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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
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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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