Waterloo Problem of the Week (E, week 29, 2024)

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[@danielbranscombe6662]

nice problem and solution. Here is how I solved it
Drop a perpendicular from J down to CD and call it K
Then we can take the area of KJFC and break it down as the sum of the areas of triangles KJG, FGJ, and FCG
area of KJFC=KC*(KJ+CF)/2=6*(9/2+3)/2=45/2
area of KJG=KJ*KG/2=(9/2)*2/2=9/2
area of FCG=FC*GC/2=3*4/2=6
thus area of FGJ=KJFC-KJG-FCG=45/2-9/2-6=12

[@HowardsMathLessons]

Cool solution!