calculating the area of hexagon without knowing its apothem

Question

I am having a problem with task b. I am afraid I can't find the area of the hexagon (and, therefore, can't find the requested value of probability) because there is no apothem for the hexagon on the picture.

Answer 1

The area of a regular $n$-polygon with side $b$ is $\dfrac 1 4 b^2 \cot \dfrac \pi n$, which you can get from any halfway decent book of formulae (I use Schaum's "Mathematical Handbook" (1968)).

$\cot \dfrac \pi 6 = \dfrac {\sqrt 3} 3$.

The area of a polygon whose distance from side to centre is $r$ is $\pi r^2 \tan \dfrac \pi n$.

$\tan \dfrac \pi 5$ you can look up.

You should have enough to finish.

Answer 2

As @YvesDaoust said, by knowing the side, we can determine the area. But if we have still problem, one can see hexagon as a union of 6 concentrated regular trigons (equilateral trianles), and sum the areas of them. Also for area of inner regular pentagon we can do similarly, but with 5 concentrated isosceles triangles, And even we can do this for left inner regular tetragon (square), with 4 ones, if we are computationary masochist!