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my teacher asked a question in exercise for calculate the perimeter of following: perimiter exercice

I get result of $144\pi$ and some of my friends got $208.2$ as result (assuming $\pi\approx3.14$).

Q: what's the correct answer of this question ?

I get $144\pi$ with this:

Since each part of circles "not used" is 90° then the rest is 270° (360°-90°)

$360° \rightarrow 2\pi r$

$270° \rightarrow x$

So,

$360x = 270 \times 2\pi 5$

$x = \frac{2700\pi}{360}$

$x = 7,5\pi$

$P = (2\times45) + (2\times12) + (4\times7,5\pi) = 144\pi$

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    $\begingroup$ How is $2 \cdot 45 + 2 \cdot 12 + 4 \cdot 7.5\pi = 144\pi$? It's $114 + 30\pi$, you can't just add $114$ and $30$. $\endgroup$ Jul 18, 2013 at 16:59
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    $\begingroup$ Is 5m the radius of the circle arc? $\endgroup$ Jul 18, 2013 at 16:59
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    $\begingroup$ $114+30\pi\approx208.2$. So your reasoning is fine, but the last line is wrong as George pointed out. $\endgroup$
    – Tomas
    Jul 18, 2013 at 17:00
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    $\begingroup$ The diagram does not make it clear what the $5$ m measures. Radius of each part-circle? Diameter? Circumference? (The last unlikely, it would be too easy.) Also, it is a little easier to note that in each circle we have one-quarter missing, and therefore our curvy bits add up to $3$ full circles. $\endgroup$ Jul 18, 2013 at 17:03
  • $\begingroup$ yes, 5m is the radius, I wrong in the sum !? oh shame :( $\endgroup$ Jul 18, 2013 at 17:19

1 Answer 1

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You're reasoning is just fine, until the last line:

Only one term in the sum has a factor of $\pi$, so the correct answer should be $$\begin{align} P & = 2\cdot 12 + 2\cdot 45 + 4\cdot 7.5\pi \\ \\ & = 24 + 90 + 30 \pi \\ \\ & = 114 + 30\pi \approx 208.2\end{align}$$

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