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Circle P shown above is centered at P. If the length of arc ABC is 40,

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Circle P shown above is centered at P. If the length of arc ABC is 40,  [#permalink]

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New post 09 May 2015, 05:14
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Circle P shown above is centered at P. If the length of arc ABC is 40, what is the area of circle P?


A. \(\frac{10'000}{\pi^{2}}\)
B. \(\frac{10'000}{\pi}\)
C. 10'000
D. 10'000\(\pi\)
E. 10'000 \(\pi^{2}\)
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Re: Circle P shown above is centered at P. If the length of arc ABC is 40,  [#permalink]

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New post 09 May 2015, 05:35
\(\frac{72}{360}\) * 2 * π * PC = 40

PC = \(\frac{100}{π}\)
Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)
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Circle P shown above is centered at P. If the length of arc ABC is 40,  [#permalink]

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New post 13 Jul 2015, 12:41
ritigl wrote:
\(\frac{72}{360}\) * 2 * π * PC = 40

PC = \(\frac{100}{π}\)
Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)


Edited

I answered my own questions :)

But now I don't understand Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)

how do we lose the second pi

we get 10,000 in the numerator (100*100) then pi^2 in the denominator. Does the pi then.... oh wait I just answered my second question... haha
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Circle P shown above is centered at P. If the length of arc ABC is 40,  [#permalink]

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New post Updated on: 13 Jul 2015, 13:00
1
DropBear wrote:
ritigl wrote:
\(\frac{72}{360}\) * 2 * π * PC = 40

PC = \(\frac{100}{π}\)
Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)


Edited

I answered my own questions :)

But now I don't understand Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)

how do we lose the second pi


The last step is :π * \((\frac{100}{π})^2\) =\(\pi\) * \(\frac{100^2}{\pi^2}\) =\(\pi\) * \(\frac{10000}{\pi^2}\) -----> you cancel out the \(\pi\) from the numerator and from the denominator to get \(10000/\pi\)

Originally posted by ENGRTOMBA2018 on 13 Jul 2015, 12:55.
Last edited by ENGRTOMBA2018 on 13 Jul 2015, 13:00, edited 3 times in total.
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Re: Circle P shown above is centered at P. If the length of arc ABC is 40,  [#permalink]

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New post 13 Jul 2015, 12:57
Engr2012 wrote:
DropBear wrote:
ritigl wrote:
\(\frac{72}{360}\) * 2 * π * PC = 40

PC = \(\frac{100}{π}\)
Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)


Edited

I answered my own questions :)

But now I don't understand Area of circle = π * \((\frac{100}{π})^2\) = \(\frac{10,000}{π}\)

how do we lose the second pi


The last step is : \(\frac{10000}{\pi^2}\) * \(\pi\) -----> you cancel out the \(\pi\) from the numerator and 1 from the denominator to get \(10000/\pi\)



Thanks for the clarification. Much appreciated!
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Re: Circle P shown above is centered at P. If the length of arc ABC is 40,   [#permalink] 13 Jul 2015, 12:57
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