Given a regular octagon $ABCDEFGH$ with side length $3$. By drawing the four diagonals $AF$, $BE$, $CH$, and $DG$, the octagon is divided into a square, four triangles, and four rectangles. Find the sum of the areas of the square and the four triangles.
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Tags: geometry, octagon, areas
MrRandomness
20.09.2021 17:21
The areas of square $=3 \cdot 3=9$ The areas of four triangles:- $\angle AO_1H=45^o$ Now, $$HO_1=AO_1=\frac{AH}{\sqrt{2}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$$$$S_{AHO_1} = \frac{3\sqrt{2}}{2} \cdot \frac{3\sqrt{2}}{2} \cdot \frac{1}{2}=\frac{9}{4}$$Areas of $4$ triangles $=4 \cdot \frac{9}{4}=9$ Sum$=9+9=18$
MrRandomness
20.09.2021 17:24
For the image please refer https://drive.google.com/file/d/1WwH3BuvGe_fgFBKiXgHRK1wDloddN-qb/view?usp=drivesdk
michchess
20.09.2021 17:24
i like to think of the 4 triangles as 4 quarters of the center square instead
MrRandomness
20.09.2021 17:25
michchess wrote: i like to think of the 4 triangles as 4 quarters of the center square instead Ya that would also work