# 2 questions of maths mock

http://upload.lsforum.net/users/public/a92491v184....

thanks for solving.

these questions belong to core maths. do you have another way to solve question 1??

### 2 Answers

- wyLv 77 years agoFavorite Answer
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.

- 螞蟻雄兵Lv 77 years ago
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