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

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In the figure, circle O has center O, diameter AB and a [#permalink]  11 Feb 2012, 14:21
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Difficulty:

45% (medium)

Question Stats:

55% (03:05) correct 44% (01:39) wrong based on 107 sessions
Attachment:

Perimeter.PNG [ 7.32 KiB | Viewed 5530 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

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Expert's post
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 5342 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}.

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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.
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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.
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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 ?
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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?
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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.
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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.
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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}.

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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 4470 times ]
Hope it helps.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]  20 Jun 2013, 05:39
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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
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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.
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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..
Re: In the figure, circle O has center O, diameter AB and a   [#permalink] 27 Sep 2013, 22:52
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