What is the area of a regular hexagon with sides 1 in. long?

• A regular hexagon of side 1 inch can be divided in to 6 equilateral triangle of side 1 inch.
  
  Area of each triangle = (root 3/4)* side^2 = (root 3/4)
  Area of 6 triangles = 6*(root 3/4)= 1.5 root 3= 2.59 square inch
• 12 years agoHelpfull: Yes(7) No(0)
• Area of hexagon= [3*3^(1/2)*t^2], where t is lem=ngth of each side.
  
  Therefore the area of 1 in. lenght is=2.59 in.^2
• 12 years agoHelpfull: Yes(2) No(1)
• area of regular polygonal =(n*a^2/4)*cot(180/n)
  n= no of sides
  a=length of side
  6*1^2/4(cot(180/6))
  6/4*cot(30) cot(30)=1.717
  (6*1.717)/4
  2.59
• 12 years agoHelpfull: Yes(1) No(1)
• Regular Hexagon has 6 equilateral triangle.
  In general Area of Triangle = (bh)/2.
  For equilateral triangle b=h. Therefore Area= b^2/2.
  1 Hexagon = 6 equilateral Triangle
  Area of Hexagon = 6(b^2/2)= 3(b^2)
  b=1, so Area of hexagon is 3 sq units.
  
• 11 years agoHelpfull: Yes(1) No(2)
• its a regular hexagon.....so,,if we divide it we get two trapeziums.....
  area of trapezium=h(each side+length of largest diagonal)/2
  now we get the angle divided as 60 degree each....
  the length of "h"=1*sin(60)=root(3)/2
  and the base will be=1*sin(30)=1/2
  the other side will also be 1/2
  the length of the diagonal will be=2 inch..
  now from the formula=(root(3)*3/4)*2 (because we have two trapeziums)
  ans will be :3root(3)/2
• 11 years agoHelpfull: Yes(0) No(0)
• (3*3^.5)/2
  
  
  its formula
• 11 years agoHelpfull: Yes(0) No(1)
• a regular hexagon can be considered as 4 equilateral triangles + 2 squares each have side 1 inch
  s0 area=4*(root3)/4*a*a+2*a*a where(a=1)
  so area=2+(root3)=3.732 in^2
• 11 years agoHelpfull: Yes(0) No(0)