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

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

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

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
Perimeter.PNG [ 7.32 KiB | Viewed 83729 times ]


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 11 Feb 2012, 15:49
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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. \((\frac{5}{3})\pi + 5\sqrt{3}\)

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



Since CD is parallel to AB then \(\angle CBA = \angle BCD=30°\) --> \(\angle CBE = 60°\) --> \(\angle COE=2*60°=120°\) (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 \(\angle CBA=30°\) then triangle ACB is 30°-60°-90° right triangle (AB=diameter means that \(\angle C=90°\)), 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.
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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 Mar 2012, 20:40
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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 radius of 5.  [#permalink]

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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 radius of 5.  [#permalink]

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

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New post 02 Mar 2012, 03:42
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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.
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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 02 Mar 2012, 03:57
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Oh, I forgot that ABS ia a right triangle. thats why I didnt get whats sqroot3/2. my fault, sorry.
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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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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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).

Answer: D.

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.

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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 14 Jul 2014, 02:41
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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
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!


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?
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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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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.

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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
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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, 15:01
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luisnavarro wrote:
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
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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?


Yes, exactly. The 3-4-5 triangle or a triangle with any of the other multiples of 3-4-5 is not a 30-60-90 triangle.

FYI, 3-4-5 triangle has angles as \(36.87^{\circ}, 53.13^{\circ},90^{\circ}\) triangle ( For GMAT, you do not have to remember angles that are not 0, 30, 45 60 or 90).

It actually follows from the fact that 30-60-90 triangle will always have sides in the ratio: \(1:\sqrt{3}:2\). The opposite is true as well that if a triangle has sides in the ration \(1:\sqrt{3}:2\), the triangle will then be 30-60-90. The same logic applies to 45-45-90 triangle as well.

Hope this clears your doubt.
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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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varunvarma401 wrote:
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?


Look at the attached picture

Yes, if you are given that "x" = 60 (in the attached picture), radius = r , then length of the arc =\(l = r*\theta\). where \(\theta\)= angle in RADIANS and not degrees that is made by the arc at the CENTER. Thus \(\theta\) in this case will be 2*60 = 120 degrees

For degree to radian conversion, use the fact that 180 degrees = \(\pi\) radians.

Thus, for the example you have quoted, \(l = r*[(120/180)*\pi] = r*2*\pi/3\)
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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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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.


CD and AB are parallel so x = 30 degrees since they are alternate interior angles.

Angle CBE is 2*30 = 60 degrees. This means arc CE subtends a central angle of 120 degrees (Central angle is twice the angle subtended at the circumference).

So arc CE is one third of the circumference of the circle.
Circumference of the circle \(= 2 *\pi * r = 10 * \pi\)

Length of arc CE \(= (10/3) * \pi\)

Since angle CBE is 60 degrees, triangle CBE is equilateral.
As discussed in the post given below:
https://www.veritasprep.com/blog/2013/0 ... relations/

The ratio of side of the equilateral triangle inscribed in a circle and the radius of the circle \(= a : r = \sqrt{3}:1\)
So \(BC = BE = 5*\sqrt{3}\)

\(Perimeter = (10/3) * \pi + 10*\sqrt{3}\)

Answer (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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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


IMPORTANT: unless stated otherwise, the diagrams in Problem Solving geometry questions are DRAWN TO SCALE.

I'll solve this question using estimation.

Since the diameter AB = 10, we can ESTIMATE the length of CB.
It looks like CB is just a little bit shorter than AB.
So, I'll say that the length of side CB is approximately 9.
This means the length of side EB is approximately 9 as well.
Finally, arc EC looks a little bit shorter than sides CB and EB, so I'll estimate it to be length 8

So, the TOTAL perimeter = 9 + 9 + 8 = 26

Now check the answer choices:

ASIDE: On test day, everyone should know the following apprximations:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2
Also, we'll say that pi ≈ 3


A. (5/3)pi + 5√3 ≈ 5 + 8.5 ≈ 13.5
B. (5/3)pi + 10√3 ≈ 5 + 17 ≈ 22
C. (10/3)pi + 5√3 ≈ 10 + 8.5 ≈ 18.5
D. (10/3)pi + 10√3 ≈ 10 + 17 ≈ 27
E. (10/3)pi + 20√3 ≈ 10 + 34 ≈ 44

Of these, it appears that D is the closest.

Aside: We can see that answer choice B is pretty close too. At this point, you have a time-management decision. You can either stick with D, and use your extra time elsewhere, or your can spend time trying to be more certain of the answer. Your choice.

That said, D is the correct answer.

Cheers,
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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 02 Mar 2012, 00:29
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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Re: In the figure, circle O has center O, diameter AB and a radius of 5.  [#permalink]

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New post 01 Jul 2012, 22:22
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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).

Answer: 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 02 Jul 2012, 01:28
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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).

Answer: D.


Look at the diagram below:
Attachment:
Circle-tr.png
Circle-tr.png [ 7.39 KiB | Viewed 68915 times ]
Hope it 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 27 Sep 2013, 04:42
Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??

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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, 04:47
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Re: In the figure, circle O has center O, diameter AB and a radius of 5.   [#permalink] 27 Sep 2013, 04:47

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