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AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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Updated on: 06 Nov 2014, 06:01
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AE  BD. AD is also the diameter. If the length of arc BCD is 4pi then what is the area of the circle?? My method: Join BA. Angle ABD is 90deg. Therefore BAD =60 Now, the formula is: Length of an arc= circumference x angle presented by the arc/360 Hence, 4pi=2.pi.r.60/360 r=12 Area=144pi.
Now could somebody please point out where I am going wrong?
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Originally posted by tejal777 on 13 Sep 2009, 22:14.
Last edited by Bunuel on 06 Nov 2014, 06:01, edited 1 time in total.
Renamed the topic and edited the question.




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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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13 Sep 2009, 23:00
 Draw a line from B to O (circle centre)  BO = OD (both radii) and now ODB is iscoceles triangle  EAD = ADB = DBO = 30 degrees  angle of arc at centre = 120 degrees (1803030)
So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi




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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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13 Sep 2009, 22:36
So I think you got the length of the arc formula wrong :
Length of Arc = C/360 (2 * 22/7 * r ) where C is the angle subtended by the arc in the centre
So join BE to intersect the diameter at center O. Now I don't how to find out angle BOD. But once you get that you have your answer.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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14 Sep 2009, 06:50
yangsta8 wrote:  Draw a line from B to O (circle centre)  BO = OD (both radii) and now ODB is iscoceles triangle  EAD = ADB = DBO = 30 degrees  angle of arc at centre = 120 degrees (1803030)
So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi thats one way of solving it ....u can also solve this by using the "angle at center theorm".....ie the angle subtended by the arc at the center is twice the angle subtended by it at any other point on the circle. arc BCD makes angle 60 deg at a : angle BAD = 60. therefore the angle made by the same arc at center :ang BOD = 120 deg. so using ur formula....4 pi = 2pi *r* *[120/306] r = 6 area = pi*r*r = 36pi.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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14 Sep 2009, 09:42
Thanks guys  both the methods are very quick ways of solving this problem.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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27 Mar 2015, 02:57
yangsta8 wrote:  Draw a line from B to O (circle centre)  BO = OD (both radii) and now ODB is iscoceles triangle  EAD = ADB = DBO = 30 degrees  angle of arc at centre = 120 degrees (1803030)
So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi How is Angle EAD = Angle ADB = 30. Which property is being used here? IF anyone could help here.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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27 Mar 2015, 04:46
earnit wrote: yangsta8 wrote:  Draw a line from B to O (circle centre)  BO = OD (both radii) and now ODB is iscoceles triangle  EAD = ADB = DBO = 30 degrees  angle of arc at centre = 120 degrees (1803030)
So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi How is Angle EAD = Angle ADB = 30. Which property is being used here? IF anyone could help here. When two parallel lines are cut by a third line, they form a system of angles. Attachment:
parallell_lines2a.gif [ 1.46 KiB  Viewed 4871 times ]
In this figure, we can see that 1 and 4 are equal, as are 5 and 8. But because they are formed by two parallel lines, they are all equal to each other. The same could be said for angles 2, 3, 6, and 7. There are many terms from geometry class you may know to describe these angles, such as “alternate interior” or “alternate exterior,” but these terms are not used on the test. For the GMAT, it is simply enough to know that all the little angles will always be equal, and all the big angles will always be equal. Additionally, you should realize that any little angle added to any big angle will always equal 180°.
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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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27 Mar 2015, 04:53
Bunuel wrote: earnit wrote: yangsta8 wrote:  Draw a line from B to O (circle centre)  BO = OD (both radii) and now ODB is iscoceles triangle  EAD = ADB = DBO = 30 degrees  angle of arc at centre = 120 degrees (1803030)
So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi How is Angle EAD = Angle ADB = 30. Which property is being used here? IF anyone could help here. When two parallel lines are cut by a third line, they form a system of angles. Attachment: parallell_lines2a.gif In this figure, we can see that 1 and 4 are equal, as are 5 and 8. But because they are formed by two parallel lines, they are all equal to each other. The same could be said for angles 2, 3, 6, and 7. There are many terms from geometry class you may know to describe these angles, such as “alternate interior” or “alternate exterior,” but these terms are not used on the test. For the GMAT, it is simply enough to know that all the little angles will always be equal, and all the big angles will always be equal. Additionally, you should realize that any little angle added to any big angle will always equal 180°. Totally Got it. However, just in reference to the above context, how do you define 'little' angle and 'big' angle among 1,2,3,4,5....



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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27 Mar 2015, 04:59
earnit wrote: Totally Got it. However, just in reference to the above context, how do you define 'little' angle and 'big' angle among 1,2,3,4,5.... One set of angles are smaller than other, so...
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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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15 May 2016, 06:31
Please Bunuel kindly illustrate your approach to solve such problem. I think it looks complex but it is simple if one recognize the concept behind it.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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15 May 2016, 06:40
hatemnag wrote: Please Bunuel kindly illustrate your approach to solve such problem. I think it looks complex but it is simple if one recognize the concept behind it. Here is a solution: aebdadisalsothediameterifthelengthofarcbcdis4pithe83893.html#p628625 Please specify which part needs to be elaborated. Thank you.
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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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15 May 2016, 08:36
Draw a line from B to O (circle centre)  BO = OD (both radii) and now ODB is iscoceles triangle  EAD = ADB = DBO = 30 degrees  angle of arc at centre = 120 degrees (1803030)
So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi I am just confused that OBD is iscoceles triangle and not right triangle with 30  30  120 angles not 45  45  90 angles.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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15 May 2016, 08:39
Hi Bunuel thank you very much for responding. I am just confused that OBD is iscoceles triangle and not right triangle with 30  30  120 angles not 45  45  90 angles.



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Re: AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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15 May 2016, 10:13
hatemnag wrote: Hi Bunuel thank you very much for responding. I am just confused that OBD is iscoceles triangle and not right triangle with 30  30  120 angles not 45  45  90 angles. OD and OB are radii of the circle, thus OBD is an isosceles triangle. The right triangle would be ABD, with right angle at B, because AD is a diameter (A right triangle's hypotenuse is a diameter of its circumcircle (circumscribed circle). The reverse is also true: if one of the sides of an inscribed triangle is a diameter of the circle, then the triangle is a right angled (right angel being the angle opposite the diameter/hypotenuse)). Attachment:
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AE  BD. AD is also the diameter. If the length of arc BCD is 4pi the
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20 Sep 2016, 07:20
I had a different approach to this and got the right answer. Not sure if it's just a coincidence though.
Drawing a line AB makes a 306090 triangle, with the 60 degree angle creating the arc length 4*pi. I used the ratio of \frac{60}{30}=\frac{4*pi}{x}, where x is the length of the arc created by the 30 degree angle. This gives x=2*pi, so the circumference is 2(4*pi)+2(2*pi) = 12*pi = pi*diameter. So the diameter is 12 and from there we get the area 36*pi.
Can someone tell me if this is also a correct approach?



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