maths 急!!!

Let MC be k
AB be h
and height ofTriangle MNC be n

Area of MNC= n*k/2
Area of MCD=k*h/2
Area of ABC=k*h
therforeArea of NCD= MCD-MNC = (k*h/2) - (n*k/2)
=k(h-n)/2

Area of MAC=Area of MCD(because they have same height and base)
Area of AMD=Area of ABC(because they have same height and base)

therfore :Area of AMN = Area of NCD=k(h-n)/2

Area of AND= Area of AMD-Area of AMN
=k*h-(k(h-n)/2)
=(k*h/2) + (n*k/2)
Areaof AND=(h-n)*2k
slove the two Area of AND
i.e. k=3n_________________________________(1)

area of ABMN=k*h- (n*k/2) ______________________(2)
sub(1)into(2)
we have k*3n-(n*k/2)
=5/2(kn)

AS above :area ofNCD=k(h-n)/2
=1(kn)
Area of NCD : area of ABMN =1(kn):5/2(kn)
=2:5
我呢個同樓上果個唔同...我用全面積黎計
用d我地最基本既野黎計
我只係show method 2比你睇=]

2008-03-26 22:00:25 補充：
我覺得第2位果位仁兄幾好真係唔錯...

Answer: 2:5

Solution:
Method 1:Analytic Approach
Refer to the figure.
(Figure: http://farm4.static.flickr.com/3251/2362975083_52f57cb127.jpg?v=0)
The equation of AC is
(y-0)/(x-a)=(2b-0)/(0-a)
-ay=2bx-2ab
2bx+ay-2ab=0
The equation of DM is
(y-0)/(x-0)=(b-0)/(a-0)
ay=bx
∵N(c,d)lies on both AC and DM.
we have 2bc+ad-2ab=0 ------------ (1)
and ad=bc --------------------------- (2)
Putting (2) into (1):
2ad+ad-2ab=0
3ad=2ab
d=(2/3)b ----------------------------- (3)
Putting (3) into (1):
a(2/3)b=bc
c=(2/3)a
The coorcinates of N are (2a/3, 2b/3).
Area of ΔNCD=(1/2){[0*0+a*(2b/3)+(2a/3)*0]-[a*0+(2a/3)*0+0*(2b/3)]}
=(1/3)ab
Area of ABMN
=(1/2){[0*(2b/3)+(2a/3)*b+a*2b+a*2b]-[(2a/3)*2b+a*(2b/3)+a*b+0*2b]}
=(5/6)ab
∴Area of NCD : Area of ABMN =(1/3)ab:(5/6)ab=2:5

Method 2: Deductive Approach
Refer to the figure.
(Figure: http://farm4.static.flickr.com/3015/2364039826_f9ca89ab6e.jpg?v=0)
Join AM. Let the area of ABCD be 12x.
∵M is the mid-point of BC.
∴Area of ΔABM=Area of ΔACM=Area ofΔCDM=1/4(12x)=3x
Consider ΔADN and ΔCMN.
∠ADN=∠CMN (alt.∠s, AD//BC)
∠DAN=∠MCN (alt.∠s, AD//BC)
∠AND=∠CNM (vert. opp. ∠s)
∴ΔADN~ΔCMN (AAA)
∴AN:NC=DN:NM=AD:CM=2:1 (corr. sides, ~Δs)
∴Area of ΔAMN:Area of ΔCMN=2:1
and Area of ΔNCD:Area od ΔCMN=2:1
∴Area of ΔAMN=Area of ΔACM*2/(2+1)=2x
Area of ΔNCD=Area of ΔCDM*2/(2+1)=2x
Area of ABMN=Area of ΔAMN+Area of ΔABM=5x
∴Area of NCD : area of ABMN =2x:5x=2:5

圖片參考：http://i292.photobucket.com/albums/mm27/cham_yan/--1.jpg
引一直角坐標系統,

設A=(0,1) , B=(0,0) , C=(2,0) , D=(2,1)

THEN M=(1,0)

SOLPE OF AC=-1/2

AC:(Y-0)=-1/2(X-2)

X+2Y-2=0--(1)


SLOPE OF MD=1

MD:(Y-0)=1(X-1)

Y=X-1--(2)

解(1) ,(2) 得N:(4/3 , 1/3)

Area of NCD=1/3

Area of ABMN=5/6

so the ratio of Area of NCD :Area of ABMN

=1/3:5/6

=2:5


SO THE ANS IS D