# regular octagon with inscribed square - area ratio question The question is shown in the picture attached. This is really puzzling to me - given that the octagon is regular with equal side length and interior angles, shouldn't the inscribed square ACEG and BDFH be exactly the same, thus same area? but the answer given is not 7/9, instead it's 9/14. How could that be possible? ThankS!

• The octagon is not given to be regular, only equilateral. The interior angles are not altogether equal. – Daniel Mathias Jan 1 at 3:42 I have added points $$I,J,K,L$$ on the axes of symmetry of the square parallel to the sides, on the mid point of the sides. Then the given proportion $$7/9$$ an be realized iff the proportion of the areas of $$\Delta AOI$$ and $$\Delta AOB$$ is $$7/9$$, i.e. $$\frac{OI}{OB}=\frac79\ .$$ (Here $$O$$ is the center of the square.)
The asked proportion is then $$\frac {\operatorname{Area}(\Delta OBM)} {\operatorname{Area}(\Delta OBC)} = \frac{OM}{OC} = \frac{9\sqrt 2/2}{7\sqrt 2} = \frac{9}{14} \ .$$ (Here, $$M=OC\cap BD$$.)