2 questions of maths mock

Let EC = 3, so BE = 7 and AD = EC + BE = 10.
Let area of triangle EFC = x.
Triangle AFD is similar to triangle EFC, so its area = (10/3)^2 x = 100x/9.
Area of triangle EFC : Area of triangle CFD = EF : FD ( triangles with same height) = EC : AD = 3 : 10
so area of triangle CFD = 10x/3.
Area of triangle AFD + area of triangle CFD = half the area of parallelogram
= area of triangle EFC + area of ABEF, that is
100x/9 + 10x/3 = x + area of ABEF
so area of ABEF = 100x/9 + 10/3 - x = (100 + 30 - 9)x/9 = 121x/9
so area of ABEF : area of triangle DFC = 121x/9 : 10x/3 = 121 : 30.

1
設四邊形ABCD面積=P
AC，BD相交於G
DE向量=(7/10)DC向量+(3/10)DB向量
=(7/10)DC向量+(6/10)DG向量
7/10+6/10=13/10
So
DF向量=(7/13)DC向量+(6/13)DG向量
GF：FC=7：6
△DFC面積=(1/4)*P*(6/13)=3P/26
△DEC面積=(1/2)*P*(3/10)=3P/20
△CEF面積=3P/20－3P/26=9P/260
四邊形ABCD面積=P/2－9P/260=121P/260
四邊形ABCD面積：△DFC面積=121P/260：3P/20=121：30

2
P(4A)+P(3A1B)+P(2A2B)
=C(9,4)/C(14,4)+C(9,3)*C(5,1)/C(14,4)+C(9,2)*C(5,2)/C(14,4)
=(126+420+360)/1001
=506/1001
=46/91