1969-05-29 — Page 15

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The diagonals of a rhombus are resp. 16 cm. ana 26 cm. Find the area of the rhombus. Sol.

As shown in figure, if ABCD is a rhombus, then diagonale

At

AC and BD bisect each other.

at right angles,

英中會考數學(三)答案 ·歐陽紹文·

S

Matha

Juggested Answer

Syllabue A, Paper III'

¡Geometry

Max. Cortificate of Ed cation Exan. English) 1969

Section A

For questions

3, put a tick "" in the box

opposite to the correct answer. For questions 4 - 8 put the answer in the space

provided on the question paper. For questions 9-10, both questions should be done at.

the beginning of the answer book..

Which one of the following expressions is equal to the sum of the market angles in Fig.

(a) 2 x 90"

By drawing parallel lines at A,B,C,D, resp.

We find a rectangle PORS,

Area of PQRS - PQ.FS - AC.BD

and area of PORS • twice the area of ABCD

Area of 1BCD • †(AC, BD).

(1,0, half the product of the two diagonale Yow the required area = 2(16 x 26) @q.cm.

208 aq.cis.

Given: St." line FAG touches the circle at A.).

DE // FG LCBA

To find / EDA."

Solution:

LEDA

LEDA

72°

DAF (alt. DE//FG) 7

B

__B = 72° ( L in alt. segment.)

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英中會考數學(三)試題

Attempt any FOUR questions from this section. Start each new question on a paw page. Proofs must be gran

in fall, Marki will be deducted for poor presentation of material,

IL

Let P be fixed point inde a circle whose centre is,"

13,

(a) State the locus of the mid-points of the chords through P.

(b) Prove this statement in two steps as followi:

0)

that every point which satisfies the conditions lies on the locus,

(i) that every point which lies on the locus satisfies the conditions.

Tir's point oude à circle with centre ◊. TA, TB are targets touching the circle at A und

Prove that TO bisects LADB and LATE.

In figure. 4. AB AC: Dis the mid-point of BC, A semicirde with centro-O touches AB and, AC, A tangent to this semicircle interacts AB and AC at M and N respectively.

Prove that LBOM = LUNU,

in Figure 5, ABCD is a quadrilater. M, N are the mid-points of AC and BD. OM//BD; ON//AC; P, Q are the mid-points of AD and AB, Prove that

{2} Quadrilateral APMQ» Quadrifpteral APOQ: (b) Quadrazteral APOQ-4 Quadrilateral ABCD

Fig.4

Fig.5

(b) 3x

୨୦୧

(0) 4

(a) 5 x

(e) 6 x 90°.

902

90o

J0000

Solution, ext.sum of an n-sided_polygon_11

4rt.Z

ABCD is a trapezium with AB//CD and ABCCD. M, N are the mid-points of AD and BC respectively. Which one of the following statements must be true?, (a) MN - DC

[o] XN = DC+#[ DC-AB)

[4) MN = \{AD+BC}

{+} \AB = MM - MN 1 DC

D

Given: Chorda AB, DC meet, when produced, at E.

AB4", BI = 6". CE = 3*

14.

Fo rinds CD

Solutions

By intersecting chlorda theorem,

EA EB - ED EC

Let CD " then

-10 x 6 - 3 (3+x),

17.

Lạt. CD . 17

7. Triangles

and Tare similar to øns another. The ratio of lengths of a pair of corr. medians

in

T2

and

T2 is 3:4

The ratio of lengths. of a pair or corr, heighte in T and T is 8:10 - 415

*

A

The ratio of lengths of a pair of corr. sides in'

and T、 in_3+5.

Solution: MN is called the median of tapezium ABCD “

By placing a congruent trapezium A'B'C1D

Bo that C'B' lies on BC (C1 coincide with B, B' coincide with C) and that D'A' and. AD are on opposite sidam of BC.

Then AD*A'D is a //gran, of which_MI__14. a median.

- 2MN AB DO!

Which one of the following sats of data wILL NOW/ necessarily imply that the triangles ABC and XYZ are congruent?:/

(a)`AB - XY, BC - TZ, AC_-_XZ}

catloof areas of triangles T. and T,

8. Given ABC with BAC

AƐ bisector

AD + BC

BAC

AC 3", AB - 4"

To find (1) EC, (11) DC,

(iii) ED

Solutions

90°

ABC is rt. at A. by Pythagoras

1 T2 = 32:52

9:25

Than

(b)_AB

ABC≈A XYZ (3.5.5.)

- XY, BC - YZ,_ |__B_- LX.

In figum 6, HE//FG; EFLFO, IF HK = KJEG, prove that

6) EK. EG

(8): LHGF-

LEGF

“ăn ngur ̈7, HG//AB/CD, EAC, EBD, FAD and GBC are straight lines. The distance between FG anil

AB is h, and that between Alt and CD is 2

ja) Prove FE EC.

16. 141

(b)

by Express the length of FG in terms of his,

by and Ar

Fig.7

in ABC, ABAC and AD It a median. By producing AD to E such that AD - DE, prove that LEAD><BAD.,

In figure 8, AD is a chord of circle O, AB = BC » CD. Using the result of (a), or by any other method, prove that are £Farc AE.

Fig. 8

本

BC-AB2

2

2

+ AC -74

3 - 5

Let EC

-

1.e.

x" then BE (5 = x) ing

E is the bisector of / BAC

AB

AC

EC

BE (Lopsector theorem)

4.2

етлой

ΓΥΠΕΓΕ

in ches

̧1) Produce the line segment both ways such that

AC 6p inches, BC - 3p inches.

With diameter AB, draw a semicifols.

At C, drop DC LAB to meet the semicircle at D. Then DC- Nie » inches,

10. Given: A regular hexagon ABCDEP of side 1.7 inches

To constructs A rent. ACXY - ABCDEF To measure 1. CX.

Then ABCA KYZ (3.4.S.)}

c) AB -

XI, LA LX. LB - LX.

A A

Then ▲ 180A XYZ (4.9.A.)

4x

15

2 5 or BC - 24 inches

in As ABC, ADC`

LBAC - LADC - 90°..

LACB = LACD (Common),

ABCDAC

(A.A.)

Corr. sides are la proportion.

DO AC

A

"ZX - 90°, BC - 12.)

B

C

Then ABCA.YZ (R,H.S.)

) AB XI, __B - ___Y, AC - X2

A

Z'

Then ABCAXYZ if LC - L4, otherwise LC + Z2 - 1800. This is what we called the ambiguous case. This fact is shown by the following figure,

In which, APQR, POS are conguent if _PRQ - LS. Otherwise PRQ + /_S` = 180°.

P

=

DC

W

or IC

- 14 5

anchas

And ED EC - DC

NOTE

ปี

1

12) inches"

12 35

inches

JE SE DE TA JE 10 13 19 1-

fou may apply the trig. ratio to both a ABC, ADC.

From rt. AABC; coac

-

ندي

Próß rt.

·AADCY cosc «

2

or DC.

14 inches

5

3

9. Given a line

segment of length p inche

To construct

a line segment AB equal in lengt to M8 p.

To measuret AB in inanes

Methodi

(1) With radius 1.7 inches, draw a circle. (2) With the given radius 1.7", divide the

circumference into é equal parts, at pi A,B,C,D,E,F,

Then ABCDEF is the required regular hexagon 67) Bide 1.7".

(3) Let p be the centre or the ciròle. Join 08\ (4). Draw XYLOE and passes through E. Then" ACXY ̧ is the required rectangle which is equal in area to ABCDEF.

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