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# Geometry problem

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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
+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
+124598
+1

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

CPhill  Apr 15, 2022
#4
+2455
+1

lol, at least we have the same answers :)

BuilderBoi  Apr 15, 2022
#2
+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 :

Apr 15, 2022