What is the area of a regular hexagon (6 sides) whose ℎ ℎ 6 ℎ ℎ ℎ ℎ 3√3 ? Show work
A) 54√3 2
B) 36√3 2
C) 108√3 2
D) 18√3 2

Step-by-step explanation:

With reference to the regular hexagon, from the image above we can see that it is formed by six triangles whose sides are two circle's radii and the hexagon's side. The angle of each of these triangles' vertex that is in the circle center is equal to 360∘6=60∘ and so must be the two other angles formed with the triangle's base to each one of the radii: so these triangles are equilateral.

The apothem divides equally each one of the equilateral triangles in two right triangles whose sides are circle's radius, apothem and half of the hexagon's side. Since the apothem forms a right angle with the hexagon's side and since the hexagon's side forms 60∘ with a circle's radius with an endpoint in common with the hexagon's side, we can determine the side in this fashion:

tan60∘=opposed cathetusadjacent cathetus => √3=Apothemside2 => side=(2√3)Apothem

As already mentioned the area of the regular hexagon is formed by the area of 6 equilateral triangles (for each of these triangle's the base is a hexagon's side and the apothem functions as height) or:

Shexagon=6⋅S△=6(base)(height)2=3(2√3)Apothem⋅Apothem=(6√3)(Apothem)2

=> Shexagon=6×62√3=216

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.