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705-805 Level|   Geometry|      
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In the figure, circle O has center O, diameter AB and a radius of 5. 18 Jul 2020, 08:45
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Bunuel


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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Hi Bunuel

We know that value of x is 30 but how did you conclude ∠C=90° even if AB=diameter. I did not get this part ∠C=90°.

Please let me know.

Best Regards
Kunal
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In the figure, circle O has center O, diameter AB and a radius of 5. 19 Jul 2020, 04:50
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Kunni
Bunuel


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.

Hi Bunuel

We know that value of x is 30 but how did you conclude ∠C=90° even if AB=diameter. I did not get this part ∠C=90°.

Please let me know.

Best Regards
Kunal
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The diameter of a circle always subtends a right angle to any point on the circle.
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In the figure, circle O has center O, diameter AB and a radius of 5. 19 Jul 2020, 13:05
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Bunuel
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Bunuel


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.

Hi Bunuel

We know that value of x is 30 but how did you conclude ∠C=90° even if AB=diameter. I did not get this part ∠C=90°.

Please let me know.

Best Regards
Kunal

The diameter of a circle always subtends a right angle to any point on the circle.
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Bunuel Thanks a lot.
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In the figure, circle O has center O, diameter AB and a radius of 5. 22 Aug 2020, 07:59
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Bunuel
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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
enigma123
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.

Look at the diagram below:
Attachment:
Circle-tr.png
Hope it helps.
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Hi!
Which property of triangle is this ?
I was unable to recognise this as 90 degrees.
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In the figure, circle O has center O, diameter AB and a radius of 5. 30 Oct 2020, 08:38
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Bunuel


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.
Show more





Are we sure that this is a gmat level question?

Caus it takes alot of time even with Bunuel's most efficient way?
Way past my time limit of 2 minutes.

FeelsBadman.
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In the figure, circle O has center O, diameter AB and a radius of 5. 15 Feb 2021, 03:41
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Bunuel


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.
Show more

hi VeritasKarishma BrentGMATPrepNow :grin:

could you pls elaborate on this part "Since CD is parallel to AB then \(\angle CBA = \angle BCD=30°\) "

if one side of triangle is parallel to the side of another triangle in circle then their angles are the same ? what is the rule ?

i hope i wont have any more questions on gmat quant :lol:

thanks!
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In the figure, circle O has center O, diameter AB and a radius of 5. 15 Feb 2021, 03:46
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Bunuel


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.





Are we sure that this is a gmat level question?

Caus it takes alot of time even with Bunuel's most efficient way?
Way past my time limit of 2 minutes.

FeelsBadman.
Show more

i got this question on MGMAT test and randomly picked correct answer , even without reading question :lol: question looked scary to solve under 2 min :lol: :lol:
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In the figure, circle O has center O, diameter AB and a radius of 5. 16 Feb 2021, 01:16
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Bunuel


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.

hi VeritasKarishma BrentGMATPrepNow :grin:

could you pls elaborate on this part "Since CD is parallel to AB then \(\angle CBA = \angle BCD=30°\) "

if one side of triangle is parallel to the side of another triangle in circle then their angles are the same ? what is the rule ?

i hope i wont have any more questions on gmat quant :lol:

thanks!
Show more

This relates to parallel lines and transversal.
CBA and BCD are alternate interior angles. They will be equal.
Check: https://www.mathsisfun.com/geometry/parallel-lines.html
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In the figure, circle O has center O, diameter AB and a radius of 5. 15 Jun 2022, 04:24
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Bunuel


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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Hi Bunuel

If AB is parallel to CD then technically, Angle DCB and Angle CBA should be equal since they are corresponding angles. That means , x=30 and Angle CBE =60. Tell me why I am stuck with this contradiction
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In the figure, circle O has center O, diameter AB and a radius of 5. 15 Jun 2022, 05:56
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Bunuel


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.



Hi Bunuel

If AB is parallel to CD then technically, Angle DCB and Angle CBA should be equal since they are corresponding angles. That means , x=30 and Angle CBE =60. Tell me why I am stuck with this contradiction
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What you've written is correct and coincides with the solutions you quote. Not sure what contradiction you are talking about.
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In the figure, circle O has center O, diameter AB and a radius of 5. 15 Jun 2022, 06:25
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enigma123


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


Show Spoiler
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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Business schools are looking for future business tycoons, not future 8th-grade geometry teachers. The MAJORITY of geometry questions of ALL difficulty levels can be answered correctly using some combination of logic and ballparking rather than geometry. It's much (much) faster, and the minimal calculations required greatly reduces the opportunities to make a silly mistake.

AB is 10. CB is a little bit shorter than AB. What, 8? 9? Let's just say 8.5.
EB = CB, so that's also 8.5.
Arc CAE looks pretty close to diameter. Let's call it 10.
Add those up. 8.5+8.5+10 = 27
Let's look at the answer choices.

A. \((\frac{5}{3})\pi + 5\sqrt{3}\)
Roughly 5 + roughly 8.5 = 13.5 Wrong.

B. \((\frac{5}{3})\pi + 10\sqrt{√}3\)
Roughly 5 + roughly 17 = 22 This is kinda close, but it's definitely not long enough. Wrong.

C. \((\frac{10}{3})\pi + 5\sqrt{3}\)
Roughly 10 + roughly 8.5 = 18.5 Wrong.

D. \((\frac{10}{3})\pi + 10\sqrt{3}\)
Roughly 10 + roughly 17 = 27 Keep it.

E. \((\frac{10}{3})\pi + 20\sqrt{3}\)
Roughly 10 + roughly 34 = 44 Wrong.

Answer choice D.


And that's without even noticing that if x=30, angle CBE is 60, which means that the cord CE is equal to the diameter, and that means that the arc CAE is just a touch longer than that. So, maybe we could have been more precise in our estimate of the length of arc CAE and called it 10.5 or so.


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In the figure, circle O has center O, diameter AB and a radius of 5. 06 Oct 2024, 03:06
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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.
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