GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jun 2019, 12:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# In the figure, circle O has center O, diameter AB and a radi

Author Message
SVP
Joined: 21 Jan 2007
Posts: 2477
Location: New York City
In the figure, circle O has center O, diameter AB and a radi  [#permalink]

### Show Tags

Updated on: 20 Nov 2013, 06:48
2
25
00:00

Difficulty:

95% (hard)

Question Stats:

56% (01:19) correct 44% (01:05) wrong based on 375 sessions

### HideShow timer Statistics

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

_________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Originally posted by bmwhype2 on 20 Feb 2008, 13:25.
Last edited by Bunuel on 20 Nov 2013, 06:48, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
SVP
Joined: 24 Aug 2006
Posts: 2080

### Show Tags

20 Feb 2008, 14:14
2
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
Director
Joined: 01 Jan 2008
Posts: 574

### Show Tags

20 Feb 2008, 14:52

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
VP
Joined: 22 Oct 2006
Posts: 1385
Schools: Chicago Booth '11

### Show Tags

20 Feb 2008, 15:39
1
1
1
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
Director
Joined: 01 Jan 2008
Posts: 574

### Show Tags

20 Feb 2008, 15:48
1
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
SVP
Joined: 29 Mar 2007
Posts: 2326

### Show Tags

20 Feb 2008, 20:50
1
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
Director
Joined: 30 Jun 2007
Posts: 674

### Show Tags

21 Feb 2008, 02:17
2
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
Math Expert
Joined: 02 Sep 2009
Posts: 55732
Re: In the figure, circle O has center O, diameter AB and a radi  [#permalink]

### Show Tags

20 Nov 2013, 06:49
4
8

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
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 11398
Re: In the figure, circle O has center O, diameter AB and a radi  [#permalink]

### Show Tags

17 Aug 2018, 08:37
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In the figure, circle O has center O, diameter AB and a radi   [#permalink] 17 Aug 2018, 08:37
Display posts from previous: Sort by