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Find the area of a regular octagon given the radius of circle circumscribing it as $r$ .

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4 \sqrt{2} r^{2}

5 \sqrt{6} r^{2}

2 \sqrt{2} r^{2}

7 \sqrt{2} r^{2}

answer is C.

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Detailed Solution

Length of

AO =

radius of circle

= r

Length of

BO =

radius of circle

= r

Regular octagon has all sides equal, Thus angle subtended by each edge of regular octagon at centre of circle is equal.
So,

∠ AOB = \frac{360 °}{8} = 45 °

Now area of

ΔAOB = \frac{1}{2} r^{2} sinθ

= \frac{1}{2} r^{2} \sin 45 °

= \frac{1}{2} r^{2} \frac{1}{\sqrt{2}}

= \frac{r^{2}}{2 \sqrt{2}}

Now, to find the area of a regular octagon, we have to find areas of all the eight triangles separately and add them all.

Area of Octagon = Area of 8 triangle = 8 \times (area of ΔAOB)

Thus, area of octagon

= \frac{8 r^{2}}{2 \sqrt{2}}

= \frac{4 r^{2}}{\sqrt{2}}

= 2 \sqrt{2} r^{2}

Thus, the area of regular octagon inscribed in a circle of given radius

r = 2 \sqrt{2} r^{2}

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