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Let [tex]A_n[/tex] be the area of a polygon with

Ok, I drew a circle with congruent triangles inscribed in it. I assumed that it was an equilateral triangle, so it has height [tex]\frac{\sqrt{3}}{2}r[/tex].

So far I have

[tex]A_{triangle}=\frac 1 2 \cdot r \cdot \frac{\sqrt{3}}{2}r[/tex]

[tex]\sin{\frac{2\pi}{n}}=\frac{\frac{\sqrt{3}}{2}r}{r}=\frac{\sqrt{3}}{2}[/tex]

[tex]A_{triangle}=\frac 1 2 \cdot r^2\cdot \sin{\frac{2\pi}{n}}[/tex]

Now I'm stuck, maybe my assumption was incorrect, and I also do not know how to incorporate the fact that it is inscribed in the circle. I know I need to take it into consideration noticing that it wants me to express the answer with the area of a circle as part of the answer. Or perhaps [tex]\pi r^2[/tex] appears through substitutions?

*n*equal sides inscribed in a circle with radius r. By dividing the polygon into*n*congruent triangles with central angle [tex]\frac{2\pi}{n}[/tex], show that [tex]A_n=\frac 1 2 \pi r^2\sin{\frac{2\pi}{n}}.[/tex]Ok, I drew a circle with congruent triangles inscribed in it. I assumed that it was an equilateral triangle, so it has height [tex]\frac{\sqrt{3}}{2}r[/tex].

So far I have

[tex]A_{triangle}=\frac 1 2 \cdot r \cdot \frac{\sqrt{3}}{2}r[/tex]

[tex]\sin{\frac{2\pi}{n}}=\frac{\frac{\sqrt{3}}{2}r}{r}=\frac{\sqrt{3}}{2}[/tex]

[tex]A_{triangle}=\frac 1 2 \cdot r^2\cdot \sin{\frac{2\pi}{n}}[/tex]

Now I'm stuck, maybe my assumption was incorrect, and I also do not know how to incorporate the fact that it is inscribed in the circle. I know I need to take it into consideration noticing that it wants me to express the answer with the area of a circle as part of the answer. Or perhaps [tex]\pi r^2[/tex] appears through substitutions?

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