# 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 '20 at 3:42

The situation can not be realized with a regular octagon, we have rather the following situation:

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, $$M=OC\cap BD$$.)

• What software did you use to make the drawing? Very nice and clear. Thanks – user526427 Jan 1 '20 at 4:55
• The software used is geogebra. It is relatively a versatile tool. The perspective used is "Geometry", the axes are added, and the "milimeter paper" too. Immediately after drawing a line one can set the color (and the thickness, and the style). For the use on this page, it is enough to Export -> Copy to clipboard, then use here a Control+G (insert graphic) and paste (the clipboard). – dan_fulea Jan 1 '20 at 5:45