10 分10 分10 分10 分10 分

21.
BAF = A

Consider ΔABE:
A + ABE + AEB = 180o ( sum of Δ)
A + ABE + 38o = 180o
ABE = 142o - A ...... (1)

Consider ΔACF:
A + AFC + ACF = 180o ( sum of Δ)
A + AFC + 46o = 180o
AFC = 134o - A ...... (2)

Consider quad. ABDF:
ABE + AFC = 180o (opp.s, concyclic quad.) ...... (3)

Put (1) and (2) into (3):
(142o - A) + (134o - A) = 180o
276o - 2A = 180o
2A = 96o
A = 48o
BAF = 48o

=====
22.(a)
Angle of major segment of AC
= 180o - 130o (opp.s, concyclic quad.)
= 50o

AOC
= 2 x 50o (s at centre)
= 100o

22(b)
Consider quad. AOCK:
AOC + CKB = 180o (opp.s, concyclic quad.)
100o + CKB = 180o
CKB = 80o

=====
23.
Join BD.
BDC = 90o ( in semi-circle)

Consider quad. ABDE:
EAC = BDC (ext., concyclic quad.)
EAC = 90o

=====
24.
Join QR.

Consider quad. QRST:
PQR = RST (ext., concyclic quad.)
PQR = 114o

Consider quad. PQRO:
POR + PQR = 180o (opp.s, concyclic quad.)
POR + 114o = 180o
POR = 66o

=====
41.
The answer is B.

XA•XB = XD•XC (intersecting chords)
6(6 + 4) = 5(5 + CD)
60 = 25 + 5CD
5CD = 35
CD = 7

=====
1.
x2 - 8x + 15 = 0
(x - 3)(x - 5) = 0
x = 3 ooro x = 5

=====
2.
3x2 - 7x - 2 = 0
x = {-(-7)+√[(-7)2 - 4(3)(-2)]}/(2x3)} or x = {-(-7)+√[(-7)2 - 4(3)(-2)]}/(2x3)}
x = (49 + √73)/6 ooro x = (49 - √73)/6

=====
3.
2x2 + 4x + a = 5
2x2 + 4x + (a - 5) = 0

The equation has no real root. Hence, Δ < 0
(4)2 - 4(2)(a - 5) < 0
16 - 8a + 40 < 0
56 - 8a < 0
7 - a < 0
a > 7
=

21. In ΔACF, ∠BAF+46=∠DFE (ext. ∠ of Δ)
In ΔDEF, 38+∠DFE=∠BDF (ext. ∠ of Δ)

Thus we have
∠BAF + 46=∠BDF - 38

On the other hand, consider ABDF,
∠BAF+∠BDF=180 (opp. ∠s, cyclic quad.)
∠BDF=180 - ∠BAF

Therefore
∠BAF + 46=180 - ∠BAF - 38
2∠BAF = 96
∠BAF = 48

22. First, consider the left circle, we have
major∠AOC=2130=260 (∠ at centre twice ∠ at ⊙^ce)

But major∠AOC+∠AOC=360 (∠ at a point)
Thus ∠AOC=360-260=100

Now consider the right circle, consider AOCK
we have ∠CKB+∠AOC=180 (opp. ∠s, cyclic quad.)
Therefore ∠CKB=180-100=80

23. Join BD, since BC is a diameter,
thus ∠CDB=90 (∠ in semi-circle)

But in ABDE, we have
∠EAC=∠CDB=90 (ext. ∠, cyclic quad.)

24. Join QR, first we have
∠PQR=114 (ext. ∠, cyclic quad.)

Now since major∠POR=2∠PQR=228 (∠ at centre twice ∠ at ⊙^ce)
and also major∠POR+∠POR=360 (∠ at a point)
Therefore ∠POR=360-228=132

**********
41. Join AD and BC, we have
∠XAD=∠XCB and ∠XDA=∠XBC (ext. ∠, cyclic quad.)
Also ∠AXD=∠CXB (common angle)
Thus ΔAXD~ΔCXB (AAA)

Now since XA/XC=XD/XB (corr. sides, ~Δs)
Thus XA*XB=XC*XD
6*(6+4)=5*(5+CD)
CD=7

Answer: B

Remark. The resule XA*XB=XC*XD in this question,
is the famous theorem called "Intersecting Secant Theorem"
which is a special case of "Circle Power Theorem".

**********
1. x^2-8x+15=0
(x-3)(x-5)=0
x=3 or x=5

2. Using Quadratic Formula

3x^2-7x-2=0

x=[7√(7^2+4*3*2)]/(2*3)
x=[7√73]/6

2008-12-13 01:04:24 補充：
2. Using Quadratic Formula

3x^2-7x-2=0

x=[7±√(7^2+4*3*2)]/(2*3)
x=[7±√73]/6

2008-12-13 01:07:50 補充：
In Q21-24, I don't know why the degree sign (°) is missing.
So in all numbers, you may add back the degree sign.
Sorry for inconvenient.

2008-12-17 16:52:15 補充：
想請問發問者點揀最佳解答，我好唔滿意！

Uncle Michael雖然做得好好，個答案都好colourful，
但佢第24題同第2題都做錯左，唔應該會係最佳解答。

你的問題是甚麼？

是不是IQ題？