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About AMC 12 Circles II Problem #29
About AMC 12 Circles II Problem #29
For question number 29 on Circles II, AMC 12, using 30,60,90 triangles, I got that the radius is sqrt(3)/2. For the next step, the answer shows that the sum of the areas of the semicircles are sqrt(3)/4*3*pi, but it should divide it by 8 instead of 4 since they are semi circles. The file for the answer is attached in this post.
Re: About AMC 12 Circles II Problem #29
That's correct, Fan, the area of the semicircles should be divided by $2$, so the combined areas of the three semicircles is be $\displaystyle \frac{9\pi}{8}$. Thus the final answer is $$\frac{\pi+6\sqrt{3}}{8}.$$
We'll update the solution on the handout. Thank you for letting us know!
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