1971-11-25 — Page 14

【頁二第張四第 BADA+EZFB, WAH KIU YAT PO

報日僑華

四期星日五廿月一十年一七九一届公年十六國民華中 育教僑華

1972

育教僑華

(25) (1. Anta, 2. termites, 3. bess, 4. butterflies;

5. houseflies, 6. cockroaches, 7. mosquitoes) are social insects, and the (8, frogs, 9. Human beings, 10. rabbite, 11. monkeys), are alos soci animals

中文中學會考試題預習專欄 G

數學科

(四)

張正邦

1972英文中學會考試題預習專欄

QUESTION 21+

第三次預習題解答

Indicate whether each of the following statements

堅道英文書院主編

生物科

(四)

BIOLOGY (4)

is true (+) or false (0)

(Answere given at the end)

The cell wall in planta la part of the protoplasm

All Amoeba will die under dry condition.

Andebs and Hydra can take in dissolved food substances from the suri anding water by diffusion.

· (1) 2 = 1 * *

其在截線同側內角之平分 線互相垂直,則此二直線平 行

A

求證

E

CUESTION I favo

Can you pick out the correct words or phrases"),

within the brackets in the following?

at the end.

(1) In general a nature (1. diest.

plant does'nt poaswam a cambium:

mongoot. J

The contractile vacuole in Amoeba 28. ary structure.

(2) Stomata are found in the (1. piliferous layer of

the root;

epidermis of the leaf,

mis of the young stem.

The Dioot. foot generally possesses (1,

·Dith.

epider-

(4) The vascular bundles in the root, stem and les of a plant (1. marve a similar function,

(5)

2. function differently.)

The root-hair in (1, a single pell,

a single cell, 3. multicellular)

(6) The elongation of the growing parts of the TOOS and the stem of a flowering plant is mainly due to the (1. increase in number of cells, 2. enlargement of the cells already formed.)

13. #pongy

(7) The (1. epidermis, 2. palisade layer,

layer,

4. Vascular bundles) in the leaf are the important parts for photosynthesìNG.

(8) Usually in a Dicot, leaf the stomata are distribu- ted more on the (1. upper, 2. lower) surface,

(9) The spidermie of a plant is composed of (1. living

2. dead) oells.

(10) The epidermis of the young ster and the leaf (1.

(1. i 2. is not) composed of a single layer. cells.

(11) The three main types of plant tissues are (1.

2. phloed, 3. cambium, 4. parenchyma,

bollenchyma, 6. salerenchyma.)

(12) The pith and the cortex of the a dioot. plant are

mainly used for (1. storage of food, 2. transport- ation of food, 3. transportation of wataz.

(13) The cambium in the dicot. plant gives rise to new

(1. zylem, 2. phloem, 3. path. 4. cortar,

5. medullary-ray) cella.

(14) The cuticle on the epidermis of the leaf and young

stem in a layer of

non-living waży: substance.)

(15) The xylem vessela are built up of (1. living,

2. dead) olls%

(16) The sieve-tube cells in the phloem are (4nti12

living, 2, dead) when mature baasuse of the (3. presence, 4. absence) of cytoplasm inside,

(7) Under normal condition the (1. xyles, 2. phloem,

3. cambium 4. pith) is capable of esil division.

self,

Orosa)

Termitea or white ants are, true ants,

The scales covering the fish form an exoskele- ton to protect the body,

7. The cell wall in animal cell is the cell

or plasma membrane.

All inaeots can fly.

Bacteria are universa in existence.

10. The scale-like leaves of Dodder cannot be used-

to perform phoynthes

11. Bauhinia de a leguminous plant whion 18

dicotyledonous.

12. Baoh Bacterium contains a definite nucleua

sar the centre of the cell body.

13. Oli membrane is equivalent to cell

111 fishes bear teeth.

There is no ayelid to protect the bony fish,

of st

Gener

he fish do not look after their

The endoskeleton or mammal a entirely comp- osed of bony tissue.

All insecte bear three paire of jointed legs.

The mature Dodder plant does not pos root.

20. The Pine tres is a monocotyledon.

21. Mung are non-green plants. which cannot perform

Photosynthesia.

The types of protein in our food are ex the same as those found in our protoplasm.

(23) 23. The matrix or sne cell wall in born.

green composed of cellulose.

24. water and air are essential to the lives of

living organisms, and they are present in immense quantities on the Earth, this show that animals surroundine conditions.

25. Green Plants do not respire in the day

(18) Mont flowering Plants exhibit

pollination.

(19) Fertilization takes place in

24 ovulaj 3. pollen grain;

Ovary

a flower

Answers to question 17

two) male

(20) Fach pollen grain contains (1 one,

gamete(s), and an ovule encloses (3, one, 4. two) #88 9011()

(21) The pericarp of a fruit is developed from the..

(1. ovary, 2, ovary wall) of the carpel of a flower, and may be divided into 3 layers, they the (3. ectoderm, 4. mesoderm, 5, on

5. endoderm. 3.epicard.. menocarp, 8. endocarp.)

(22) The seed coat called (1, testa, 2. testis) In

developed from the outer part of the £3. ovule, 4. ovary

endos per

(23) The food substances in the seed are usually stored

in the 1. teata, 2. cotyledon,

plumule, 5. radiole)

(24) Ine parts responsible for the vegetative re

ion in flowering plants are the (1. zoot 3. leaves. 4. flowerTS..

(1).

(9)

ི་

(4)

(7)

(11).

(13)

(14)

(15)

(16)

(17)

(18)

(19)

(20)

Cka 21=22.

KMIHM 求證 ABUCD 證明

<1+23+25=180 (=A) 形内角和等於180°* > KMLHM (B40)

<5=90° (垂線構成

直角)

21+23+90=180 (De)itxus)

21+23=90 (129)

6 CKHB=21+22

21=62

8-2KHB=21721

-241

9 <HKD=243 (7 ZKHB

同理可求) <KHB+<HKD=2(

2 (4+13)

2+9量加等量)

<KHB+ZHKD÷2x90

(Q^^O).

12. <K HB tz HKD=180°

(化簡)

Eko AE=EB BF÷FC CG=GD DH=HA

EFGH為一平行四 邊形

證明引AC

·AE= EB BF = FC (BAP)

? EF == AC XEFIL.AC

(三角形兩邊中點联線

13 AB//CD ()**=

直線所截,若所成之同侧 内角互補,則此两直線平行

「行第三週目寺が第三

HG==AC, HGIAC

(A] @)

EF=HG(23代换) EFIL AG (EPAS- 直線之两直線平行) EFGH 平行四遷形 ・中還形有一対遷平行亘 相寺則高平行四建形)

QED (4)設通過專腰三角形ABC 底邊船上一點D畫二腰 之平行線各過腰於吸 -F B] DF + DE AC-

(2) 32 ABIED B|2|+22+23.

360

BAR ABIED

Zike AC=BC

DEIIBC DFIIAC

I DF +DE=AC 證明

AC-BC (za): <1=22 (18-19-13

面等)

DEABC (ZAR)

L3=22 (平行線同位

41=43 (O@nt)

6

AE÷DE (980k@

等則其腰亦

DE BC DF#/AC (zko) •

EDFC 平行四形

(四邊形两组对邊平行則

為平行四邊形?

*85 21+22+25=360 9 DF = EC(19) PELASA

PER LE DC AB KS F..

ABIED (CAD)

43+44=180°(平行線

“同側内角互補)

22=25+248=275

外角等於內對角和)

4 <|+45=180° (12-

側鄰角互補)

5 <3+44 +22+el+25

=188+25+24+1.80.

( 0 + 0 + 0 )

<1+22+<3=360°

(化簡)

QED

117

10

AE + EC = AC (11

各分量之和)

!!|_ DE + DF=AC

(0 OKAO)

QED

第四次預習題

解下列各方程式、

2{3(4 (51−1)-81-201·

++ 2 = 2

(+)-3-2-2x

4 (22−1)(32−1)(4x+1)(EX

+2)=0

{(3) WILD X2EĦE 1= 19] 08 A3 5 [(a+b) X-<l=[(a-b)X

各中點え直線成一平行

A

+c]2

6. [2(2x+34)=3(2X-34)+10

141-34-4(64-20)+3

|ax+by=a2+20+b2

bx+ay=a2+2b+b2

8.1 x + y = 1 + 2

(未完轉入第五張第四頁)