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育教儒韻第三集張四第日四十月四年酋巴歷夏 WAH KIU YAT PO
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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.)
四期星日九廿月五年九六九一曆公年八十节国足难中
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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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買四第張四第1日四十月四年西己饜
WAH KIU YAT PO