Question 1108613
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MY BET ABOUT THE IMAGE:
{{{drawing(350,200,-1,13,-1,7,
line(0,0,12,0),line(3,5.196,9,5.196),
line(0,0,3,5.196),line(12,0,9,5.196),
red(line(9,5.196,7.5,2.598)),
red(line(4.5,2.598,3,5.196)),
red(line(4.5,2.598,7.5,2.598)),
red(line(9,0,7.5,2.598)),
red(line(4.5,2.598,3,0)),
locate(-0.1,0,A),locate(11.9,0,D),
locate(2.9,5.8,B),locate(8.9,5.8,C)
)}}}
If that is how it is, we know some segment lengths:
{{{drawing(350,200,-1,13,-1,7,
green(rectangle(9,0,9.3,0.3)),
green(triangle(9,0,9,5.196,12,0)),
line(0,0,12,0),line(3,5.196,9,5.196),
line(0,0,3,5.196),line(12,0,9,5.196),
red(line(9,5.196,7.5,2.598)),
red(line(4.5,2.598,3,5.196)),
red(line(4.5,2.598,7.5,2.598)),
red(line(9,0,7.5,2.598)),
red(line(4.5,2.598,3,0)),
locate(-0.1,0,A),locate(11.9,0,D),
locate(2.9,5.8,B),locate(8.9,5.8,C),
locate(8.9,0,E),locate(2.9,0,F),
locate(9.1,2.8,green(h)),
locate(5.9,5.2,6),locate(5.9,0.6,6),
locate(1.4,0.6,x),locate(10.4,0.6,x),
locate(1.5,2.6,6),locate(10.2,2.6,6)
)}}}
We want to know height {{{green(h)=CE}}} and {{{x=DE=AF}}} .
The large trapezoid and the four smaller trapezoids are similar figures.
So, {{{x/6=6/(6+2x)}}} .
{{{x(6+2x)=6*6}}}
{{{2x^2+6x=36}}}
{{{2x^2+6x-36=0}}}
{{{x^2+3x-18=0}}}
{{{(x+6)(x-3)=0}}}
{{{x=3}}} because a negative length does not make sense.
Then,
{{{green(h)=sqrt(6^2-3^2)=sqrt(36-9)=sqrt(27)=3sqrt(3)=about5.196}}}
That makes the area of ABCD the same as the area of AGCE
{{{drawing(350,200,-1,13,-1,7,
green(rectangle(9,0,9.3,0.3)),
green(rectangle(0,0,9,5.196)),
line(0,0,12,0),line(3,5.196,9,5.196),
line(0,0,3,5.196),line(12,0,9,5.196),
red(line(9,5.196,7.5,2.598)),
red(line(4.5,2.598,3,5.196)),
red(line(4.5,2.598,7.5,2.598)),
red(line(9,0,7.5,2.598)),
red(line(4.5,2.598,3,0)),
locate(-0.1,0,A),locate(11.9,0,D),
locate(2.9,5.8,B),locate(8.9,5.8,C),
locate(8.9,0,E),locate(2.9,0,F),
locate(9.1,2.4,green(h)-3sqrt(3)),
locate(5.9,5.2,6),locate(5.9,0.6,6),
locate(1.2,0.6,x=3),locate(10.4,0.6,x),
locate(1.5,2.6,6),locate(10.2,2.6,6),
locate(-0.1,5.8,G)
)}}}
{{{(6+3)*3sqrt(3)=9*3sqrt(3)=27sqrt(3)}}}
Although your teacher may prefer
{{{BC+6}}} , {{{AD=6+2x=6+2*3=6+6=12}}} , {{{green(h)=3sqrt(3)=about5.196}}}
{{{area[ABCD]}}}{{{"="}}}{{{(BC+AD)*green(h)/2}}}{{{"="}}}{{{(12+6)*3sqrt(3)/2}}}{{{"="}}}{{{9*3sqrt(3)=27sqrt(3)}}} .