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In the circle above, CD is parallel to diameter AB and AB

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fozzzy
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In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   Updated on: 11 Jul 2018, 11:21
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In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π

(B) 7π/3

(C) 2π/3

(D) 4π/3

(E) 8π/3

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E

Originally posted by fozzzy on 22 Dec 2012, 03:39.
Last edited by Bunuel on 11 Jul 2018, 11:21, edited 3 times in total.
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   22 Dec 2012, 04:36
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In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π
(B) 7π3
(C) 2π3
(D) 4π3
(E) 8π3

Since CD is parallel to diameter AB, then <BCD=<ABC=30.

Next, Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

Minor arc \(BD=\frac{60}{360}*circumference=\frac{8\pi}{3}\).

Answer: E.

Identical question from GMAT Prep: in-the-circle-above-pq-is-parallel-to-diameter-or-93977.html
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   22 Dec 2012, 04:54
Any other way to solve this problem? if you didn't know the central angle theorem.
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   22 Dec 2012, 04:57
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fozzzy wrote:
Any other way to solve this problem? if you didn't know the central angle theorem.


Central Angle Theorem is a must know for GMAT. Check more here: math-circles-87957.html
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   22 Dec 2012, 07:27
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Bunuel wrote:
fozzzy wrote:
Any other way to solve this problem? if you didn't know the central angle theorem.


Central Angle Theorem is a must know for GMAT. Check more here: math-circles-87957.html


So I made changes to the diagram based on what I understood so <1 = 60, <2=30. Correct me If i'm wrong.
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   23 Dec 2012, 06:54
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fozzzy wrote:
Bunuel wrote:
fozzzy wrote:
Any other way to solve this problem? if you didn't know the central angle theorem.


Central Angle Theorem is a must know for GMAT. Check more here: math-circles-87957.html


So I made changes to the diagram based on what I understood so <1 = 60, <2=30. Correct me If i'm wrong.


It's correct.

Check here: math-circles-87957.html
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 Sep 2013, 01:19
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I've added a modified image hope it helps and O is the center.

the portion in red is the arc with the respective angles

Angle COA = Angle DOB = 60 degrees

So the sum of the angles needs to be 180 degree so angle COD is 60 degree

after that apply the formula 2 pi R * 60 / 360 -----> we are given diameter 16 so then radius becomes 8

solve the equation we will get option E.
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 Sep 2013, 22:30
Central angle/360 = arc/circumference
So An inscribed angle is exactly half the corresponding central angle.
Inscribed angle is 30, then central angle is 60 so E….
Circumfrence = 2pir , so r= d/2, then 16pi
So 8pi/3, E….
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   06 Aug 2016, 00:03
Learning for me : central angle is double the inscribed angle....

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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 May 2017, 02:00
Bunuel wrote:

In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π
(B) 7π3
(C) 2π3
(D) 4π3
(E) 8π3

Since CD is parallel to diameter AB, then <BCD=<ABC=30.

Next, Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

Minor arc \(BD=\frac{60}{360}*circumference=\frac{8\pi}{3}\).

Answer: E.

Identical question from GMAT Prep: https://gmatclub.com/forum/in-the-circle ... 93977.html


Hi Bunuel!

Would you be so kind and explain this point in more details please?
Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

As I understand, the central angle has to be twice more than other inscribed angle. Please clarify. This point.
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 May 2017, 02:16
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michaelkalend wrote:
Bunuel wrote:

In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π
(B) 7π3
(C) 2π3
(D) 4π3
(E) 8π3

Since CD is parallel to diameter AB, then <BCD=<ABC=30.

Next, Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

Minor arc \(BD=\frac{60}{360}*circumference=\frac{8\pi}{3}\).

Answer: E.

Identical question from GMAT Prep: https://gmatclub.com/forum/in-the-circle ... 93977.html


Hi Bunuel!

Would you be so kind and explain this point in more details please?
Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

As I understand, the central angle has to be twice more than other inscribed angle. Please clarify. This point.


Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle, so the central angle is twice the inscribed angle. You can check for more here: https://gmatclub.com/forum/math-circles-87957.html
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 May 2017, 03:34
Bunuel wrote:
michaelkalend wrote:
Bunuel wrote:

In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π
(B) 7π3
(C) 2π3
(D) 4π3
(E) 8π3

Since CD is parallel to diameter AB, then <BCD=<ABC=30.

Next, Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

Minor arc \(BD=\frac{60}{360}*circumference=\frac{8\pi}{3}\).

Answer: E.

Identical question from GMAT Prep: https://gmatclub.com/forum/in-the-circle ... 93977.html


Hi Bunuel!

Would you be so kind and explain this point in more details please?
Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

As I understand, the central angle has to be twice more than other inscribed angle. Please clarify. This point.


Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle, so the central angle is twice the inscribed angle. You can check for more here: https://gmatclub.com/forum/math-circles-87957.html



Yes, that is absolutely clear! But how is that applied to the following:

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

How did you arrive with <BOD and does it mean that angle BOD is 120, while angle BCD is half which means it equals 60?
And if I am getting above correctly, how did you arrive that angle BOD is 120?

Please explain.
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 May 2017, 04:50
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michaelkalend wrote:
Bunuel wrote:

In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π
(B) 7π3
(C) 2π3
(D) 4π3
(E) 8π3

Since CD is parallel to diameter AB, then <BCD=<ABC=30.

Next, Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

Minor arc \(BD=\frac{60}{360}*circumference=\frac{8\pi}{3}\).




Yes, that is absolutely clear! But how is that applied to the following:

Consider, O to be the center of the circle, then according to the central angle theorem above <BOD=2<BCD=60.

How did you arrive with <BOD and does it mean that angle BOD is 120, while angle BCD is half which means it equals 60?
And if I am getting above correctly, how did you arrive that angle BOD is 120?

Please explain.


Check the image below:


Blue (central angle) and Green (inscribed angle) subtend the same Red arc DB, thus (Blue angle) = 2*(Green angle).

H ope it's clear now.


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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   25 May 2017, 11:28
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Length of the arc CDB = 2 pi * 4 * \(\frac{120}{360}\) =\(\frac{8 pi}{3}\)
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   11 Jul 2018, 11:00
fozzzy wrote:
Attachment:
circle_ABCD.png
In the circle above, CD is parallel to diameter AB and AB has length 16. What is the length of arc DB?

(A) 2π

(B) 7π/3

(C) 2π/3

(D) 4π/3

(E) 8π/3


CD and AB are parallel .
Which means angle DCB and angle ABC are alternate angles
since alternate angles are equal
angle.DCB=angle.ABC=30
also ,
angle subtended by an arc at the centre of a circle = 2 * angle subtended by the same arc on any other point on the circle.

length of arc = 2pir@/360 (@= angle subtended by arc at the centre of the circle)
length of arc.BD=2*pi*8*60/360
8pi/3
E
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink] New post   27 Feb 2023, 03:04
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Re: In the circle above, CD is parallel to diameter AB and AB [#permalink]
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