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If the circle above has center O and circumference 18m, then the perim
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Updated on: 21 May 2017, 11:49
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Originally posted by carcass on 21 May 2017, 11:39.
Last edited by Bunuel on 21 May 2017, 11:49, edited 3 times in total.
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Re: If the circle above has center O and circumference 18m, then the perim
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21 May 2017, 11:48



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If the circle above has center O and circumference 18m, then the perim
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21 May 2017, 19:06
carcass wrote: If the circle above has center O and circumference 18π, then the perimeter of sector RSTO Is (A) 3π + 9 (B) 3π + 18 (C) 6π + 9 (D) 6π + 18 (E) 6π + 24 Circumference = 2\(\pi\)r = 18\(\pi\) r = \(\frac{18}{2}\) = 9 Perimeter of sector RSTO = Arc Length + r + r Arc Length = 2\(\pi\)r\(\frac{C}{360}\) (C is central angle of the sector) C = \(60^{\circ}\) Arc Length = 2\(\pi\)r\(\frac{60}{360}\) = 2*9*\(\pi\)\(\frac{60}{360}\) = 3\(\pi\) Therefore Perimeter of sector RSTO = 3\(\pi\) + 9 + 9 = 3\(\pi\) + 18Answer B..._________________ Kindly press "+1 Kudos" to appreciate



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Re: If the circle above has center O and circumference 18m, then the perim
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22 May 2017, 13:20
Let's solve it part by part Arc makes 60 deg i.e. 1/6 of 360 deg that is 1/6 of circumference = π.18/6 = 3.π Now we know the formula of circumference = π . D where D is diameter C = π• 18 Observe OR + OT = diameter = 18 Answer B Sent from my Redmi Note 3 using GMAT Club Forum mobile app



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Re: If the circle above has center O and circumference 18π, then the perim
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22 Nov 2017, 05:44
Answer is B. 3*pi + 18.
We can find the radius from the circumference which is equal to 2*pi*r.
And the length of the arc is (Angle/360) X Circumference.
Hence the answer.
B
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If the circle above has center O and circumference 18π, then the perim
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23 Nov 2017, 15:02
Bunuel wrote: If the circle above has center O and circumference 18π, then the perimeter of the sector RST is (A) 3π + 9 (B) 3π + 18 (C) 6π + 9 (D) 6π + 18 (E) 6π + 24 Attachment: 20171121_1032_002.png Perimeter of sector RST = RST arc length + 2*(radius) Sector RST is what fraction of the circle? \(\frac{SectorAngle}{Circle}=\frac{60°}{360°}=\frac{1}{6}\) Sector RST = \(\frac{1}{6}\) of the circle. RST arc length? \(\frac{1}{6}\) of circumference \(\frac{1}{6}*18π = 3π\) Radius length? From circumference: 2πr = 18π r = 9 Perimeter = (3π + 2*9) = 3π + 18 Answer B
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Re: If the circle above has center O and circumference 18π, then the perim
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27 Nov 2017, 12:25
Bunuel wrote: If the circle above has center O and circumference 18π, then the perimeter of the sector RST is (A) 3π + 9 (B) 3π + 18 (C) 6π + 9 (D) 6π + 18 (E) 6π + 24 Attachment: 20171121_1032_002.png We see that arc RST corresponds with a central angle of 60 degrees; thus, arc RST is 60/360 = 1/6 of the circumference of the circle. Thus, arc RST = ⅙ x 18π = 3π. Since the circumference = 18π, the radius of the circle = 18π/(2π) = 9. We see that RO and TO are radii, so each is equal to 9. Thus, the perimeter of sector RST is 3π + 9 + 9 = 3π + 18. Answer: B
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If the circle above has center O and circumference 18m, then the perim
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04 May 2018, 07:52
Bunuel what is the concept underneath this line "Since angle ROT is 60 degrees then minor arc RT is \(\frac{60}{360}*circumference=3\pi\);" Please clarify how you're getting 3pi from 60 degree single angle ROT. Thanks.



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Re: If the circle above has center O and circumference 18m, then the perim
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04 May 2018, 08:06
sadikabid27 wrote: Bunuel what is the concept underneath this line "Since angle ROT is 60 degrees then minor arc RT is \(\frac{60}{360}*circumference=3\pi\);" Please clarify how you're getting 3pi from 60 degree single angle ROT. Thanks. Length of arc of the circle is calculated by \(\frac{The Angle Arc Makes With Center}{360}\) * Circumference. In this case its, \(\frac{60}{360}\)*\(18pi\)= \(3pi\). Hope its clear.
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Re: If the circle above has center O and circumference 18m, then the perim
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10 Sep 2018, 06:44
carcass wrote: If the circle above has center O and circumference 18π, then the perimeter of sector RSTO Is (A) 3π + 9 (B) 3π + 18 (C) 6π + 9 (D) 6π + 18 (E) 6π + 24 The circle has circumference 18πcircumference = (2)(radius)(π) So, (2)(radius)(π) = 18π Solve to get: radius = 9 So, OR = 9 and OT = 9Now, we'll deal with arc RST Here the sector angle = 60° 60°/360° = 1/6 So, the arc RST represents 1/6 of the ENTIRE circle Since the ENTIRE circle has circumference 18π, the length of arc RST = (1/6)(18π) = 3πSo, the perimeter of sector RSTO = 9 + 9 + 3π= 18 + 3π Answer: B
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Re: If the circle above has center O and circumference 18m, then the perim
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14 Sep 2018, 09:30
carcass wrote: If the circle above has center O and circumference 18π, then the perimeter of sector RSTO Is (A) 3π + 9 (B) 3π + 18 (C) 6π + 9 (D) 6π + 18 (E) 6π + 24 Since the circumference of circle O is 18π, the radius is 9. Therefore, OR = OT = 9. Furthermore, since angle ROT is 60 degrees, which is 1/6 of 360 degrees, arc RST is 1/6 of the circumference. Therefore, arc RST = 3π. So the perimeter of sector RSTO is 9 + 9 + 3π = 18 + 3π. Answer: B
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Re: If the circle above has center O and circumference 18m, then the perim
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14 Sep 2018, 11:42
[quote="carcass"] If the circle above has center O and circumference 18π, then the perimeter of sector RSTO Is (A) 3π + 9 (B) 3π + 18 (C) 6π + 9 (D) 6π + 18 (E) 6π + 24 first find the arc, 60/360=x/18π, x=3π. then find radius, 2πr=18π,so r=9. 9+9+3π=18+3π. Answer B.




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