# Getting a Circular Crown's area and perimeter

Okay, this is really bugging me:

My Math book has this practice where I need to get the area and perimeter of the next Circular Crown:

$R = 3$cm , $r = 1.75$cm.

Well, I do it. But my results are simply different than the ones the book tells me (on the back of the book, there is an answers sheet). Why? The process to get the area and perimeter are relatively simple. The instructions are on the book. I follow them, and I fail!

Now, my Math teacher told me it was because [put reason here], but it was ages ago, and I can't remember. I remember the book is right, but I forgot why and how.

The book says that the area is $5.94\pi^2$ and the perimeter is $9.5\pi$. However, I get the area is $\frac{95}{16}\pi$ and the perimeter is $\frac{19}{2}\pi$.

What am I doing wrong?

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? $9.5 = 19/2$ and $95/16 \approx 5.94$ –  user17762 Feb 28 '11 at 23:47
It is not 5.94, it is 5.94π^2. Why is that? –  Zol Tun Kul Feb 28 '11 at 23:51
Given outer radius $R = 3$ and inner radius $r = 1.75$, the area is the difference between circular areas (area of bigger circle, minus the area of the hole): $\pi R^2 - \pi r^2 = \pi ( R^2 - r^2 ) = \frac{95}{16} \pi = 5.9375 \pi \approx 5.94 \pi$. The perimeter is the sum of the circumferences: $2\pi R + 2 \pi r = 2 \pi ( R + r ) = \frac{19}{2} \pi = 9.5 \pi$. It looks like the "$\pi^2$" above is a typo (either in the book or your posting); apart from that, it looks like the book is simply opting for rounded decimals instead of exact fractions. (Personally, I prefer your answers.) –  Blue Feb 28 '11 at 23:54
I see..... thank you. It still confuses me a bit, so I will ask my teacher tomorrow =D.... umm.... what do I do now? Close this? Because technically my question was answered through comments haha. –  Zol Tun Kul Mar 1 '11 at 1:29
A thought about the "typo": Neither you nor I bothered to put units on our answers. As NASA's Mars lander engineers can attest, units are important! Here, they may be the crux of your problem. Perimeter has units of length; in this case, centimeters: $\frac{19}{2}\pi \;\; {\bf cm}$. Area has units of ... well ... area; in this case, square-centimeters: $\frac{95}{16} \pi\;\;{\bf cm}^2$. So, perhaps you've confused the exponent on "cm" for an exponent on $\pi$. (If you need help understanding the difference between length units and area units --centimeters vs square-centimeters-- just ask.) –  Blue Mar 1 '11 at 9:27

As stated in the comments, your results are correct and to be preferred over the rounded results in the book. The area is $\frac{95}{16}\pi\text{ cm}^2$, not $\frac{95}{16}\pi^2\text{ cm}^2$.