The lengths, in order, of four consecutive sides of an equiangular hexagon are 1, 7, 2 and 4 units, respectively. what is the sum of the lengths of the two remaining sides?

Let ABCDEF be an equilanqular hexagon with consecutive sides 1,7,4,2, respectively. All angles of this hexagon are equal to 120° (because total anglea sum is 720°).

Draw lines: CG parallel to ED, PG and AH parallel to BC and EG parallel to CD. You obtain three parallelograms CGED, PGEF, ABCH and one trapezoid APGH.

Since CGED is parallelogram, CG=ED=2, GE=CD=4.

Since ABCH is parallelogram, AH=BC=7, AB+CH=1. Then GH=CG-CH=2-1=1.
Since PFEG is parallelogram, PF=EG=4, EF=PG. Let's find GP. APGH is equilateral trapezoid, then AP=GH=1 ahd thus FA=1+4=5 and GP=AH-0.5-0.5=7-0.5-0.5=6.

All sides have lengths 1,7,4,2,6,5, respectively.