The distance between two opposite vertices of the dodecagon is 2. Find the area of the dodecagon.
Is there any way to do this without trigonometry?
And could you include a proof also? :O
$\endgroup$ 33 Answers
$\begingroup$Obviously we have to assume it is a regular dodecagon, otherwise its area is indeterminate. I follow your idea of taking a circle through the vertices. We have $\angle A_1OA_2 = 30^o$, so $\angle A_1OA_3 = 60^o$. Also $OA_1 = OA_3$, so the triangle $A_1OA_3$ is equilateral. We are given that $OA_1=1$, so $A_1A_3 = 1$. $OA_2$ intersects $A_1A_3$ at $M$ the midpoint of $A_1A_3$. So $A_1M = \frac12$.
The area of triangle $A_1OA_2$ is half its base $OA_2$ times its height $A_1M$ and hence $\frac14$. There are 12 triangles, so the total area of the dodecagon is 3.
$\endgroup$ $\begingroup$This animated gif gives a proof without words that the area of a regular dodecagon is three times the square of its circumradius:
Another form of the area for a regular dodecagon that doesn't explicitly contain trigonometric functions, is $A = 3R^2$ where $A$ is area and $R$ is the circumradius (distance from center to vertex).
Therefore you have that the circumradius = $1$ (half the distance between opposite vertices) hence the area $A=3\cdot1^2=3$.
You can prove this by a cute tiling argument:
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