An octagon is inscribed in a circle with a radius of 6cm. Find the area of the region outside the octagon.
HElp please, thank you.!
$\displaystyle A_{\textrm{circle}} = \pi r^2$
$\displaystyle = \pi (6\,\textrm{cm})^2$
$\displaystyle = 36\pi\,\textrm{cm}^2$.
The octagon (I'm assuming it's a regular octagon) can be broken into 8 isosceles triangles. The two equal sides are each $\displaystyle 6\,\textrm{cm}$ and the angle between them is $\displaystyle \frac{360^\circ}{8} = 45^\circ$.
So $\displaystyle A_{\textrm{triangle}} = \frac{1}{2}ab\,\sin{C}$
$\displaystyle = \frac{1}{2}(6\,\textrm{cm})(6\,\textrm{cm})\sin{45 ^\circ}$
$\displaystyle = 18\cdot \frac{\sqrt{2}}{2} \,\textrm{cm}^2$
$\displaystyle = 9\sqrt{2}\,\textrm{cm}^2$.
Therefore $\displaystyle A_{\textrm{octagon}} = 8\cdot A_{\textrm{triangle}}$
$\displaystyle = 8\cdot 9\sqrt{2}\,\textrm{cm}^2$
$\displaystyle = 72\sqrt{2}\,\textrm{cm}^2$.
Therefore, the "leftover area" is
$\displaystyle A = A_{\textrm{circle}} - A_{\textrm{octagon}}$
$\displaystyle = 36\pi\, \textrm{cm}^2 - 72\sqrt{2}\,\textrm{cm}^2$
$\displaystyle = 36(\pi - 2\sqrt{2})\,\textrm{cm}^2$.