Respuesta :
The sides of the regular hexagon ABCDEF can be posed as a. If so, the area of ABCDEF should be 6 times the are of the interior angle in the hexagon, considering there are 6 equilateral triangle that can fit in this regular hexagon,
Area = 6( a * a * sin 60 ) / 2,
Area = ( About ) 2.6 sq units
Now applying cosine for triangle ABC -
AC^2 = AB^2 + BC^2 – ( 2*AB*BC*cos 120 ),
a^2 + a^2 – ( 2a^2 * ( - 0.5 ) ) = a^2 + a^2+a^2 =3a^2,
AC = a√3
The area of ABC should thus be the following -
( a√3 * a√3 * sin 60 )/2 = 1.299038106 sq units
As you can see, the area of ABC is half the area of ABCDEF, thus the ratio of the area of ABC to ABCDEF is 1 : 2
