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Prove that the area of an octagon with side a can be computed by the formula: $$S=2a^2(1+\sqrt 2)$$ I tried to prove this by the area of 8 isoscles triangles inside the octagon, but I didn't manage to find the area of those triangles.

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2 Answers

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Divide the octagon into $8$ isosceles triangles each having their apex at the centre of the octagon. Each apical angle is $45^\circ$ and each base angle is $\frac{180^\circ-45^\circ}{2}$

The side length $s$ of each such triangle can be found with the sine rule.

$s = \frac{a}{\sin 45^\circ}\sin \frac{180^\circ-45^\circ}{2} = a\sqrt 2 \cos \frac{45^\circ}{2}$

The area of each triangle is then $\frac 12 s^2 \sin 45^\circ = \frac{a^2}{\sqrt 2}\cos^2 \frac{45^\circ}{2} = \frac{a^2}{\sqrt 2} \frac{1 + \cos 45^\circ}{2} = \frac 14 a^2 (1 + \sqrt 2)$ where the second last step used the half angle formula.

The area of the octagon is therefore eight times that, which is $2a^2(1 + \sqrt 2)$

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Join all the vertices of the regular octagon to the centre. This would give us 8 isosceles triangles, each with base $a$ and the angle at the vertex would be $45$ degrees. Drop a perpendicular from the vertex and now: $tan(\frac{45}{2}) = (\frac{a}{2})(height)$. Area of the triangle = $(\frac{1}{2})(a)(height)$ = $(\frac{1}{2})(a)$ $\frac{tan(\frac{45}{2})}{\frac{a}{2}}$ Now multiply this by 8 and you will get the required expression.

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