1968-06-10 — Page 15

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教育僑華

頁三第四鰽

日五十月五年中戉言

WAH KIU YAT PO #

B

英中會考數學(三)答案

<PAQ=*< BPE(

(續)

歐陽鋊文

SUGGESTED ANSWERS

MATHEMATICS

PAPER

GEOMETRY

Section 8 ( 60 marks)

« FQA” « PBE{="90"),

PARE PEB. YAQ-PB

期星

Corr 4, PEIAZ,

A.A.S

(il)" Given:" "St." "I'mes AB, AC_intersect ̄at*54",

To Constructia a perp PQ such that PQL AC

( where P lies on AB, Q lies on AC) and that AP + AQ=35-

To... Messum

AQ

日十月六年八六九一座公年七十五國民幕中

くくし

*#$

%$#

S#ES

*#$%

$#$%$#3%$#&%$#$%$#$%$#$%$%$#*#*#*#$%$#578#3%%$#$%$$%$#3%#%$# $#$%$#$%$##5%$#&%8 #$%$#$%$#$%$#5%8 #3%$#$%$#$%$#$%$#$%$#

英中會考中史科重考試題解答(續完)

H

盧梅庵,

79 粗

Given: ABCDEFGH is a regular octagon

"To Prove:

Proof

12,

And

inscribed in circle, centre PB. PE O

O

tangents at B, E resp.

BCG EDG QM

BQ = P Q

st lines

A B C D E F G M 14 a regular octagon (Give)

$ of (2x8~*}* ŷ o * = { 3 5

Each int 4 C

of 360° = 45*

£4, 160* = 45" (ragular

tangent

Each ext

4B0C

GP is F

800 of 45°

← alt. Seg.

BC=ED 2 C

(for «QCD# <QPC )

28-88

CD / BE

BCDE A

an

equal rätis

isos trapezium

<EBP < & BQ + b

PB. PE

* + * = <> P

Subst

Given

Neasurment: AQ ZA93"

Given⋅ In 00, AB is a diameter

+

chard PQ cũk AB at T Tasawwego aƒ»x*

To find the as of PAQ in terms of To prove RQ > AQ > AP

Proct

Bin of

Denote « POT. -dot by

radii OP # 04 LOMQ==OQPmx° baseas. In*a OPQ 1_0, +0 ̧≈ 180°*~ - OPQ ~ -ORP

tangents

isos

length of tang

[IN MOTQ

= 674*

ame

PB PE APSE

སྙོན, 11: the vest. < EPB is 180°– 67£*{

A PBQ 3 A PEQ (225)

† - £ - BPE = aa4*,

BQ * FR

Hence

Hent

`ssible gym wie's

= 180°- 45". x*

=== {38°- X"

• Sum of

Q = ( 180*~ 2 x *) − ( 158°~ a *)

= 41°- 1'.

is the centre of the airclt.

・ザー

- 226-4

ZAPA == ± of On (ast)

{Ix^c_APQ,

INNONG

by

of a st

= ± of (138*- x*). *671*-*

•2 Sum of 4', we find «PAQ = 180° - ( <PQA + £APG)

₩ 180° - (6742 +

= 186* - (94°-4

(5) BELLE BESLUTKDEZEA !

WE

OR PUTEA 206HEL · CHR› › #-## › S 將紙辛,始同機號,將自然,大床而大月氏,搜終不潯裹而選,後叉建交所採 可奴有坪,武帝遺駕車主鳥品·梵烏來,從遞關將至大苑,大月氏,無,于

WE » PASIUNE : ATLEE* •

MAPEHUB• 777SHUKLA MEREÆRZTE MESIN

• EF*· · · · paste, VA AN; PAZAR · KRÓ 器等先,其多署於物流之耳塞,文化之流者,並非海鮮也*

(6) BEREZ±4 · VNUEVARRAZEM !

筆:驗(七六九——八二四)字退之,南人,其先世潔,故世稱輸自黎,爲人性明鏡 1志行堅正•德宗時舉進士,累官吏部侍郎在朝藥廠,上極爺宮市之興:際樂時族迎 轉彎,管坐眨,所玄有政、潮州刺史時,民尤襁之,帶年五十有七,餘日文,國艾 誼交公,還有有「韓峊定全條」

韓承於六經百樂,原不實,蔉文然換本源,問深奧街,卓然自成一諛,後人推爲層、宋 AATZEŻE · EREBRNEKAZE]· EZERREA - HEALETU · S FAT › -DATESI › KANN› TEST.#0##› AHTSIREE › E 「文以數濫」,排不斷偶,不遺急。以說:「非三代两漠之零不敢觀,非聖人之 存」苦心確,不無關,於交還不變,而古文一證,逐城大盛。其後宮古文者,我 VER · MEEX4TSSLEZATZAZE'

T

(7) 王莽篡演欲,即託言放制,但結果失敗,試簡述其改制之內容及失敗之原因!

答:王明野心,更民跟僵,比之周公師代滿,於是托古政局,厲行政,其改革者約有

BOR:

(一)土地之問有也:六下田畝盡有,名曰「生的」,授予拼老而不符買賣*

(二)榮制之改革也:廢適用鉄感,尤均等品為代,謂之「寶貨』。

(三)沉均六臂之推行也;於風安、洛頭、邯鄲、臨酯、宛、成都路大都市殻五均市官,

以平抑物價;而以隱、誠、酒、山陶子及郷錢、除黨等收費別痎•

(四)賃資之雞等也:國家以下均屬您與寬爲保金,貸臘民之欲營

年利十之一;如與自己之,則息子口

(我)奴婢解放也令天下之奴婢改命法,確止買沒口

者,年息不過

王菲之改制,旨在均貧富,抑供,用密原非不错,我所以終令款者,殆由於求成急 ,既無大無實施之對面,讓證實先後之步驟,使民難以適憾世·確全之政治組織,固付 關妮,卽爆行之中下人員,刃感不足,致政治類雜費現世。從古簿向其,會都不 通民情,不切世務,認竄倒攬,然而不為,但使民慎趣也。改制能遭立法而您於執行, 以您「制作宗而天下平」,於朝令夕改,敏感或觸法望也•夷慮識失當,匈奴谳 句鏴相鬰振新,新兵進軍遠征,師老無功,民日困,於是新制未見其效,社會經 *DAMAR •

(&) 何旟「戊戌新政」?說明其內容及其重要之影響!

鬱;甲午,有識之士,深知非別謀復興之锋,不足以救宇,於是新法之起出澗 主張暴力者,則南海憲有用也時密宗三復親政,于術世家,兩欲發奮理:第嶼顧朝臣 *潔皆顯守舊,無足與謀者,於是滿格擺用度有關,以新重任。光绪二十四年(一 八九八),戊戌四月,正式下詔豬新,史稱「虎成新政!.

自是新政部甜,凡數十下,如:有關政治者之崧汰員及校座,就勵民上職,許線 *者批評時政;有軍事之廢刀而欲試桕泡,今次邂奮,厲行保甲;有教育者之 緊杁發而放試論,開立學堂,教牾館,辦報紙;有關賞弟老之開辦銀行,設立續發

、潀陷、長、工、南等局,無勵製造發明,一時需属行,頗有新燃也。

*新政推行涂,襲克艾多反對,魏泰右鄰太后復出欲,因德宗,捕殺新黨,靡

遠東減,新政蹇靜,而影響大:(一)新政失敗,識者已深悟胥轴無能之清廷,决不 朗再者以旋乾轉坤之希望,欲免瓜分之軀,舍武力革命,另立新政府不爲功,因而有孫中 山先生創建中哪民國之積極行業也*(二)康、梁站亡源,陳長慈稍太后之拖霖新政! 於是組織保皇黨,擁謨憶京主城,我有後來君主立意之施行也。(三)縻、榮之相打 外,鱗於獎、日之協助,列賬至反對滑望之立,致無際太后及滿州話遇貴一致仇外,因 FFBOR · #H^DBIŻA** __(5)

ES

H

Given: AABC with pt. H. Kon BC

Such that

BH= KC (BH<BK) Circle. AHK Cuts AB: at E. AC at E.

BE BACF CA

To Prove : 0)

Un

Proot. Son 6:

Tom C

if £F #80_then B6 = CF.

BE- BA - BH. BK

84 = CA

BR = CH + HRK

BE BACK CH CA CH= OF· CA BEBAS CE CA

Secondly, if EF II BC, then

Einftstefing Chowa) (given)

(subst.)

Cintersecting Chardi

Subst

m 4480:

LIBE

> PRA

In a▲, the greater « has the greater

side opposite to it"

PQ > AQ > AP

** SHÚC + A PÁS

DAC

DE

1.

ABCD

ADAC

DE

}

LADOX BEY L DE and BXIDX (Given), EY # BX

JA A, DAY, DEY !

*- b

!com.4 EY//EX,

<D is in common ADRYADBX

DA

A. A Similar A,

!

ty fence,wBrad ABCD

ADAC

EY

16,

AE =

EB

AE DEB AFFEC

Fe

AB EG

AL FC

From (1) By division.

AS FC

8 #

AC - eA

# A

AB- FC BEBA

F-CA

SF CA

E

ВЕ

<F

85

Of

12

BE

Given: As shown in the tig

To prove Proof

PB = AQ

a. I am a me

a1 = e PAPE PAQ, BIE

PAM PL

E

Prop for GFARC

B

Given Qund. ABCD with ding, meet at E

BX LCD.

(To Prove: (i)

EYJCD

* AB¢ : «JĄC = BE · DE

is Quad. ABCD › A DAC - BX: EY Proof : Drop AM LBD, CN I BD.

AABE - AM - BE AADE +AM. DE

SALE » GE

• ADE

[Similarly.

a CBE

ta

Given As shown in the fig

To Prove: (4) BK, T, H 24 consyalië, dw $f PA = QA = AB then;

<HTK = 16′′

AB= AT

Proof) ABHQ is a*cyclo{quad Given.

And

a = <H Q = 1

H

BHTK OM concu

AP AB = AG

Pext.c, cycki good, 4, in some ing

extint app

Given

(4.0)

Gifre

OM6% PEHLAC

C/;

1 Sides opp = d3

proved

• col = ± ON DE

ACBE

BE

H

OCDE

DE

From <a> and <b>

Ac

DADE

ADE

ARBE + A CBE GADE + A CDE

DE

- BE (Ratio)

BE

все

ABC

fromtis, we find

ABAC

.DE.

ABBC ADRC

*

+

Centre A, with radius AP, doaut a civoli: Then, the circle passes though P. Bland &') of which

#

PAQ is a diàmet, #_{\PAQ in H) PBQ is a semicircle

BHYK is a cyclic quod proved

LHTK -

PER-90*

QTH is a st. line

<PTQ+

PTH = 180°

in semicicle

ext.4. Gyak

guest

given

proved

4PTH = 90*

<PTQ=</BQ = 90°)

PBT, Q. Are comoyolic eliks on the semicirle PBQ

AB AT

radit

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