,

**E AE ACT BE BD- •AB

(謖題)

(証明)週作EFLAB, BC,

LACB=Mt. L (#83 #1614)

3.. rt. A ACB cs at a AFE (<A ✩*)

D

BABAE AC, AƑ (‡a× #j *£) · AE·AC=AB-AF

外同理可証 BEBDBA BF (14)

6. AE-ACIBE BDAB-AF + AB· BF÷AB(AF+BF)=AB*(•)

Q. E.D.

MAE AC-AB-AF B, C, E, F XE MAZ

DEC=LF (A)*

A 4ABB=LDEC (4)

Q. E. D

(3)任意四边形两对边中央联线,不大於他两对边之和之半

(已知)E,F分别為AB,CD之中英

(*) EF Aπ ± (BC+AD)

(作图) 1.A.C並平分之柊口奌

2.0 OE AOF

'di. E AB + 1⁄2 (8*~)

2.0E-1 BC (AQP414)

3. * OF

4.

AD.

A

AOEFT, DE+OF YEFIA BLÅ, AH V =L)

Béc.

Definition of its unit

A dyne is the force which gives an döceler. ation of lem./neb./set. to a wet of 1 gå.

1 erg of energy in that which can perform the work of moving a body through 1 cm, distance, in direction of 1 nyne of force.

Rate of working of 1 erg/ant

Power

Pover is the rate of doing work.

SEA

Pat-

axis (1) The component of 20 lb.wt. which lies along the

parallel to the plane, P, is the effective force causing the body to slide;

P- 20 sin 30°

10 lb.wt,

Iriction opposing its mation;

t = 20% of 20 it..

- 4 lb.vt.

Total fores moving the body davw the plane i

- (10 - 4) 15.wt.

6 lb.vt. or 16 = 32) pd1.

- gāz

rácdeleration í) » sa.

• (6 * 32)/20

9.6 ft./sec./sec.

time of the flight 2.5 sec.

distance travelled

* * 9.6 × (2)o

- *30 ft.

(1) The force down the plane in 6 lb.vt.

(ii) The acceleration of the body is 9.6 ft./sec./se. (iii) The distance travelled is 30 ft.

24

(11)

• Force

P

Д

T

(BC+ AD)

(111)

}. £ 8c+ ± é > EF (4th) & £F<$(BCTAD) 討論)肴の EF域上剣=054 故 不大+(BC+AD)

"(4) @ ▲ A B C & F 1Z ABC MATS D, 11 DE LAB, OF LACTI

AE=AP =÷(ABTAC).. BE=CFS(AB MAC). ($4·10).

Ane.

(証明) 11AD BD, CD (两点定一直傳)

2. 21=22 (HVX2 B®Â¥)

3. rt. AADE certa ADF (-923), did)

14. DE=DF, AE-AF (ĦALL)

5. X# ADEB DECE

BD= CD (ØD.

6. MADEB SAMADFC (RHS)

7. BE=CF (RE)

d. AB== AE+ BE, ACEAF-OF (f**240)

9. AB+AC=AE+BE+ AF - CF = 2A6=2AF (7§ davo

• AE = APL(AB+AC)

10 AB-AC

C = AE+ BE - (AF-C^)=286=2CF ($£40%) BE≈ OF −4 (AB-AC)

Q.E.D.

1. 本题中D奌為△ABC中之<A平分线與B中岳线送交奌,故生 D至 AB, AC地躲足 E及F一在AB边上,一在AG延线上合

2如岳E尾店的錯在延上剣嶌読法隊用

以引数証明“任意三角形必等腰ż錯誤结票

(5)四边形内楼於同兄外切於他国,则两对造切奌錢必 互相垂直

(已知) ABCD外切於0园,又内接栏

ABC, E, F, G, HA Auto

(###) EGLFH

(STR). THE, EF, FG

(两类决定一直线)

1.443=26=21,

(弦切角等於央弧时对园周角)

314A+24+£6=180.

1x2 +63 + 25 = 180 (==) Auto 180°)

4. <A+ a<2+<c+24=360 ($£**** 1€ **

#180(內楼四边形對角相補!

AC BD延長銭相刻控國外区美時本題仍適合店?語】Atic 者可式作証目

第八次練習題

1)P為定角XOY内定奌,求遇ㄗ引直线交OX, OY於A.B而使PA, PB蕎定比(参者側)

2)求作一直线使典三角形一边垂直,而將之分属两徊等積部分 (3)PQ分別為两同心圓之大园及小园上任意类,大园直狸 AB

CD. SE PC+PD-QA+QB

پر

2 62+24) = }}0 ($m)

LEPF-90a (amā sve že 29)

2qFH (11*2+1)

QE.D

"(6) 0,P MERI A, 8. £LE A 12-£W CDAABY OLD:A

使Ò定長线段老並討論其解香

(4)設0萬△ABC內任意奌;AO, BO, CO延長线文三角形各边(已粒) OP炳园安 AB KP, Q, R. XI

OQ

(5)ABCD 正方形延長CDEP益CD上任意解日本 工作其垂线PQLADE平分线交於Q、試証PB=PQ. (6):中美正三角形外一奌,由ㄗ至三顶美聯线中。盖两聯线 秘等於第三线,則P真在此三角形之外接园上。(參考例三) (B

1

(##) LAU ÎI CD PAR

* CD-

(分析)讀出暠所求之直线 分割遇已半作CD大会 OE,PF,則

¿CEEA, AR=FO

(税抜)

た

OPII 201¥## POI DE HOLI PQ=88; 2PQ0mst. 2)

由得作

(作法)雞p【盐以之高直程作半阅,

(P(RO)BE,AL*£*#*#***E*Z(#2)

-

- ut int

{u - 01

When the body is at rest on the top of the plane, it posÑESAÜS

Owing

a potential energy of 300 ft-lb.t. (30 x 30 min 30.),

to the roughness of the plane 120 ft.lb.t. (0.2 x 20x30) 63 energy is used to overcome it. Thus, the kinetic energy possessing by the body when it reaches the "bottom is. 180 ft.lbst. 00 - 120)

Solution

Kinetic casrgy to the energy which a body bas by reason itą motion. It is a staler quanifty, While as for mencašu is a qusätity of motion, mod is measured by the product of 1 mavo and its volucity, It invester quastity.

Kinetis energj...»

Potential energy

(dyne-sao.)

When a number of objecta collide or interact with each other, their totai momentum remains constant, provided no external forces. are acting. This is known as the law of Conservation of Momentum,

Let

be the man of the truck A

By be the mass of the truck B. stationary, one.

P.E. of track A

E.E. of truck 1

.

+320 .80

Energy 16SS =

- 6400 It.ib,c,"

*** 320 x

...

30 ft.par.

† x 320/32 × 303 \zt.1b;wt;) 4500 ft.lb; urba

(6400 × 4900) It.lb.wt.

1900 ft.lb wg j

By the Law of Conservation of Mementus,

Sum of moneata before collide

+ g

where a

ولا

' of moneata after impact

- vel. of, truok A before inpset

30 ft./soc.

40 (at rest)

comson valocity after 'collision-

12 ft./sec.

thus, 320 30 + 0°

- 12 (320

480

Ans. Energy lost on the way is 1900 ft.lb.vi.

Weight of the stationary truck is 480 lb,vz

Questions to be univered

1. (a) A 200-gm, Body starts from rest and slides down a smooth, inolined pläne, If it travais 120 cm, during the "third: moond, what is the angle of inclination of the plans? (b) A block weighing 8 lb. resting on a horizontal surface 1.

connected by a cord passing over a light frictionless pulls to a hanging block weighing 8 lb. The coefficient of friction between the block and the horizontal surface 1. 0.5. Find the tension in the cord and the acceleration of ench block.

2, (a) Give the meaning of each of the terms mechanio al advantage,

velocity ration, efficiency, of a simple machine, without mentioning the other two. Derive the relation between thear (b) Explain the principle of the hydraulic press or jack,

illustrating -your answer by a diagram.

(a) In a form of hydraulic press the bare of the pump is 1in,

and that of the main cylinder 1 ft. The pump is operated

lover afvelocity ratió 5. What is the velocity ratio of the whole preser 'If the officiency in 85 per cent, what is the greatest forse the press can exert when a force of

by

5 lb.wb. is applied to the end of the lever?

3. 34 TQ (M0Q'}}

4.1 A15 CD (C'D) A£FIFA PQ (#0Q")

(###) (1) OP>*LEZULTAN (2)_0P÷÷2 0# – **

Ju) op <+2#t, **

(注意)本题中CD线級 何時為氨長?