奥术几何面积的题.如图E在矩形ABCD外,连接E与矩形顶点,EB交AD与F.已知：S△EAF=18S梯形BCDF=50,S△EDC=8求S△DEF?

Question

奥术几何面积的题.
如图 E在矩形ABCD外,连接E与矩形顶点,EB交AD与F.已知：S△EAF=18 S 梯形BCDF=50 ,S△EDC=8 求S△DEF?

Answer 1

△ECD中,设CD边上的高长度为h,则S△ECD = h * CD/2,即h * CD = 16,由CD=AB,得h*AB=16
E到CD的高为h,则E到AB的高为h+AD
∴S△EAB = (h+AD) * AB / 2
又因为S△EAB = S矩形ABCD-S梯形FDCB+S△AEF = AD*AB - 50 + 18 = AD*AB-32
∴(h+AD)*AB/2 = AD*AB - 32
h*AB = AD*AB - 64,故AD*AB = 16+64 = 80
所以S△ABF = S矩形ABCD - S梯形FDCB = AD*AB - 50 = 30
即AF * AB / 2 = 30, AF*AB = 60
所以AF*AB / (AB*AD) = 6/8,即AF : AD = 3 : 4,即AF : DF = 3 : 1
三角形EAF和EFD在底边AF和DF上同高
所以S△EAF：S△EFD = AF：DF = 3:1,最终得S△DEF = S△EAF/3 = 6