Geometry problem #2 *Old rule: no calculator!*

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Geometry problem #2 *Old rule: no calculator!*

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Area of the polygon-12 * area of the 1 cm side equilateral triangle. Am I missing something?
I think that's it :)

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You could find the area of the entire polygon by breaking it into triangles and multiplying the area of that triangle by 12: |dw:1338109350047:dw| If you divide that triangle in half, it becomes a right triangle so you can use the magic of trigonometric functions to find the height. |dw:1338109449920:dw| Once you have the height, you can just find the area of the triangle using "A=1/2 bh", and then multiply by 12 to get area of the regular 12-sided polygon.
Hmm why is everyone viewing it?

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