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標題:
數學不懂做3T^T
發問:
Q18:http://i59.tinypic.com/fc4miu.jpg Q42:http://i61.tinypic.com/votxxf.jpg Q33:http://i59.tinypic.com/2ljt1mx.jpg
最佳解答:
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 ==== 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 ==== 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
其他解答:
數學不懂做3T^T
發問:
Q18:http://i59.tinypic.com/fc4miu.jpg Q42:http://i61.tinypic.com/votxxf.jpg Q33:http://i59.tinypic.com/2ljt1mx.jpg
最佳解答:
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 ==== 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 ==== 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
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