In the figure, circle O has center O, diameter AB and a radi : GMAT Problem Solving (PS)
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# In the figure, circle O has center O, diameter AB and a radi

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CEO
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In the figure, circle O has center O, diameter AB and a radi [#permalink]

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20 Feb 2008, 12:25
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In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

A. (5/3)pi + 5√3
B. (5/3) pi + 10√3
C. (10/3) pi + 5√3
D. (10/3) pi + 10√3
E. (10/3) pi + 20√3

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-circle-o-has-center-o-diameter-ab-and-a-127386.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 20 Nov 2013, 05:48, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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20 Feb 2008, 13:14
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I like D

x= 30 degrees since alt interior angles.

the angle then is 60 degrees from the center the arc.

60/360 * 10 pi = 10pi/6

multiply by 2 = 10pi/3

drop a line from C to A to form a 30-60-90 triangle. side opposite angle CAB = 5 sqrt3

multiply it by 2

you get

10 sqrt3 + 10pi/3
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20 Feb 2008, 13:52

the circle part CE is (2*pi*r)/3 = 10*pi/3
BC = BE = sqrt(5^2+5^2+2*5*5*(1/2)) = 5*sqrt(3)

10*pi/3 + 2*5*sqrt(3) -> D
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20 Feb 2008, 14:39
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Why isn't the perimeter of the arc 10 PI / 6 or 5PI/3 ?? I got B but I guess I am wrong

The Degree measure is 60 degrees 60/360 * 10PI
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20 Feb 2008, 14:48
terp26 wrote:
Why isn't the perimeter of the arc 10 PI / 6 or 5PI/3 ?? I got B but I guess I am wrong

The Degree measure is 60 degrees 60/360 * 10PI

angles:
COE = 2*COA=2*(2*CBA) = 4*CBA = 4*BCD = 120 degrees = 360/3 -> 10pi/3 not 10pi/6
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20 Feb 2008, 19:50
bmwhype2 wrote:
In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

(5/3)pi + 5 sqrt3
(5/3)pi + 10 sqrt3
(10/3)pi + 5 sqrt3
(10/3)pi + 10 sqrt3
(10/3)pi + 20 sqrt3

Good problem.

We have 10pi as the circumfrence. so 60*/360* --> 1/6*10pi --> 5/3pi but we must not forget we had an insribed angle so we must multiply this result by 2.

10/3pi

Now just draw a line from A to C. The hypotenous of this new triangle is 10. Thus the small side is 5 and the larger side 5sqrt3

10/3pi +10sqrt3
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21 Feb 2008, 01:17
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Applying parallel lines cut by traversal rule: X = 30
Now applying the central angle to Inscribed angle rule
Central angle = 2 * Inscribed angle
Central Angle = 2 * 60 = 120

CAE perimeter = (120 / 360) * 2*pi*5 = 10pi/3
Sides BC and BE are equal and also angle B is equal to 60, so other two angles are equal to 60.
Now apply 30-60-90 rule: BC = 5 sqrt3
Total perimeter = CAE + BC + BE
= 10pi/3 + 5 sqrt3 + 5 sqrt3
= 10 pi/3 + 10 sqrt3

Answer: 10 pi/3 + 10 sqrt3
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Re: In the figure to the right, circle O has center O, diameter [#permalink]

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20 Nov 2013, 03:56
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Re: In the figure, circle O has center O, diameter AB and a radi [#permalink]

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20 Nov 2013, 05:49
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In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

A. (5/3)pi + 5√3
B. (5/3) pi + 10√3
C. (10/3) pi + 5√3
D. (10/3) pi + 10√3
E. (10/3) pi + 20√3

Since CD is parallel to AB then <CBA=<BCD=30 --> <CBE=60 --> <COE=2*60=120 (according to the central angle theorem) --> length of minor arc CE is $$\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi$$;

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus $$\frac{BC}{AB}=\frac{\sqrt{3}}{2}$$, (BC is opposite to 60 degrees so corresponds to $$\sqrt{3}$$) --> $$\frac{BC}{10}=\frac{\sqrt{3}}{2}$$ (AB=diameter=2r=10) --> $$BC=5\sqrt{3}$$;

Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-circle-o-has-center-o-diameter-ab-and-a-127386.html
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Re: In the figure, circle O has center O, diameter AB and a radi   [#permalink] 20 Nov 2013, 05:49
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