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师 香港學校朗誦節特
賓四第張七第 22
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日九廿月一十年七七九一腊公年六十六國民篩中育教儒攀
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2. (2x
1)2
The
教僑華
T
20 T
- 4(0) + 17
學能推理練習專欄
17 (Ans)
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選出下列棒選最適當的答案,並在它的下面 糖糖。
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The required probability
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The equation is
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(북)=
18x
+ 7 = 0 (Ans.)
數
學
(九) 文長波
Mathematica 9
Solution to exercise 3 Section B
4 GME=ZEMK=90°
EM-EM (common) A GEMFAKEM (A,S,A.)
GM-MK (Corr, sides) Similarly HMM..
GHKI, is a rhombua, (Diaga, bisect each
other and cut at rt.L.) Given: BX and EY are
altitudes of BCD and EDC'
respectively
To prove: (a) ABC: DAC 4ABC:6DAC-BE÷DE.
quad' ABCD: ^DAC=BX÷EY
3. AB, BC and CA are the
tangents of the circle,
therefore,
AX AZ, HX - BY and CY = CZ
Let BXx en
BY
[CY=[CZ] BO
AX-
AB
ཝ
The perimeter of the triangle
11 +13+
32 (em)
2x + 2(13-1) + 2(11-x)
(Ans.)
SECTION B..
9. Let P(n) be the statement
(2n + 1) > 0
for all positive integers n>2.
When n-3
p(3) ► 23 - (2)(§) + 1
-20
It is true for n = 3.
Assume P(n) is true for nuk, therefore
(2k + 1)
2k - 17.0
Consider when n k+1
P(k + 1)
BX - Bem (Ans.
下列各题有五個項目,其中一個和其餘四個 敢不同題,然選出這個不同類的项目
Since AB is the diameter,
therefore
8. P
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PL
R
RA
S*
TR
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16. P
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R
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90
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以下各题是一則謎語,試選出每則謎語兼遇
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新數學(九)
HODERN MATHEMATICS (9)
Answers to Test Four
BECTION A
.<
Multiply both sides by (x-2)2
(2x-3)(x-2) (x-2)2 - (2x-3)(x−2) – (x-2)2 (x-2)(2x-3)-(x-2)] >
(x − 2)(x − 1 )?
xler x 2 (Ana (2x + 1)2 + 4y2
4)2+4y2
2
The centre of the circle 18 (1, 0)
The required circle is :
(x
or
1)2 +
1)2+
52
- 100
榮家
k+1
· 2 (k+1)
2k
since 2k
2(k+1)
2k
1)+ (2 = 3)
- 2k1 0.
Also, k is a positive
integer greater than 2,
عار
- 3 must greater than 0;
therefore
P(k + 1) > 0
Hence, it is true for n=k+1.
ince, P(n) is true fom
and if p(k) is true,
then it is also true for
p{k+1},
true for all positive inte-
gera
greater than 2,
therefore, it is
2 > 2n + 1
for all positive integers n
greater than 2.
10. Since sino and con0ré
the roots of the equation
3x
Therefore,
2
the product of roots =(sine)(cose)
Kindc ose -
the sum of roots
-nine cose
(1)
sino + co80 ■
(2)
From (2)
2
11. Solution:
Sundries Total
Material
Labour
Rent
Original
after changed
20 units
50 units
25 units
5 units.100 units.
20
()units 50()units 25()units 5 units 106,6 units
-17,bunite
=54 units
30 units.
The percentage change in the running cost
106.6-100
x100%
100
=0,6%
12. Solution:
volume of the rectangular block
4x6x8cu.cm,
-192cu.cm.
volume of one solid sphere
*r ?
r(1) eu.cm.
4.19cu.co.
Number of solid spheres that can be made
192
4.19.
45.8
Number of complete solid spheres that can be made is 45. 13. Given: ABCD is a cyclic quad,
GF bisects LAFD.
To prove: (a) x=y.
(b) If EL bisecte
ZAEB, GHEL
is a rhombus.
Proof:
(a) Draw BP, DOLAC.
Since SABC, DAC have the same base AC.
4ABC; ADAC=BP: DQ. Z DEQ= [PEB (vert.
opp. 4)
ZDQE ZEPB➡irt4.
BP, DOLAC)
ZODETM 4EBP(3rd 4 ofA)
aare similar
DEQ
BEP
(Equiangular
As)
(b).
BP: DQ-BE; ED (Corr.
sides, similar 45):
AABC ADAC-BE: ED ABC+ DAC); DAC -(BE+ED):ED
i.e, quad ABCD: ADAC
-BD: ED 4EDY-BDX (Common) ¿DYE=4DXB=1rt4.
(*.*BX, EY-DG).
DEY-4DBX (3rdofs)
DEY ME
*DBX
fare similar
(Equiangularis)
BD:ED-BX:EY.
corr. sides, similar
4s)
quad ABCD; 3DAC=BX÷EY,
Solution:
Tet hm he the height of the hill
AF-hcosec28 m AX-hcosec12 m. AE=hcosech”m.
AF
COS/FAX AX
hcosec28 hcosec
LFAX~63°43′ COLFAE AF
AE
hoosec28" hoosec5o
FAE-79 18 4XAE-79 18-63°43
-15°35
- X
+8
10
(x2-x-9)+1__ (x2-x69)~1
2
3x2
3x2
x-9
(x+5)+3 (x+5)-3
-
-3x-27-x+
4x
+
32 - 0
- 0
(3x+8) (x-4) *--23
on 1 - 4 (Ang. ).
(2x-
(Ana.)
7. Since 2 + 31 is the solut-
ion of the quadratic
equation
Therefore,
(2+31)2.
+8x + b = 0
+ a(2+31) + b
4101 +912)+28+
(4 +121
0
+ b = 0
0
4+ 121 9+ 2a +Jai+b
(2a + b −5)+(Jn +,12)i
2a + b 5 = 0
3a + 12
- 0
(Ans.)
b. 13
8. The area of the outer
cimele
T(10)2
= 100 T
-
= 0
sin" +cos ̃0 + 2ain8c os ✪
16
+2sin8co88 = 16/9
#inecose -7/18
From (1)
k * 3sin@conÒ
k-
7/6 (Ans.)
(b) Let the required equation
be
2
X + px + ( ́ = · 0
qproduct of roats
(tane) (tane)
I..
p-sum of roots
1
- tano +
sine
tano
Wing
Proof:
(a) x=¿GFD+ZBCF,
y=LGFA+LGAF
(Ext.4. ofA)
ZBCF=ZGAF (Ext. 4 cyclic
quad.)
*GFD=4GFA (*.* GF
bisects (AED)
.x-y (Subst.)
(b) ZAEL-BEL (*.* EL
bisects LAEB)
.*. LGMF-LEMK (3rd 2 of 4)
Since ZGME+ZEMK+180°.
(Adj zy on at. line)
Solution: AB-DO-CA-24cm, Area of 4ABC
(2x)(2r)sin60s
22
4q.co
-1.732r sq.cm. Area of sector AFE
2 T
2
◓qcem
sq.CD,
Area enclosed' between the circles
-(1.732x2-3x2)sq.cm
2
-0.161r sq.cm,
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