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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 \(ADO\)

 Apr 15, 2022
 #1
avatar+2455 
+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:

 

 

 Apr 15, 2022
 #3
avatar+124598 
+1

You beat me to it, BuilderBoi  !!!!!   {LOL!!!}

 

cool cool cool

CPhill  Apr 15, 2022
 #4
avatar+2455 
+1

lol, at least we have the same answers :)

BuilderBoi  Apr 15, 2022
 #2
avatar+124598 
+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 :

 

 

cool cool cool

 Apr 15, 2022

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