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In the figure, circle O has center O, diameter AB and a radius of 5.

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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 27 Sep 2013, 23:52
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.. :-D
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 27 Apr 2014, 08:09
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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).

Answer: D.


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?
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 27 Apr 2014, 11:03
Ok Got it..it will not be a circle but an ellipse..My bad..
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 12 Jul 2014, 17:52
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
Image


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

Answer: D.

Hi Bunuel,
How did you find that BC and BE are equal?
Thanks!
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 13 Jul 2014, 01:35
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 correct Answer is D.
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)
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 03 May 2015, 21:23
I must be missing something. Can anyone point me in the right direction to read about parallel lines in Circles? i don't understand why the angles are equal. Thanks!
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 05 Jul 2015, 13:45
Bunuel wrote:
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°.
Image
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.


Hi Bunuel,

Could you help me with this?:

The right triangle with hypotenuse 10 is not supposed to be a multiple of the (3 - 4 - 5) right triangle??? Then if it were a multiple of this (3 - 4 - 5) it would be (6 - 8 -10); then CB would be 8 instead of 5√3.... I am confused why this results are different. I was thinking that whatever hypotenuse of 10 with 30 - 60 - 90 right triangle had to match with the 6 - 8 - 10 shortcut... Could you help me?

Thanks a lot.

Regards.

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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 05 Jul 2015, 14:51
Engr2012 wrote:
luisnavarro wrote:

Hi Bunuel,

Could you help me with this?:

The right triangle with hypotenuse 10 is not supposed to be a multiple of the (3 - 4 - 5) right triangle??? Then if it were a multiple of this (3 - 4 - 5) it would be (6 - 8 -10); then CB would be 8 instead of 5√3.... I am confused why this results are different. I was thinking that whatever hypotenuse of 10 with 30 - 60 - 90 right triangle had to match with the 6 - 8 - 10 shortcut... Could you help me?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


You are not looking at this correctly.

\(1:\sqrt{3}:2\) is only applicable for triangles having angles as \(30^{\circ},60^{\circ}, 90^{\circ}\)

2nd 'special' type of right triangles are the \(45^{\circ},45^{\circ} , 90^{\circ}\) , in which the ratio of the sides becomes \(1:1:\sqrt{2}\)

Right triangles that do not fall under the above 2 cases should not be mixed with the above rules. These triangles have sides that vary with the relative angles inside the triangle. Triangles with sides 3:4:5 or any multiples of this such as 6:8:10 or 9:12:15 etc do not come under the 2 rules mentioned above and thus you will not get the same proportions for the sides.

Hope this clears your doubt


Thanks a lot... That means that 3 - 4 - 5 right triangle is not a 30 - 60 - 90 right triangle?
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 21 Jul 2015, 11:38
Please help me where I am going wrong with this approach.


I got the length of BC =5[square_root]3

Then for calculating the length of the arc I used the angle CBE = 60 with BC and BE.

The length of the arc would be 1/6*2*pi*5[square_root]3

The perimeter = 10[square_root]3 + 1/6*2*pi*5[square_root]3
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 21 Jul 2015, 11:49
varunvarma401 wrote:
Please help me where I am going wrong with this approach.


I got the length of BC =5[square_root]3

Then for calculating the length of the arc I used the angle CBE = 60 with BC and BE.

The length of the arc would be 1/6*2*pi*5[square_root]3

The perimeter = 10[square_root]3 + 1/6*2*pi*5[square_root]3


For calculating the length of the arc, the angle MUST be the center angle (or the angle that the arc makes at the center and NOT at the circumference).

Any angle made by an arc on the circumference = 0.5* Angle made by the SAME arc at the center.

Thus, instead of 60 degrees, you need to calculate the length of arc made by 120 degrees. You will correct answer after that. (1/6) above will become (1/3). Also the entire red equation above is incorrect.

It should be, \(\frac{120}{360} * 2\pi*5 = \frac{10\pi}{3}\)
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 21 Jul 2015, 12:03
Thank you for the explanation Engr2012.

But if we consider a circle with the radius of BC and the sector with angle 60, then the arc length could be calculated right?
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 01 Feb 2016, 05:50
Why did we not add length of Diameter AB in the perimeter of the shaded region?
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 01 Feb 2016, 06:29
devbond wrote:
Why did we not add length of Diameter AB in the perimeter of the shaded region?


Due to the fact that in order to calculate perimeter of any area, you only need to consider OUTER bounds of the said area. What is inside, does not matter.

Thus the perimeter for the question asked will be = CB+BE+ minor arc(EAC)

Hope this helps.
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 22 Oct 2016, 20:27
Hi, if we were to find the area of the shaded region, what would the solution be.

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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 07 Oct 2018, 04:25
<x = alt. <BCD = 30 degrees.
<CBE = 60 degree

O be the center and the angle subtended at the center = <COE = 120 degree

Length of the arc(CAE) = (120/360)*pi*5^2

Now considering triangle OCB, OB = OC = 5 units.
<COB = 120 degree
5/sin30 = BC/sin120
BC = 5sqrt3

Total perimeter = (120/360)*pi*5^2 + 10sqrt3 (D)
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 18 Jan 2019, 14:40
Could someone provide insight about what this would look like drawn out and labeled? I am trying to grasp it. Thank you so much!

enigma123 wrote:
Image

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. \((\frac{5}{3})\pi + 5\sqrt{3}\)

B. \((\frac{5}{3})\pi + 10\sqrt{√}3\)

C. \((\frac{10}{3})\pi + 5\sqrt{3}\)

D. \((\frac{10}{3})\pi + 10\sqrt{3}\)

E. \((\frac{10}{3})\pi + 20\sqrt{3}\)


Attachment:
Perimeter.PNG


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.
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In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 19 Jan 2019, 01:27
Hello Bunuel,

I imagined that point B is the centre of another circle say P.
now we know the radius of the circle as BA; now BC = BE = 10
From the figure, we know that x = 30

we have a radius and an angle, we can find arc length.

the answer should be (10pi/3) + 20.

please clarify where am I missing.

Thank you.
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 19 Jan 2019, 01:44
Bala0801 wrote:
Hello Bunuel,

I imagined that point B is the centre of another circle say P.
now we know the radius of the circle as BA; now BC = BE = 10
From the figure, we know that x = 30

we have a radius and an angle, we can find arc length.

the answer should be (10pi/3) + 20.

please clarify where am I missing.

Thank you.


I think the figure below might help:
Attachment:
Untitled.png
Untitled.png [ 10.28 KiB | Viewed 346 times ]

So, if B is the center and BA is the radius, then you'd get above circle. For that you can see why your further logic is not correct.
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.   [#permalink] 19 Jan 2019, 01:44

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