求周界,help me....
2011-06-28 6:39 am
回答 (4)
2011-06-29 1:06 am
✔ 最佳答案
(1+1)^2+1^2=AB^2 (1+1)^2+1^2=BC^2√5=AB √5=BC
(1+1)^2+1^2= AD^2 (1+1)^2+1^2=CD^2
√5=AD √5=CD
AB+BC+AD+CD = 周界
√5+√5+√5+√5 = 4x√5 = 周界
ANS = 4√5
用左pythagrous theorm.
參考: me
2011-06-29 3:48 am
sqrt(2^2+1^2)x4
=4sqrt5
=4sqrt5
2011-06-28 7:19 am
求周界ABCD....SolAB=BC=CD=AD=√(1^2+2^2)=√5周界=4√5
2011-06-28 6:44 am
4CM~~~~~~~~~~~~
收錄日期: 2021-04-30 16:29:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110627000051KK01221