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 \(ADO\)
+2668 I solved this problem using co-ordinate geometry.
Let the bottom-right corner be the origin. The equations for lines \(AN\) and \(DM\) are \(y = {1 \over 2}x\) and \(y=-2x+1\)
Letting these 2 equations equal each other, we have: \(0.5x=-2x+1\).
Solving, we find \(x = 0.4\)
Thus, the area of \(\triangle AMO\) is \(1 \times {2 \over 5} \div 2 = \color{brown}\boxed{1 \over 5}\)
Here is the diagram:
+130071 Let A = (0,0) D = (0,1) N= (.5, 1) M = (1,.5)
The slope of the line sebment AN = [1 -0] / [ .5 - 0 ] = 2
The equation of the line containing this segment is
y = 2x
The slope of the line segment DM = [ .5 - 1 ] / [ 1 - 0 ] = -.5 = -1/2
The equation of the line containing this segment is
y = (-1/2)(x - 1) + .5
y = (-1/2)x + 1
Setting the equations = and solving for x will give us the x coordinate of O = the height of ADO
2x = (-1/2)x + 1
2.5x = 1
x = 1/2.5 = 10/25 = .4
So....the area of ADO = (1/2) AD * .4 = (1/2) (1) (.4) = .2 units
Here's a pic :
