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what is the area of a hexagon with the sides being 1.8m

2007-06-11 10:57:47 · 7 answers · asked by Chillin 1 in Education & Reference Homework Help

7 answers

use area of a regular polygon= 1/2aP
where a is the apothem and P is the perimeter
P would be 6(1.8) which is 10.8
since 360/6= 60, and 60/2=30, the right interior triangle of a hexagon is a 30-60-90 triangle, so use the scale factoer 1-root3-2
1.8/2 =.9, so set up the proportion 1/.9 = root3/a
you get a= .9root3
so plug the variables in
1/2(.9root3)(10.8)
you end up with a decimal answer of about 8.42 square meters.

2007-06-11 11:18:12 · answer #1 · answered by Anonymous · 0 0

http://i24.photobucket.com/albums/c3/rascalflattsgurl43/Untitled.jpg

I'm not sure which way you were taught, but we were taught to use the formula, A=1/2ap where a is the apothem and p is the perimeter. To get the apothem, you would draw a line from the center to the outside corner 'the radius'. Then draw a line from the center to the flat side. This makes a 30-60-90 triangle. Then use the 30-60-90 formula to figure out a. Then get the perimeter and put it in the formula.

2007-06-11 11:10:18 · answer #2 · answered by rascalflattsgurl43 2 · 0 0

i do no longer understand of any formulation, yet you could wreck a hexagon up into 6 triangles. basically connect the corners with a line. The formulation for the portion of a triangle is (one million/2)*b*h. If each and every of the aspects are an identical length, multiply this by making use of 6. no count number if it incredibly is an outstanding hexagon, you ought to discover each and every of the triangle areas and upload them mutually.

2016-10-07 07:50:46 · answer #3 · answered by emilios 4 · 0 0

Let s be the length of a side
You have a rectangle in the middle with the dimensions of s by sqrt(3)*s, so its area is s^2*sqrt(3).
The triangles from one side can be stacked on the other side to make another rectangle with the dimensions of .5s by sqrt(3)*s, so that area is s^2*sqrt(3)/2.

Adding them s^2*sqrt(3) + s^2*sqrt(3)/2 = 3*s^2*sqrt(3)/2

2007-06-11 11:02:06 · answer #4 · answered by dogsafire 7 · 0 0

Let r = the side of the triangle;
then the area of each triangle is
(1/2)r^2*sin(pi/3) = (1/2)r^2*sqrt(3)/2 = (1/4)r^2*sqrt(3).

For six such triangles the area = (3/2)r^2*sqrt(3)

2007-06-11 11:01:50 · answer #5 · answered by nellymegs 2 · 0 0

the formula to find the area of any regular polygons is:

Area = 1/2 * apathem * number of sides * side length

2007-06-11 11:10:27 · answer #6 · answered by      7 · 0 0

Area=3a^2sqrt(3)/2... hope this is helpful...a is the length of the sides.

2007-06-11 11:07:34 · answer #7 · answered by sheri641 2 · 0 0

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