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Math Help - Octagon!

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    Octagon!

    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.!
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  2. #2
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    Quote Originally Posted by reiward View Post
    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.!
    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.
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