The following trigonometric approach is less elegant (and somewhat harder) than the ones that have been suggested. Its main advantage is that something like it works for any regular polygon.
Let $P$ be the point at the centre of the octagon. Join $P$ to all the vertices. We have divided the octagon into $8$ triangles. Now we find the area of one of these triangles, and multiply by $8$.
Let $PAB$ be one of our $8$ triangles. Draw a perpendicular from $P$ to $AB$. Suppose that this perpendicular meets $AB$ at $Q$. Then the area of $\triangle PAB$ is $\frac{1}{2}(AB)(PQ)$. We know that $AB=10$, and we need to find $PQ$.
Angle $APQ$ is $360^\circ/8$, that is, $45^\circ$. So the two equal angles $PAQ$ and $PBQ$ add up to $180^\circ-45^\circ$, that is, $135^\circ$. So each of them is $67.5^\circ$.
Note that $\frac{PQ}{AQ}=\tan(\angle PAQ)=\tan(67.5^\circ).$ But $AQ=5$, and therefore $PQ=5\tan(67.5^\circ).$ So the area of $\triangle PAB$ is $\frac{1}{2}(10)(5\tan(67.5^\circ))$. Multiply by $8$, and simplify. We get that the octagon has area $200\tan(67.5^\circ))$. If you want a decimal approximation, the calculator will give you one.