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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 $ABMO$.

 Jan 26, 2021
 #1
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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

 

 

cool cool cool

 Jan 26, 2021

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