Mathematics Question and Solution
1. The proof that triangle EGH is equilateral.
Let the side length of the regular hexagon be 1 unit.
a = ⅙(180(6-2))
a = 30*4
a = 120°
a is single interior angle of the regular hexagon.
b² = 1²+0.5²-2*0.5cos120
b = 1.3228756555 units.
b is EG.
(1.3228756555/sin120) = (0.5/sinc)
c = 19.1066053509°
c is angle DEG.
d² = 2-2cos120
d = 1.7320508076 units.
d is BF.
e = ½(d)
e = ½(1.7320508076)
e = 0.8660254038 units.
e is FH.
g² = 1²+0.8660254038²
g = 1.3228756555 units.
g is EH.
Notice!
EH = EG = 1.3228756555 units, which is two equal sides.of triangle EGH.
tanh = 0.8660254038/1
h = 40.8933946496°
h is angle HEF.
i = 120-40.8933946496-19.1066053509
i = 60°
I is angle GEH.
j² = 1.3228756555²+1.3228756555-2*1.3228756555*1.3228756555cos60
j = 1.3228756555 units
j is GH.
Observing the above solution;
EG = GH = EH = 1.3228756555 units.
Therefore, triangle EGH is equilateral.
2. Calculating ratio area EGH/area ABCDEF.
Area EGH is;
0.5*1.3228756555²sin60
= 0.7577722283 square units.
Are ABCDEF is;
2(0.5sin120)+(1*1.7320508076)
= 2.5980762114 square units.
It implies;
Area EGH ÷ Area ABCDEF
= 0.7577722283÷2.5980762114
= 0.2916666667
= 7/24
