I have been looking through my book for an example and I'm not able to find one. May someone help me.
The question is: What is the area of a regular hexagon whose sides are each 12 inches long? (Round to the nearest square inch.)
For a regular polygon, if given the length l of the sides, the area is given by:
A = [(l^2)*N]/[4tan(pi/N)]
where l is the length of a side and N is the number of sides.
For a regular hexagon:
N = 6
=> A = (6*l^2)/(4tan(pi/6))
now replace l with the length of one side
For more formulas and a cool interactive utility, see Regular polygon area formula
I think here's a way to work it out without that formula (btw, this might form basis of the forumla I'm not too sure)
Since it's a regular hexagon each interior angle is 120 deg.
Split the hexagon into its 4 traingles, find the area of each traingle and x4 for the total area.
area of any triangle = 1/2 ab Sin c
we have 12" and 12" for two sides and 120 deg for the other angle (30 deg ea for rest since 12 and 12 are same). This is sufficient information to use this formula.
so area of this triangle = 62.35382907
Area of hexagon = 4 x 62.35382907
is that correct? I may have gone wrong in workings so...
How can you split the hexagon into 4 triangles, each nhaving an angle of 120?
I would split the hexagon into 6 equilateral triangles, having the side 12.
So, area of any triangle = 1/2 ab Sin c
Thanks for enlighting me
I thought the answer was a bit dodgy