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il_mares Not too shy to talk


註冊日: 2005-07-03 發表數: 30
| 圓 |  | 半圓O內切於四邊形ABCD,圓心O在邊BC上,若邊AB=12,邊CD=18, 求邊BC=? |
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2008-10-23 11:34 |  |
jsir520 Just popping in

註冊日: 2008-05-06 發表數: 6
| Re: 圓 |  | 如果ABCD是梯形,也就是說線段AD平行線段BC的話就好算了 |
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2008-10-24 15:26 |  |
wanghp Quite a regular


註冊日: 2006-09-10 發表數: 42
| Re: 圓 |  | 既然他只給兩個數字
所以答案大概只有這幾種可能
0 1 兩數之和 兩數之積 算數平均 幾和平均 調和平均
按照填充題的精神
這題答案應該是30
_________________ Simple
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2008-11-07 21:39 |  |
colfulus Not too shy to talk


註冊日: 2008-08-29 發表數: 27
| Re: 圓 |  | AD is tangent to the circle at X. Then ABO and AXO are congruent, and so are DXO and DCO. So AB=AX, CD=DX, AD=30. Project A onto CD to get A'. Then AA'D is a right triangle; A'D = 18-12 = 6 and AD = 30, so AA' = 12*(6^0.5). AA' = BC so BC = 12*(6^0.5). |
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2008-11-08 11:17 |  |
2266 Not too shy to talk


註冊日: 2007-05-17 發表數: 23
| Re: 圓 |  | 題目條件不足 A.B.C.D.四點共圓 或AD//BC 此題才有解=30. |
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2008-11-13 21:27 |  |
小Y Just popping in

註冊日: 2008-11-19 發表數: 18
| Re: 圓 |  | 這個答案是:BC=AB+CD=30 1. 取BC一點P,使得CD=CP,連OA,OD,PA,PD 2. 由切線等長性質,可證角DAO=角DPC,則APOD共圓 3. 可推得角BPA=角BAP,即AB=PB,又CD=PC得之 |
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2008-11-19 02:28 |  |
lochihsin Quite a regular


註冊日: 2009-12-28 發表數: 63
| Re: 圓 |  | 12根號6 因為內切 且圓心在邊上 所以四邊形ABCD是梯形 AD=12+18=30 AOD是直角三角形 設半圓O半徑為a a^2+12^2+a^2+18^2=30^2 a=6根號6 2a=12根號6 |
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2010-02-18 21:39 |  |