Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 03 Oct 2015, 21:23

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 553
Location: United Kingdom
GMAT 1: 730 Q49 V40
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 43

Kudos [?]: 1471 [3] , given: 217

In the figure, circle O has center O, diameter AB and a [#permalink]  11 Feb 2012, 14:21
3
KUDOS
13
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

51% (03:54) correct 49% (02:08) wrong based on 359 sessions
Attachment:

Perimeter.PNG [ 7.32 KiB | Viewed 17080 times ]
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

I can think of few pointers such as

The value of x will be 30 degrees because CD is parallel to diameter. Angle COA and AOE will each be 60 degrees as the central angle is double the inscribed angle. But I am still struggling to get through the question.
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [6] , given: 8043

6
KUDOS
Expert's post
6
This post was
BOOKMARKED
enigma123 wrote:
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
Attachment:

Perimeter.PNG [ 7.32 KiB | Viewed 16672 times ]

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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

_________________
Intern
Joined: 03 Dec 2010
Posts: 22
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  01 Mar 2012, 19:40
I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.

Pls explain.
Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [0], given: 8043

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  01 Mar 2012, 20:18
Expert's post
priyalr wrote:
I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.

Pls explain.

The red parts are not correct.

The point is that $$AB=diameter=2*radius=2*5=10$$, not 5,. You've done everything right after this step but got different numerical values because of this one mistake. It should be: $$AC=\frac{10}{2}=5$$ --> $$CB=5\sqrt{3}$$ --> $$CB+BE=5\sqrt{3}+5\sqrt{3}=10\sqrt{3}$$.

Hope it's clear.
_________________
Intern
Joined: 03 Dec 2010
Posts: 22
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  01 Mar 2012, 23:29
priyalr wrote:
I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.

Pls explain.

thnx a ton. My bad !

I want to give my exam by APril end. I have studied the quant notes for this specific topic and leter solve the PS q, for the given topic. The problem is I get stuck at the start or in middle of my solving. I seem to not apply the concepts learned on to the q. What shld i do ?
Current Student
Joined: 23 Oct 2010
Posts: 386
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 15

Kudos [?]: 221 [0], given: 73

Bunuel wrote:

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) --> $$BC=5\sqrt{3}$$;

I didnt get this part. could u please explain sqroot3/2, or give some links where I can read theory?

btw, Bunuel, is there another way to solve this question?
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [1] , given: 8043

1
KUDOS
Expert's post
LalaB wrote:
Bunuel wrote:

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) --> $$BC=5\sqrt{3}$$;

I didnt get this part. could u please explain sqroot3/2, or give some links where I can read theory?

btw, Bunuel, is there another way to solve this question?

Triangle ACB is a 30°-60°-90° right triangle. AC is opposite the smallest angle 30°, BC is opposite 60° angle, and hypotenuse AB is opposite the largest angle 90°.

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

So, for our question BC (the side opposite 60° angle) and AB (the side opposite the largest angle 90°) must be in the ratio $$\sqrt{3}:2$$ --> $$\frac{BC}{AB}=\frac{\sqrt{3}}{2}$$, and since AB=diameter=2r=10 then $$\frac{BC}{10}=\frac{\sqrt{3}}{2}$$ --> $$BC=5\sqrt{3}$$.

For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
_________________
Current Student
Joined: 23 Oct 2010
Posts: 386
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 15

Kudos [?]: 221 [0], given: 73

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  02 Mar 2012, 02:57
Oh, I forgot that ABS ia a right triangle. thats why I didnt get whats sqroot3/2. my fault, sorry.
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Manager
Joined: 02 Nov 2009
Posts: 141
Followers: 2

Kudos [?]: 80 [0], given: 97

Bunuel

This is the only part that alvvays troubles me

right triangle (AB=diameter means that <C=90)

Hovv I knovv angle at center subtended by diameter is 90 but cant form it here please explain

Bunuel wrote:
enigma123 wrote:
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
Attachment:
Perimeter.PNG

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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [0], given: 8043

Expert's post
venmic wrote:
Bunuel

This is the only part that alvvays troubles me

right triangle (AB=diameter means that <C=90)

Hovv I knovv angle at center subtended by diameter is 90 but cant form it here please explain

Bunuel wrote:
enigma123 wrote:
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
Attachment:
The attachment Perimeter.PNG is no longer available

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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

Look at the diagram below:
Attachment:

Circle-tr.png [ 7.39 KiB | Viewed 14201 times ]
Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [1] , given: 8043

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  20 Jun 2013, 05:39
1
KUDOS
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

_________________
Intern
Joined: 23 Jul 2013
Posts: 22
Followers: 0

Kudos [?]: 6 [0], given: 63

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  27 Sep 2013, 03:42
Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [0], given: 8043

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  27 Sep 2013, 03:47
Expert's post
ishdeep18 wrote:
Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??

Thanks

Arc CE is a part of the circumference --> central angle COE is 120 degrees --> a central angle in a circle determines an arc.

For more check here: math-circles-87957.html

Hope this helps.
_________________
Intern
Joined: 23 Jul 2013
Posts: 22
Followers: 0

Kudos [?]: 6 [0], given: 63

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  27 Sep 2013, 22:52
Bunuel wrote:
ishdeep18 wrote:
Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??

Thanks

Arc CE is a part of the circumference --> central angle COE is 120 degrees --> a central angle in a circle determines an arc.

For more check here: math-circles-87957.html

Hope this helps.

Yes definitely, I was thinking why 60* s not considered , Now I am clear that central angle is considered while calculating the length of the arc.

Thank You Bunuel..
Senior Manager
Joined: 17 Sep 2013
Posts: 379
Location: United States
Concentration: Marketing, Strategy
GMAT 1: 690 Q48 V37
GMAT 2: 730 Q51 V38
GPA: 3.24
WE: Analyst (Consulting)
Followers: 12

Kudos [?]: 160 [0], given: 136

Bunuel wrote:
enigma123 wrote:
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
Attachment:
Perimeter.PNG

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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

TO calculate the length of the arc CAE...I considered B as the center of a larger circle with radius 5\sqrt{3}..and the arc substending an angle of 60 at the center..But I got a different answer...The legth of arc comes out to be 5\sqrt{3}/3

Dunno what did I do wrong here?
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Senior Manager
Joined: 17 Sep 2013
Posts: 379
Location: United States
Concentration: Marketing, Strategy
GMAT 1: 690 Q48 V37
GMAT 2: 730 Q51 V38
GPA: 3.24
WE: Analyst (Consulting)
Followers: 12

Kudos [?]: 160 [0], given: 136

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  27 Apr 2014, 10:03
Ok Got it..it will not be a circle but an ellipse..My bad..
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Manager
Joined: 12 Feb 2011
Posts: 58
Followers: 0

Kudos [?]: 8 [0], given: 1814

In the figure, circle O has center O, diameter AB and a [#permalink]  12 Jul 2014, 16:52
Bunuel wrote:
enigma123 wrote:
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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

Hi Bunuel,
How did you find that BC and BE are equal?
Thanks!
Intern
Joined: 12 Jul 2014
Posts: 10
Location: India
Concentration: Operations, Technology
GMAT Date: 11-06-2014
GRE 1: 303 Q159 V144
Followers: 0

Kudos [?]: 7 [0], given: 40

In the figure, circle O has center O, diameter AB and a [#permalink]  13 Jul 2014, 00:35
enigma123 wrote:
Attachment:
Perimeter.PNG
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

I can think of few pointers such as

The value of x will be 30 degrees because CD is parallel to diameter. Angle COA and AOE will each be 60 degrees as the central angle is double the inscribed angle. But I am still struggling to get through the question.

The perimeter of the shaded region is the sum of lengths of CB + BE + arc EC.
Now let us find out the individual lengths.
1. arc EC
Length of an arc in a circle making an angle X with center O and radius 'r' is given by
[X][/360]*2*pi*r
In the given question we need to find out angle COE which is X in the above formula.
Since CD||AB, angle DCB = angle CBA = 30
Now consider the triangle BOC. This is an Isosceles triangle with OC = OB = 5 and thus the angles in the triangle are as follows:
angle OBC = 30 (given), angle OCB = 30 (property of an Isosceles triangle) and angle BOC = 120 (180-(30 +30))
Now consider the parallel lines CD and AB with CO as the transverse. angle DCO = 60 which is equal to angle COA i.e., 60.
Now the total angle X = angle COE = angle COA + angle AOE = 120
Hence length of arc CE is [120][/360]*2*pi*5 = [10][/3]*pi

2. CB = BE
Consider triangle ACB which is 30-60-90 triangle where angle ACB = 90, angle ABC = 30 and angle CAB = 60.
Now, Sine (angle CAB) = [opposite side][/Hypotenuse] i.e., Sine 60 = [BC][/AB] i.e., sqrt(3)/2 = BC/10 which gives BC as [10*sqrt(3)]/2

Now Perimeter of the shaded region = arc CE + (2 * CB) = [10][/3]*pi + 10*sqrt(3)
_________________

Regards,
Bharat Bhushan Sunkara.

"You need to sacrifice what you are TODAY, for what you want to be TOMORROW!!"

Intern
Joined: 12 Jul 2014
Posts: 10
Location: India
Concentration: Operations, Technology
GMAT Date: 11-06-2014
GRE 1: 303 Q159 V144
Followers: 0

Kudos [?]: 7 [1] , given: 40

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  13 Jul 2014, 00:47
1
KUDOS
Dienekes wrote:
Bunuel wrote:
enigma123 wrote:
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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

Hi Bunuel,
How did you find that BC and BE are equal?
Thanks!

Hello Dienekes,
Triangles ACB and AEB are Similar Triangles with a common side AB. As per the property of Similar Triangles, ratios of sides are equal, hence BC = BE. Hope this clarifies your doubt.

Regards,
Bharat.
_________________

Regards,
Bharat Bhushan Sunkara.

"You need to sacrifice what you are TODAY, for what you want to be TOMORROW!!"

Math Expert
Joined: 02 Sep 2009
Posts: 29679
Followers: 4888

Kudos [?]: 53108 [1] , given: 8043

Re: In the figure, circle O has center O, diameter AB and a [#permalink]  14 Jul 2014, 01:41
1
KUDOS
Expert's post
Dienekes wrote:
Bunuel wrote:
enigma123 wrote:
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: $$\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}$$.

Hi Bunuel,
How did you find that BC and BE are equal?
Thanks!

BC and BE are deviated by the same degree from the diameter AB, thus they are mirror images of each other around AB, therefore they must be equal.

Does this make sense?
_________________
Re: In the figure, circle O has center O, diameter AB and a   [#permalink] 14 Jul 2014, 01:41

Go to page    1   2    Next  [ 32 posts ]

Similar topics Replies Last post
Similar
Topics:
7 The diameter of the circle with center O is 18m and AB is pa 6 07 Jun 2013, 12:15
73 In the figure above, point O is the center of the circle and 34 16 Jan 2011, 15:56
24 In the figure above, the radius of circle with center O is 1 15 27 Aug 2009, 11:03
20 In the figure above, the radius of the circle with center O 17 11 Sep 2008, 04:49
10 In the figure, circle O has center O, diameter AB and a radi 8 20 Feb 2008, 12:25
Display posts from previous: Sort by