數學不懂做3T^T

18.
The answer : D. 3 : 23

AE is produced to cut the produce of DC at H.

∠ABC = ∠BCH (alt. ∠s, AB // DH)
∠AEB = ∠HEC (vert. opp. ∠s)
BE = CE (given)
Hence, ΔABE ≅ ΔHCE (ASA)
Area of ΔABE = Area of ΔHCE

Let BC = 2h, then BE = BC = h
Let AB = DC = 3k, then DF = k and FC = 2k

Area of ΔABE = (1/2) × AB × BE = (1/2) × 3k × h = 1.5hk
Area of ΔHCE = Area of ΔABE = 1.5hk
Area of ΔBFC = (1/2) × BC × FC = (1/2) × 2h × 2k = 2hk

Let a be the area of ΔBEG.

∠AGB = ∠HGF (vert. opp. ∠s)
∠ABF = ∠HFB (alt. ∠s. AB // DH)
∠BAH = ∠FHA (alt. ∠s. AB // DH)
Hence, ΔABG ~ ΔHFG (AAA)
(Area of ΔABG) : (Area of ΔHFG) = AB² : FH²
(Area of ΔABE - Area of ΔBEG) : (Area of BFC - Area of ΔBEG + Area of ΔHCE) = (3k)² : (2k + 3k)²
(1.5hk - a) : (2hk - a + 1.5hk) = 9 : 25
9(3.5hk - a) = 25(1.5hk - a)
31.5hk - 9a = 37.5hk - 25a
16a = 6hk
a = 0.375hk

Area of ABCD = Area of ΔABE + Area of BFC - Area of ΔBEG + Area of ADFG
2h × 3k = 1.5hk + 2hk - 0.375hk + Area of ADFG
Area of ADFG = 2.875hk

Area ΔBEG : Area of ADFG = 0.375hk : 2.875hk = 3/8 : 23/8 = 3 : 23


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42.
The answer : B. -8 ≤ k ≤ 22

Circle : x² + y² +2x - 2y - 7 = 0 ... [1]
St. line : 3x - 4y + k = 0 ... [2]

From [2] :
x = (4y - k)/3 ... [3]

Put [3] into [1] :
[(4y - k)/3]² + y² +2[(4y - k)/3] - 2y - 7 = 0
(4y - k)² + 9y² + 6(4y- k) - 18y - 63 = 0
16y² - 8ky + k² +9y² + 24y - 6k - 18y - 63 = 0
25y² - (8k - 6)y + (k² - 6k- 63) = 0

As they intersect, the discriminant ≥ 0
(8k - 6)² - 4(25)( k² - 6k- 63) ≥ 0
64k² - 96k + 36 - 100k² +600k + 6300 ≥ 0
-36k² + 504k + 6336 ≥ 0
k² - 14k - 176 ≤ 0
(k + 8)(k - 22) ≤ 0
-8 ≤ k ≤ 22


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33.
The answer : C. 5/12

Total number of combinations = 6² =36

Combinations fulfilling the requirement :
(2) : 1+1
(3) : 1+2, 2+1
(5) : 1+4, 4+1, 2+3, 3+2
(7) : 1+6, 6+1, 2+5, 5+2, 3+4, 4+3
(11): 5+6, 6+5

Number of combinations fulfilling the requirement = 15

Required probability = 15/36 = 5/12