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數學題啦又,成日唔識添
回答 (2)
2011-01-23 4:52 am
2011-01-23 4:57 am
作直角三角形△ABC ﹐∠ABC是直角，BC = 7，內接圓的半徑是 2。
再作由圓心O至AB,BC,AC之垂線﹐交點記為D,E,F
則BD = BE = 2﹐CE = CF = 7 - 2 = 5。記AD = AF = x
則(x + 2)^2 + 49 = (x + 5)^2
x^2 + 4x + 4 + 49 = x^2 + 10x + 25
6x = 28
x = 14/3
故AC = AF + CF = 29/3
再作由圓心O至AB,BC,AC之垂線﹐交點記為D,E,F
則BD = BE = 2﹐CE = CF = 7 - 2 = 5。記AD = AF = x
則(x + 2)^2 + 49 = (x + 5)^2
x^2 + 4x + 4 + 49 = x^2 + 10x + 25
6x = 28
x = 14/3
故AC = AF + CF = 29/3
收錄日期: 2021-04-23 23:23:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110122000051KK01088