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 \(ADO\)

Guest Apr 15, 2022

#1**+1 **

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:

BuilderBoi Apr 15, 2022

#2**+2 **

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 :

CPhill Apr 15, 2022