# Area of a polygon on the unit circle

• Apr 5th 2011, 11:41 PM
dondonlouie
Area of a polygon on the unit circle
I'm having a hard time solving this problem:
Find the area of a regular dodecagon whos vertices are 12 equally spaced points on the unit circle.

what i did was split the entire polygon into 12 smaller triangles
A = 1/2absin(theta)
A = 1/2 (1)(1)sin(pi/4)
A = pi/4 / 2 = (1/sqrt(2))/2 = 1/2sqrt(2) - for 1 triangle

for the entire polygon i did 12(1/2sqrt(2)) = 3sqrt(2)
but then the answer says it's 3
what am i doing wrong?
• Apr 5th 2011, 11:54 PM
earboth
Quote:

Originally Posted by dondonlouie
I'm having a hard time solving this problem:
Find the area of a regular dodecagon whos vertices are 12 equally spaced points on the unit circle. <--- OK

what i did was split the entire polygon into 12 smaller triangles
A = 1/2absin(theta)
A = 1/2 (1)(1)sin(pi/4)
The central angle of one triangle is calculated by: $\frac{2 \pi}{12}=\frac16 \pi$.
... and $\sin\left(\frac16 \pi\right)=\frac12$
A = pi/4 / 2 = (1/sqrt(2))/2 = 1/2sqrt(2) - for 1 triangle

for the entire triangle i did 12(1/2sqrt(2)) = 3sqrt(2) <--- the value in red is wrong. See above!
but then the answer says it's 3
what am i doing wrong?

...
• Apr 6th 2011, 12:04 AM
dondonlouie
Quote:

Originally Posted by earboth
...

how did you get the central angle? i thought it was pi/4 cuz the two sides each have a radius of 1 D:
• Apr 6th 2011, 12:14 AM
earboth
Quote:

Originally Posted by dondonlouie
I'm having a hard time solving this problem:
Find the area of a regular dodecagon whos vertices are 12 equally spaced points on the unit circle.

...
what am i doing wrong?

Quote:

Originally Posted by dondonlouie
how did you get the central angle? i thought it was pi/4 cuz the two sides each have a radius of 1 D:

Make a sketch!

With 12 points on the unit circle you produce 12 isosceles triangles whose legs include an angle of $\frac1{12} \cdot 2 \pi = \frac16 \pi$
• Apr 6th 2011, 12:16 AM
dondonlouie
Quote:

Originally Posted by earboth
Make a sketch!

With 12 points on the unit circle you produce 12 isosceles triangles whose legs include an angle of $\frac1{12} \cdot 2 \pi = \frac16 \pi$

ahhh okay i see it now (: thank you so much!