The lengths, in order, of four consecutive sides of an equiangular hexagon are 1, 7, 2 and 4 units, respectively. what is the su
m of the lengths of the two remaining sides?
1 answer:
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.
Answer: All sides have lengths 1,7,4,2,6,5, respectively.
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.
Answer: All sides have lengths 1,7,4,2,6,5, respectively.
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