1.设EF长度为x,则 OG = x, OF = GE = √5 - x
由于OC = AO = OF + AF = OF + FE = √5 - x + x = √5
得正方形的对角线长度为2√5.由此可得,边长是√10.
2.
原理是证明AC^2 + BC^2 = AB^2
设正方形b与CA交与E, a与CA交与F,则CA = CE + EF + FA
同理设BC上的2个交点为G和H,则BC = CG + GH + HB
在三角形AEK中,根据比例求AI (设为x),BJ(设为y)
a : (x+a) = (b – a) : b
x = a^2 / (b-a)
AC^2 + BC^2 = [CE^2 +CG^2] +[EF^2] + [FA^2] + [GH^2] + [BH^2]
= [b^2] +[(b-a)^2 + a^2] + [a^2 + x^2] + [(b-c)^2 + c^2] + [c^2 +y ^2]
AB^2 = [x ^2] + a^2+b^2+c^2 + [y ^2]
计算过程省略了.键盘上太难敲了,多多包涵.
3.过D做BE//AC,交BC的延长线于E,则
ED=AC=6,
BD=8,
BE=10
三角形BDE是直角三角形,
BE边上的高为4.8（即梯形ABCD的高）
梯形ABCD的面积 = (2+8)*4.8/2 = 24