Qing asked in 科學及數學數學 · 7 years ago

2 questions of maths mock

Update:

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

2 Answers

Rating
  • wy
    Lv 7
    7 years ago
    Favorite 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.

  • 7 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

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