1. ## Octagon!

An octagon is inscribed in a circle with a radius of 6cm. Find the area of the region outside the octagon.

2. Originally Posted by reiward
An octagon is inscribed in a circle with a radius of 6cm. Find the area of the region outside the octagon.

$A_{\textrm{circle}} = \pi r^2$

$= \pi (6\,\textrm{cm})^2$

$= 36\pi\,\textrm{cm}^2$.

The octagon (I'm assuming it's a regular octagon) can be broken into 8 isosceles triangles. The two equal sides are each $6\,\textrm{cm}$ and the angle between them is $\frac{360^\circ}{8} = 45^\circ$.

So $A_{\textrm{triangle}} = \frac{1}{2}ab\,\sin{C}$

$= \frac{1}{2}(6\,\textrm{cm})(6\,\textrm{cm})\sin{45 ^\circ}$

$= 18\cdot \frac{\sqrt{2}}{2} \,\textrm{cm}^2$

$= 9\sqrt{2}\,\textrm{cm}^2$.

Therefore $A_{\textrm{octagon}} = 8\cdot A_{\textrm{triangle}}$

$= 8\cdot 9\sqrt{2}\,\textrm{cm}^2$

$= 72\sqrt{2}\,\textrm{cm}^2$.

Therefore, the "leftover area" is

$A = A_{\textrm{circle}} - A_{\textrm{octagon}}$

$= 36\pi\, \textrm{cm}^2 - 72\sqrt{2}\,\textrm{cm}^2$

$= 36(\pi - 2\sqrt{2})\,\textrm{cm}^2$.