Question 390645
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A circle with radius 5 cm. has a circumference *[tex \Large 2\pi r\ =\ 2\,\cdot\,\pi\,\cdot\,5\ =\ 10\pi]


Since *[tex \Large \frac{5\pi}{3}] as a fraction of a whole circle is *[tex \Large \frac{\frac{5\pi}{3}}{2\pi}\ =\ \frac{5}{6}] of a circle.


So if the arc length of the whole circle is *[tex \Large 10\pi], the arc length of the arc that subtends an angle of *[tex \Large \frac{5\pi}{3}] must be *[tex \Large \frac{5}{6}] of the whole circle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{6}\ \times\ 10\pi\ =\ \frac{25\pi}{3}]


Which is the exact answer.  If you want to put that into your calculator and round the result to a two decimal place approximation, be my guest.


By the way: "pie" is a dessert item, usually with a flaky crust, and frequently with a fruit filling. "pi" is the English language representation of the Greek alphabetic character *[tex \Large \pi] representing the ratio of a circle's circumference to its diameter.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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