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Director  Joined: 29 Nov 2012
Posts: 685
In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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13 00:00

Difficulty:   45% (medium)

Question Stats: 67% (02:05) correct 33% (02:20) wrong based on 206 sessions

### HideShow timer Statistics 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

Attachment: circle_ABCD.png [ 7.77 KiB | Viewed 11471 times ]

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59722
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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3
4 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 $$BD=\frac{60}{360}*circumference=\frac{8\pi}{3}$$.

Identical question from GMAT Prep: in-the-circle-above-pq-is-parallel-to-diameter-or-93977.html
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Director  Joined: 29 Nov 2012
Posts: 685
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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Any other way to solve this problem? if you didn't know the central angle theorem.
Math Expert V
Joined: 02 Sep 2009
Posts: 59722
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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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Director  Joined: 29 Nov 2012
Posts: 685
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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.
Attachments circle_new.png [ 15.6 KiB | Viewed 9535 times ]

Math Expert V
Joined: 02 Sep 2009
Posts: 59722
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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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Director  Joined: 29 Nov 2012
Posts: 685
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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1
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.
Attachments Circle modified.png [ 8.44 KiB | Viewed 8977 times ]

Manager  S
Joined: 25 Mar 2013
Posts: 225
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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….
Manager  Joined: 21 Jun 2016
Posts: 70
Location: India
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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Learning for me : central angle is double the inscribed angle....

Sent from my Lenovo A7000-a using GMAT Club Forum mobile app
Intern  Joined: 21 Mar 2017
Posts: 16
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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 $$BD=\frac{60}{360}*circumference=\frac{8\pi}{3}$$.

Identical question from GMAT Prep: http://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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59722
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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 $$BD=\frac{60}{360}*circumference=\frac{8\pi}{3}$$.

Identical question from GMAT Prep: http://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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Intern  Joined: 21 Mar 2017
Posts: 16
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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 $$BD=\frac{60}{360}*circumference=\frac{8\pi}{3}$$.

Identical question from GMAT Prep: http://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?

Math Expert V
Joined: 02 Sep 2009
Posts: 59722
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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?

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.

Attachment: Untitled.png [ 14.66 KiB | Viewed 6521 times ]

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Manager  S
Joined: 23 May 2017
Posts: 231
Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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Attachment: FullSizeRender (2).jpg [ 22.98 KiB | Viewed 5355 times ]

Length of the arc CDB = 2 pi * 4 * $$\frac{120}{360}$$ =$$\frac{8 pi}{3}$$
Director  G
Joined: 20 Feb 2015
Posts: 737
Concentration: Strategy, General Management
Re: In the circle above, CD is parallel to diameter AB and AB  [#permalink]

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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]

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_________________ Re: In the circle above, CD is parallel to diameter AB and AB   [#permalink] 22 Nov 2019, 01:33
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