麻煩師兄幫忙解決以下4條MC 謝謝!
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ANS:
47.3 C 47.4 D 47.5 B 47.6 D
THX!
MATHS MC 4條
2013-03-13 5:39 am
回答 (4)
2013-03-13 7:55 am
✔ 最佳答案
47.3) 3DP = 2PCDP : PC = 2 : 3在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQPS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.4)△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12CD² = (12 - 8)² + 15²
CD = √241
選 D. 47.5) 6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4 △AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B) 47.6)△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC1/2 : 3/4 = 2 : 3
選 D.
2013-03-13 15:10:04 補充：
47.3)
方法二 :
DP : PC = 2 : 3
在 QB 上取點 R 使 CR // AK , 則 QK = QR ,
而 DK : KR = 2 : 3 , 故 DK : KQ = 2 : 3/2 = ４：３
下同...
2013-03-16 5:41 am
47.3)
3DP = 2PC
DP : PC = 2 : 3
在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQ
PS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3
故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.3)
方法二 :
DP : PC = 2 : 3
在 QB 上取點 R 使 CR // AK , 則 QK = QR ,
而 DK : KR = 2 : 3 , 故 DK : KQ = 2 : 3/2 = ４：３
下同...
47.4)
△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12
CD² = (12 - 8)² + 15²
CD = √241
選 D.
47.5)
6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4
△AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24
故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B)
47.6)
△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC
1/2 : 3/4 = 2 : 3
選 D
3DP = 2PC
DP : PC = 2 : 3
在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQ
PS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3
故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.3)
方法二 :
DP : PC = 2 : 3
在 QB 上取點 R 使 CR // AK , 則 QK = QR ,
而 DK : KR = 2 : 3 , 故 DK : KQ = 2 : 3/2 = ４：３
下同...
47.4)
△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12
CD² = (12 - 8)² + 15²
CD = √241
選 D.
47.5)
6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4
△AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24
故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B)
47.6)
△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC
1/2 : 3/4 = 2 : 3
選 D
2013-03-15 2:20 am
47.3)
3DP = 2PC
DP : PC = 2 : 3
在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQ
PS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3
故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.4)
△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12
CD² = (12 - 8)² + 15²
CD = √241
選 D.
47.5)
6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4
△AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24
故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B)
47.6)
△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC
1/2 : 3/4 = 2 : 3
選 D.
3DP = 2PC
DP : PC = 2 : 3
在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQ
PS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3
故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.4)
△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12
CD² = (12 - 8)² + 15²
CD = √241
選 D.
47.5)
6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4
△AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24
故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B)
47.6)
△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC
1/2 : 3/4 = 2 : 3
選 D.
參考: 唔知
2013-03-13 9:29 am
47.3)
3DP = 2PC
DP : PC = 2 : 3
在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQ
PS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3
故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.4)
△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12
CD² = (12 - 8)² + 15²
CD = √241
選 D.
47.5)
6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4
△AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24
故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B)
47.6)
△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC
1/2 : 3/4 = 2 : 3
選 D.
3DP = 2PC
DP : PC = 2 : 3
在 AC 上取點 S 使 PS 丄 AC , 則 QS = (2/5)QC
故 AS = (1 + 2/5)AQ = 7/5 AQ
PS = 7/5 KQ
3/5 DQ = 7/5 KQ
DQ : KQ = 7 : 3
DK : KQ = 4 : 3
故 △DPK : CPKQ
= △DAP - △DAK : △PAC - △KAQ
= (2/5)2 - 4/7 : (3/5)2 - 3/7
= 8 : 27
選 C.
47.4)
△AEB = 15*8/2 - 24 = 36
故 DE : EB = 24 : 36 = 2 : 3 = AD : BC = 8 : BC
得 BC = 12
CD² = (12 - 8)² + 15²
CD = √241
選 D.
47.5)
6AM = 4MN = 3ND
AM : MN = 2 : 3 and MN : ND = 3 : 4
AM : MN : ND = 2 : 3 : 4
△AMB = 4 , 故 ABCD = (2+3+4)/2 * 4 * 2 = 36 , 則MBCN = 36 - 4 - 8 = 24
故 △VMN + 24 = △VMN * ((2+3+4) / 3)²
△VMN = 3
選 A. (應不是B)
47.6)
△ABE = (1/2)△ABC
△BCD = 3/(3+1)△ABC = (3/4)△ABC
1/2 : 3/4 = 2 : 3
選 D.
收錄日期: 2021-04-21 22:28:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130312000051KK00297