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Suppose that a regular pentagon circumscribes a circle of radius $r$. Show that the area of the pentagon is $5r^2\tan(36°)$.

I know that the area of a triangle is $\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. But how do I solve this problem and prove this theory?

This was my attempt and I'm stuck.

$5r^2\tan(36°) = \frac{1}{2}bh \cdot \sin(36°)$

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  • $\begingroup$ The result must be $5r²\cos(36)\sin(36)=2.5r^2\sin(72)$ $\endgroup$ – newzad May 20 '16 at 19:11
  • $\begingroup$ No the result has to be 5r^2tan(36°). I agree with the cos(36)sin(36) part but they want proof that I can get to the actual area I posted. $\endgroup$ – crhodes May 27 '16 at 3:43
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The central angle of each of the five triangles that make up the circumscribed pentagon is $72^{\circ}$. Half of this is $36^{\circ}.$

Cut each of these triangles in half to make ten scalene right triangles, with angles $36, 54, \text{ and } 90$ degrees.

The distance from the center of the circle (and of the pentagon) to the edge of the circle is $r$. This is also the distance to the centers of each of the sides of the pentagon.

Half the length of the pentagon's side by trigonometry, then, is $x = r \tan 36^{\circ},$ making the area of each of the right triangles

$$A = \frac{rx}{2} = \frac{r \cdot r \tan 36^{\circ}}{2}.$$

There are $10$ such triangles in the pentagon, so $10A = 5r^2 \tan 36^{\circ}$ is your answer.

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  • $\begingroup$ "The perpendicular length from the center of the circle (and pentagon) to the edge of the circle" is not $r$, it is $r\cos{36}$. The radius of circle is $r$. $\endgroup$ – newzad May 20 '16 at 20:59
  • $\begingroup$ @nikamed Edited that paragraph; probably wasn't terribly clear. $\endgroup$ – John May 20 '16 at 21:14

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