= Solution
Join the centre of the <unit circle> to adjacent vertices of the <regular polygon>. Each of the $n$ resulting <triangles> has two sides of <length> $1$ and included <angle> $2\pi/n$. Bisecting one gives a <right triangle>, so each polygon side, a <chord of a circle>, has <length> $2\sin(\pi/n)$. Thus the perimeter is $2n\sin(\pi/n)$.
Each of these <triangles> has <area> $\tfrac12\sin(2\pi/n)$, so the polygon's <area> is $\tfrac n2\sin(2\pi/n)$. Comparing the two expressions gives
$$
\boxed{m=2,\qquad k=2.}
$$
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