A unit square $ABCD$ is such that $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. A straight line is drawn from $A$ to $N$ and another from $D$ to $M$. This two lines $AN$ and $DM$ meet at $O$ which is inside square $ABCD$. Find the area of $ABMO$.
+130511 See the following image
The slope of the line containing AN is (1/.5) = 2
And the equation of the line containing this segent is y = 2x (1)
The slope of the line containing DM is ( 1 - .5) /( 0 -1) = -.5 = -1/2
And the equation of the line containing DM is
y = (-1/2)x + 1 (2)
Equating (1) and (2) we have
2x =(-1/2) x + 1
2.5x = 1
x =1/2.5 = 2/5 = .4 =the xcoordinate of O
And the y coordinate of O is y = 2(.4) = .8
So triangle DCM has a base of 1 and a height of .5 = 1/2
So its area = (1/2)(1) (1/2) = 1/4
And triangle DOA as a base of 1 and a height of .4 = 4/10 = 2/5
So its area is (1/2) (1)(2/5) = 1/5
So [ABMO ] = area of the square - [DCM ] - [ DOA ] =
1 - 1/4 -1/5 =
1 - .25 - .20 =
1 - .45 =
.55 = 55/100 = 11/20
