# Region of an Octagon +-- {.image} [[RegionofanOctagon.png:pic]] > The shaded area has the same value as the perimeter of the regular octagon. What is this value? =-- ## Solution by [[Isosceles]] [[Right-angled Triangles]] +-- {.image} [[RegionofanOctagonLabelled.png:pic]] =-- With the points labelled as above, $D A$ is [[perpendicular]] to $E D$, and $E B$ is at $45^\circ$ to $E D$, so triangle $E D F$ is an [[isosceles]] [[right-angled triangle]]. Let $E D$ have length $a$, then triangle $E D F$ has area $\frac{1}{2} a^2$. The same applies to triangle $A B F$, so the shaded region has area $a^2$. The perimeter of the octagon is $8 a$, so $8 a = a^2$ and hence $a = 8$, so the common value for the area of the shaded region and the perimeter of the octagon is $64$.