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If the circle above has center O and circumference 18m, then the perim  [#permalink]

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Difficulty:   15% (low)

Question Stats: 81% (01:14) correct 19% (01:42) wrong based on 193 sessions

HideShow timer Statistics  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

Attachment: 6p5jvHu.jpg [ 10.49 KiB | Viewed 5309 times ]

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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.
Edited the question.
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Re: If the circle above has center O and circumference 18m, then the perim  [#permalink]

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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

Attachment:
6p5jvHu.jpg

The perimeter of of sector RSTO is (radius OR) + (radius OT) + (minor arc RT).

The circumference is $$18\pi$$: $$2\pi{r}=18\pi$$ --> $$r=9$$;

Since angle ROT is 60 degrees then minor arc RT is $$\frac{60}{360}*circumference=3\pi$$;

So, the perimeter of sector RSTO is (radius OR) + (radius OT) + (minor arc RT) = $$9+9+3\pi=3\pi+18$$.

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If the circle above has center O and circumference 18m, then the perim  [#permalink]

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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

Attachment:
6p5jvHu.jpg

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$$\pi$$ + 18

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Re: If the circle above has center O and circumference 18m, then the perim  [#permalink]

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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

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Re: If the circle above has center O and circumference 18π, then the perim  [#permalink]

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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.

B

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If the circle above has center O and circumference 18π, then the perim  [#permalink]

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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:
2017-11-21_1032_002.png

Perimeter of sector RST =

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
$$\frac{1}{6}*18π = 3π$$

From circumference:
2πr = 18π
r = 9

Perimeter = (3π + 2*9) = 3π + 18

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Re: If the circle above has center O and circumference 18π, then the perim  [#permalink]

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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:
2017-11-21_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.

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If the circle above has center O and circumference 18m, then the perim  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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

Attachment:
6p5jvHu.jpg

The circle has circumference 18π
Solve to get: radius = 9
So, OR = 9 and OT = 9

Now, 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) =

So, the perimeter of sector RSTO = 9 + 9 +
= 18 + 3π

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Re: If the circle above has center O and circumference 18m, then the perim  [#permalink]

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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

Attachment:
6p5jvHu.jpg

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.

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Re: If the circle above has center O and circumference 18m, then the perim  [#permalink]

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[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.

2πr=18π,so r=9. Re: If the circle above has center O and circumference 18m, then the perim   [#permalink] 14 Sep 2018, 11:42
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