INVES
1968
台大初月四申慶買
英文中學會考試題預習專欄
生物科
(廿七)
•
廖百酵。
Biology (27)
..
Answer to the exercise of last week:
1. This is a diagram of a single cell of the Spirogyra,'
WAH KIU YAT PO
{Qamations for this wel
1. (a) What do you understand by regetative
propagation?
(b) Give a brief account of advantages and
disadvantages of vegetativa propagation.
2. State the type, rood-storage and method of
四湖县
日二月五年八六九一慶公年七十五國民中
1968
言教事
式題颈習書滴
數學科 數學科
(廿七
(**)
·喬仲強。
番伸弘
reproduction in sach of the following vegetative" reproductive organs iliustrated in the sketches and label the parts in saab sketes indicated by, letters.
(x)
第廿六次預習題解答
(1)以任意三角形ABC的两边 AB, AC边向形外作正方形
ABDE & ACFG,
{$L$&•£* H)
BE+ C&*— EQ2+ BC*
An
(a) label the parts from a to 0.
F
(b) State briefly, the nutrition of the Spriregyja,
1 A. Nucleus, 8. Cellulose cell-wa 11,
C. Spiral chloroplast D. Prenoid, E. Cytoplasm F. Cytoplasmic strand G. Vacuole
(b) The spirogyra has a holophytic type of
nutrition. It obtains its water supply and
dissolved carbon dioxide from the water in which / it lives. The spiral chloroplasts contained in each sell of this plant enables it to carry out the process of photosynthesis in which carbol, ka hydrates are made from vater and carbon dioxide, In addition to the building up of ourbokytra tas, the plant is also able to mke its own proteins by combining the dissolved mineral salts obtained from the water and the photosynthetie product.
The drawing below is a winter twig of a deciduous
1 B
{
A
做
Type
Food-storage
Method of Reproduction
[(註明))
1. <BAE= st.2,
<CAG=rt.4, AE=AB,
AG=AC
(正方形各角為
直角,各论相等)
B
2. BE*— A6*+ A8—2A6?
Label:
୫ C
D
Type
Food-atóra ge
Method of Reproduction
Label:
CQ*— A&*+ AC”—ZAG^
(華氏理代換)
3. "<BAC+ 28AE+LE AQ+<CAG — 360° ( MAŁ¤)
4 4BAC+LEAG 1am 180° (@<8AE+¿CAG =90°)
5. #4BACT. BC=AB+AC-ZAB× AC on < BAC
★AEAG☀, EGTMTM AE*+AG*ZAE»AGON LEAG (#k£4)
6.
CH LBAC == - Cot LEAG ( § 4 BAC=180*~ LEAG)
7. BC= AE*+ AG”+ 2AE× AG CLEA@ (***}
8
9.
EG2+BC= ZAE*+2AQ' ($1»<«)
BE*+¢6*— Ea2+Bc2 (RÌ
Q.E.D
(2) OA, OBŽO§ M#141, CDŽ☀ AB PÍTÓ93E,S
OA, 08 ERF. ✯ CE*+DF*+ CD-EF= =A8*
(Zkm) OA10B, CDILAB
(**) CEaDF2+ CD × 6F =A8,
((STE®8) 1.1% 04$ OG LCD, $1
CG = DG =Ź CD.
(因心至弦之瘗线,甲分比弦)
2. 4A48=45*
【等膳三角形角材等又
LA08=90°)
3 LOEF=LA=45
LOFE. B za 44.5*
QUAB, #4ƒ1⁄4À'è@#A¥)
GE ·OF (**A LOEF = 40FE=45") OG H LEOF 27SW (¶Ñ‹Ã££¿Â FÓTIA).
BY LEOG¿F0Q ==== 45*
6. OG EA=FG (H*AZIZ)
7. CG-EG-DA-FG
(等量相減)
(*).
the parts indicated by guide lines.
Explain the meaning of each of the following terms:
(1) a terminal búd
(141) & lateral bud
(V) a node and internode.
4. Terminal bad esale, 8. Torginal but
(11),an axillary bud
(2x) u dost naár
Type
HONG KONGY
6. Intermede,
D.
bud
S. LONG BORE,
F. Lentiɑel.
G. Vascular strand,
Girdle agar left on
the node.
b) (1) A bud which grows at the apex of a sten or
(11) 4 bid which grows between a leaf and shom, (111) A bud which grows at the axil of a stem or
(19) 10000 4. laur falls and leaves a space cu tan
twig, and this space is called leaf soar, (v) Á mëdo is that spion of the stem from whian,
branabas, leaves, flowers, fruits and buda arise. An internode is the space of the stem be tareas two nodes.
Compare în a table form, the differences betareen
stens and a root.
Ston
(a)
Root
The root bears no lea VES or flowers,
(a)
1. The stem bears isaves
and flowers.
2. Nodes and internodað
are present.
3. It bears axillary buds.
4. It usually grows above
Nodes and internod
are absent.
There is no axillary bud.
It usually grows below
the surface of the ground, the surface of the ground,
5. Its epidermis covered
by a cuticle layer.
6. Stems of herbaceous are
usually green.
7. The stem hairs, if
present, are
multicellular.
. Stomta or lenticels
are present,
9. The growing point is
protected by bud scales,
10. The branches of the stem
-originate from the -
internal buda sugaro ficially,
There is no autisïe layer on the spidermia,
The root is never green,
The root hairs are unicellular,
No stomata or lenticals,
The growing point is protected by the root cap.
The side roots originate from the persoyale internally,
Food-storage
Method of Reproduction
Туре Food-store go
E
F
By
CE= DE
8 oc. 41 OC2= OG*+ CG' (#&%*)
OC*= 04*+ ( CE + EG)*
EG+ CE*+ 66a + 2C8 × E4
— CE2+ZEG (EG+CE)
R$Y TIE
BRAKI
CE*+ 2EG X CA
CE*+ CD × 64
OC*+oD":
OD*— DÉ*+ CD*FG (8),
*== CE*+ CD × EG + DF*+CD=FG (¶¶nde)
CE*+DF2+ CD ( EG+FG)
= CE2+DF+ CD = EF
#1. AB** OÀˆ+OB2 = OC2+OD* (442¤ ®a¥13)
12.
6*+DF*+CD• EF = AB*
Q. E.D.
(3)老H晢△ABC 的垂心,又為△ABC的外擭园直往,則
VERS LIBE
Hethod of Reproduction
Туре
bež/
Food-storage
Method of deproduction
Label:
A
A
C
C
AH+BC=BH*+ CA= CH*+ A8*= d3 (設証如題 (証明)1.聨BO並延長之
0园於K,又联AK及CK,則
<BCK = rt, ¿ (对直径园周角茑直角
2. ADLBC
F
(@HAABC #10)
3. AHH KC(ARI BC)
4 同理可証 CHIKA.
(223)
B
5. AHKC (mínidlė 11)
6 AH=CK (~*2**)
‚7. £ 8CK $, CK*+ BC*===
A
BK*(畢氏定理)
8. AH*+BC2 = dˆ (H★★ ̧ BK Ž 0 B Í 43 )
K
9. *TE BH2+ CAˆ=d3‚_CH*+ AB2=d2 (↳ 128)
10. AH*+BC*— BH°+ CÁˆ200
• CM'+ ABˆ— d2 (#1+#A#)}]
Q.E.D
(註)本題亦可過口作BC士垂线以証明之
̈¦ AE : @ D-= ( AB+ AC); BC
\($£$£•£1
(*) 1. ¡AE: ED= AB: BP,
BD: DC AB: AC.
(△内角平分线,分对地典鄰地成比例)
'2.' BD+DC ASTAC
BD
BBC: 8D=()
(合比)
AS
3, AB: BD=(AB+AC): 80 {****
'AE : ED➡ (AB+AC): 80 (KB3"
E
(5) A 20-21, K£ QA ± 4 - § PAPORE 108 ***k 4* PA.
(EK) AŽOE«-£4
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