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 \(ABCM\) ?

CalculatorUser Feb 9, 2019

#1**+2 **

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

CPhill Feb 9, 2019