Triangle ABC is drawn inside regular hexagon ABCDEF. What is the ratio of the area of triangle ABC to the area of the hexagon?

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

10

step-by-step explanation:

3 qrs + 3/2 xy hope this : /

step-by-step explanation: