1969-03-18 — Page 25

REFERENCE LIBRARY

18 MAR 1969

育教僑華買三第張六第日一初月二年酉己屦复WAH KIU AT POHALL報日橋

%$#$%$#$%$#$#3#%$%&#3#Z4%#$%$#$%$##

3#$%$#$#5#%3%$;

二期星日八十月三年九六九一公年八十五國民中

夏僑

HO試顆預켱

英文科 (二十)

桂枝

#$%$# 5 % $ #$%$%$#$%$!

$%$#$%5##$%S #SKS #5%$#$%$#$%8#3%$#3%$#8%$# 65 #5%$ JEXSE:

1969

數學科

(二十) *歐陽銘文·

MATHEMATICS (20)

LESSON 21 Mid-pts, & Intercept Theorems

Theorems:

(A) Mid-poines inburen:

The st. line joining the mid-noints of two sides of a triangle is

(a) parallel to the third side; and

(b) is equal to half the third side

Intercept Theorem:

The st. line drawn (a) through the mid-point of

olde, disects the andrd side parallel to another

third såde.

Or, the intercept theorem may be generalised as:

If three or more parallel st. lines make equ intercepts on a given transversal, they make equal intercepts on any other transversal.#

ExampleEn

Example 1: D is the mid-point of the side BC of 4 ABC;

CA is produced to E. If BR is the perpendicu- lar from B to- the hi sector of Z RAR. ·nrove that - DR == k(AB÷AC)

Proof: Produce BR to cut CA produced at F

In

ARB ARF:

AR=AR

Gồn common

·AS.A.

LAR BILARF (x+4)

ARB SAARF

VRB=RE

DOIDC, BR=RT

DRA CF

CF CA+AF

=CA+AB

• Given, proved

Mid-point Thecrem

: AF#AB, corr, sides of DR == (ACTAB)

Example 2 Prove that, the st. line joining the mic

points of the diagonals of a trapezium is parallel to the bases and equals half the difference of the two bases.

Given Trap. ABCD with

AD // BC & AD< BC

K-EC

BF-FD

To prove (1) EP // AD // BC

Proofi

(2) EF=2(BC-AD).

Let G be the mid-point of AB. Join GE,GF (Suppose that G,E,F, are not. collonsar and we will show here the contrary,

A ABOU

GE are the mid-points of ABAC

GEBC

A ABD

mid-pt. Theorem

G F are mid-pt. of AB, BD rësp.

OF ZAD

mid-pt. There'm Since, AD // BC (Given) and

GE #1 8C

"GE/AD: proved

GE IGF

and F. are collinear

EFI AD # BC

GEBG,

GF-LAD

EF = 66 - GF

EF = =CBC - Abu

ENGLISH (20)

Answer to Paper XIX

Travel is a part of a young man's education.. When abroad, therefore, he should try to see every- thing worth seeing, to familiaries himself with the manners and customs of the people, and to enlarge the circle of his acquaintance,

To make this possible, he should know something of the language before he sets out, and, if practicable, should be accompanied by a tutor familiar with the country. He should carry letters of recommendation to influential people. “

When abroad, he should keep constantly on the move, and wherever he goes visit all placén of interest; watch the people at work and recreation and study their commerce, military and naval aquipment, and legal administration. He should mix freely with them in their social functions and eschew the company of his own countrymen. He should endeavour to meet important people and gleanboke knowledge from then, but he must be careful to avoid chance acquaintances likely to lead to brawls, A record of all these AIDATÍANNAR shħuld ha kant in diary:

When he returns home, he should correspond with. где

friends he has made. He should refrain from telling sensational stories and aping foreign fav The result of his travel should be improved

conversation and manners,

Paper A

I.Q. standa for Intelligence Quotient' which 10. a measure of a person's intelligence found by means of an intelligence test. Before marks gained in such a test can be useful as information about a person, they must be compared with some standard, or nora. It is not enough simply to know that a boy of thirteen has scored, say, ninety marks in a particula test. To know whether he is clever, average, or dull, his marks must be compared with the average achieved - hoys of thirteen in that test,

In 1906 the psychologist, Alfred Binet (1857- 1911), devised the standard in relation to whion intelligence has since bean assessed. Binet was, asked to find a method of selecting all children in the Echools of Paris who should be taken out of ordinary classes and put in special classes for defectives. The problem brought home to him the need for a standard of intelligence, and he hit upon the very simple concept of 'mental age'..

First of all, he invented a variety of tests and put large numbers of children of different ages through them. He then found at what age each tast was passed by the average child. For instance, he found that the average child of seven could count backwards from 20 to 1 and the average child of thres could repeat the sentences "He are going to have a good time in the country.' Binet arranged the various tests in order of difficulty, and used them as a scale against which he could measure every individual If, for example, a boy aged twelve could only do tests that were passed by the average boy of nane Binet held that he was three years below average, anu that he had a mental aze of nine...

The concept of mental age provided Binet, and through him, other psychologists, with the required standard. It enabled him to state scores in intelligence tests in terms of a norm. At first, 11 waa usual to express the result of a test by the

difference between the 'mental and the chronological age. Then the boy in the example given would be three years retarded" Soon, however, the 'mental ratio' was introduced; that is to say, the ratio of the mental age to the chronological age. Thus a boy of twelve with a mental age of nine has à mental ratio of 0.75

The mental age was replaced by the intelligence quotient' or 'I.Q.!. The '1.Q. is the mental ratio multiplied by 100. For example, a boy of twelve with a mental age of nine has an *I.Q. of 75. Clearly, since the mental age of the average child is equal so the chronological age, the average I.Q. is 100.

For each of the words taken from the passage chooBS; from the four words or phrases, the one nearest masningas used in the passavet

By norms 119 what usually happens

(ii) a standard with which something in

compared

(iii) usual

(A) everyday

defectiveS E

concep

(a) average children

(11) poor children

(iii) children of low intelligence

\(iv) very intallizant children

(i) device

(11) tric

thought

(iv)

d) mentals (1) monthly

ii) intelligence (iii) unintelligent

(iv) concerning the mina

(1) delayed in progress (11) advanced in progresH (iii) distorted

(iv) backwards

For each question choose the pest or the four answers given i

à) To judge a person's standard his marks in a

test must be compared with.

(i) marks gained by other boya or una mano age, (ii) mazke gained by older boys and younger boys (iii) marks gained hw a number of have seed

thirteer

(iv) marks gained by the same boy when no was

younger and when he is older

A psychologist 18.

(a) an educationist

ii) a school inspector

(iii)

a kind of doctor -

(iv) a person who studies the working of the

mind

6) A child of seven who can count backwarde fron

20 to 1 has a mental age of

(1) seven (ii) three tij twelve

nine

4) A boy of eine, wie is "inree years retardadi

has a mental age of

(i) six. (ii) nine

iii) twelve (iv) fifteen

The ".." 18..

i) the mental age divided by the chronological

and multiplied by 100-

ii) the mental age multiplied by the

chronological age and divided by 100 (iii) the chronological age divided by the

mental age and multiplied by 100 (iv) the chronological age multiplied by the

mental age and multiplied by 100

Example 4: ABER, ACGH are squares outside AABC, AM

is a median and produced to meet FH at O Prove that (1) M2 — FH, (2) AM=;

Proof Produce" 4M to P

such that MP AM.

Then, ACPB is a gram

diag. bisect each other)

BPACAH

opn. side // gram.

L ́ABP + 2 BAC=14<

Example3:The diagonala AC,BD of the square ABCD

intersect at K. The bisector of ← BAC cuts BK at X and cuts BC at Y. Prove that GY=2×X. Proof: Draw KH/BC: C Cuts AY at H)

In A AYC AK=KL

<XAB

AM-HY

diag. square

Constr."

intercept Th.

=LC) Mid-H. Th.

ACH BP)

2 FAH + ZBẠC = 204,

ABK

ABP

AB= AF BP AH

FAI

Droved

SAS

2)(a) 25=270.2 −8{ZA+4B)

«G=2rt.« −1(ZC+&D)

(b) IF ABGD is a gram

then, 1{A+.B)=1 rt.. Hence ZE = 1rt!?! (c) If ABCD is a rect.

then ADF * 4 BCH,

As shown in Tig. AMJ 19 1308. QPR=DPA DAC Ext,« of a noc is also isos. such that

QDC=45*+90′′-135′′

And.**

2DQR = 3(180°-135′′) RPC=RDQ=22 <RCP=RQD #225

RP=RC

And

XBA Exist

ding of a square

AKH = < ACB (=14) con, as KH #CB

ZA KHX

< KNX + LUKK 4 =XKH =>

ΚΑΙ ΚΗ

H&=CY ( proved ›

proved

side's opp equal d's

And But

AM== AP=FL

BAF

BAT

HINTS. & ANS. TO EX. 19

As shown in fig. Let AC meets HF at 0. Join AF,

HO

Since AHE ≈ ▲ CFG (AAS)

AHCF is a gram (AH 4 FC:))

0 is then the M.P. of HP & of AC:

AC

19

a common 'alagonal

of grans ABCD, APOX & QYOR.

ABC

<ADC -

APQ+ 4 QYC+ #gram PBXQ ·

AXQ+ 4QRO+ #gram XORD

TAABC → ∞ ADC, <APQ = ▲ AXQM@YEMBR

#gram XQRD =/gram PBYO..

EXERCISE 20

In ABC; 6=2B, AD LBC. If Mis the mid-

point of BC, prove that DM = AC.

2) Q, R are the mid-points of the 1sues au, AB of

4BC AD is the perpendicular from A to BC. Prove that RDQ =➡ BAC,

3) In ABC, AB> AC; CT is the perpendicular from

C to the line bisecting BAG D is the mid-point of BC. Prove that (a) DTW BA, (b) DT=}(AB-AC)

Dłut. Produce CT to meet AB at K 1

Br, CQ are the bisectors of 25, «t of ABC; AH, AK are the perpendiculars from A to BP, Drove that KH is parallel to BC..

Page 25Page 26

十裏四第張六第一日一初月二年期己腦夏 WAH KIU YAT PO