Re: [解题] 国二数学下学期期末考

楼主: thepiano (thepiano)   2016-06-23 15:14:34
※ 引述《swyn (茵)》之铭言:
: ABCD梯形 AD平行BC AC垂直BD BD线段等于20
: AD+BC=25
: 求ABCD梯形面积
作 AE 垂直 BC 于 E,DF 垂直 BC 于 F
AC * 20 = 25 * AE
令 AC = 5x,AE = DF = 4x
25 = AD + BC = BF + CE = √(400 - 16x^2) + 3x
x = 3,AE = 12
ABCD = 150

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