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Find the shaded area enter image description here

Here is the equation that i've made \begin{align*} S&=\pi R^2\\ S_1&=\pi {R_1}^2\left(\frac{24}{360}\right)\\ S_2&=\pi{R_2}^2\left(\frac{24}{360}\right) \end{align*}

then, \begin{align*} S_1-S_2&=\text{ shaded area}\\ S_1-S_2&= \pi {R_1}^2\left(\frac{24}{360}\right)-\pi {R_2}^2\left(\frac{24}{360}\right)\\ &= \pi 4^2\left(\frac{24}{360}\right)-\pi 1^2\left(\frac{24}{360}\right)\\ &= 16\pi \left(\frac{24}{360}\right) - \pi \left(\frac{24}{360}\right)\\ &= \frac{16}{15}\pi -\frac{1}{15}\pi\\ &=\pi \end{align*} Answer: $\pi$

I feel my equation is wrong, but I don't know what's wrong with it.

Also, i am wondering the way to do this kind of question is changing the area degree to radiuss first or we dont need to.

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  • $\begingroup$ sorry I forgot to upload my picture, now here it is! $\endgroup$ – Momoko Jan 29 '14 at 20:57
  • $\begingroup$ @RossMillikan My bad, thanks! I am always bit worried while I am doing some edit, that I will miss something. Sometimes I just go through few questions in my spare time. Should have paid more attention to what I was doing! $\endgroup$ – quapka Jan 29 '14 at 22:52
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You have done fine. What do you think is wrong?

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – M Turgeon Jan 29 '14 at 21:55
  • $\begingroup$ @MTurgeon: OP asked what was wrong. I think this is an answer to the question. $\endgroup$ – Ross Millikan Jan 29 '14 at 21:58
  • $\begingroup$ Fair enough. I guess it's just a matter of personal taste, but I feel this could have been simply a comment. $\endgroup$ – M Turgeon Jan 29 '14 at 22:06
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    $\begingroup$ @MTurgeon: Then the question stays in the unanswered queue and gets brought back up by the robot. $\endgroup$ – Ross Millikan Jan 29 '14 at 22:07
  • $\begingroup$ Sure, whatever. But you still asked for clarification ("what do you think is wrong?") and I thought it was MSE policy to ask for clarification through comments, not answers. $\endgroup$ – M Turgeon Jan 29 '14 at 22:09

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