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# The area of the circle above is and the perimeter of sector OABC is .

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The area of the circle above is and the perimeter of sector OABC is .  [#permalink]

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Updated on: 18 Jul 2019, 12:09
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Difficulty:

35% (medium)

Question Stats:

74% (02:03) correct 26% (02:56) wrong based on 35 sessions

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The area of the circle above is 49$$\pi$$ and the perimeter of sector OABC is 14 + $$\frac{35{\pi}}{12}$$. What is the value of t?

(A) 56

(B) 72

(C) 75

(D) 80

(E) 85

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Originally posted by AbdurRakib on 10 Jun 2016, 04:18.
Last edited by Bunuel on 18 Jul 2019, 12:09, edited 2 times in total.
Edited the question.
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Re: The area of the circle above is and the perimeter of sector OABC is .  [#permalink]

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10 Jun 2016, 08:00
Answer is 75 : C

Given Area = 49*pi, we know that radius is 7. Therefore given perimeter of sector OABC is 7 + arc ABC + 7 = 14 + 35*pi / 12.

So if angle of arc ABC is 360 * Arc/circumference.
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Re: The area of the circle above is and the perimeter of sector OABC is .  [#permalink]

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22 Jun 2016, 10:11
AbdurRakib wrote:

The area of the circle above is 49$$\pi$$ and the perimeter of sector OABC is 14 + $$\frac{35{\pi}}{12}$$. What is the value of t?

(A) 56

(B) 72

(C) 75

(D) 80

(E) 85

Area =49$$\pi$$
$$\pi$$(7)^2----------->radius=7
perimeter=OA+OC+arc ABC
7+7+2$$\pi$$*7*c/360=14+$$\frac{35{\pi}}{12}$$
C=75

Ans C
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Re: The area of the circle above is and the perimeter of sector OABC is .  [#permalink]

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22 Jun 2016, 10:44
AbdurRakib wrote:

The area of the circle above is 49$$\pi$$ and the perimeter of sector OABC is 14 + $$\frac{35{\pi}}{12}$$. What is the value of t?

(A) 56

(B) 72

(C) 75

(D) 80

(E) 85

$$\pi$$$$r^2$$ = 49$$\pi$$ ; So r = 7 = AO = OC

Perimeter of sector = 2$$\pi$$$$r$$*$$\frac{t}{360}$$ + $$2r$$

Or, 2$$\pi$$$$7$$*$$\frac{t}{360}$$ + $$14$$ = 14 + $$\frac{35{\pi}}{12}$$

Or, 14$$\pi$$*$$\frac{t}{360}$$ = $$\frac{35{\pi}}{12}$$

Or, $$\frac{t}{15}$$ = $$5$$

Or, t = 75

Hence answer will be (C)

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Abhishek....

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Re: The area of the circle above is and the perimeter of sector OABC is .   [#permalink] 22 Jun 2016, 10:44

# The area of the circle above is and the perimeter of sector OABC is .

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