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# Cross-Shape Geometry Problem (AMC 10A)

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Consider the \(12\) sided polygon \(ABCDEFGHIJKL\) below, Each of its sides has the length of \(4 \), and each consecutive sides form a right angle. What is the area of \(ABCM\) ? Feb 9, 2019

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Maybe easier ways to do this....but....

Let H =   (0, 0)     A = (0, 12)   C = (4, 8)    and  G = (4, 0)

The slope of the line containing  AG   =  [ -12/4] = -3

And the equation of the line on which AG lies is   y = -3x + 12   (1)

The slope of the line containing HC is  8/4 = 2

And the equation of the line on which HC lies is   y = 2x   (2)

So....setting (1) = (2) to find the x coordinate of M, we have that

-3x + 12 = 2x

12 = 5x

12/5 = x

And y = 2(12/5)  = 24/5

Connecting LC  and calling the intersection of LC and  AG   =  N

Using similar triangles

Triangle ALN is similar to triangle AHG

So LN /AL = HG / AH

LN /4 = 4 / 12

LN /4 = 1/3

LN = 4/3

So.....NC is the base of trapezoid ABCN  = LC - LN =  4 - 4/3  = 8/3.....and the area of this trapezoid is

(1/2)BC ( AB + NC) =   (1/2)(4)(4 + 8/3)  = 2 ( 20/3) = 40/3 units^2      (3)

And triangle NCM  has a base of NC =  11/3   and a height of  8 - 24/5 =  16/5

So...the area of this triangle is  (1/2)(11/3)(16/5) =  88/15  units^2    (4)

So...[ ABCM ]   =   (3) + (4)   =   [ 40/3 + 88/15 ]  =  [288/15] units^2   =  19.2 units^2   Feb 9, 2019