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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?

Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the [#permalink]
13 Sep 2009, 22:00

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This post received KUDOS

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- 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 (180-30-30)

So 120/360 (2* Pi * r) = 4 * Pi solving you get r = 6 Area = 36 Pi

Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the [#permalink]
14 Sep 2009, 05:50

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 (180-30-30)

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.

Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the [#permalink]
05 Nov 2014, 22:19

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Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the [#permalink]
27 Mar 2015, 01:57

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 (180-30-30)

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?

Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the [#permalink]
27 Mar 2015, 03:46

1

This post received KUDOS

Expert's post

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 (180-30-30)

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 1135 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°. _________________

Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the [#permalink]
27 Mar 2015, 03:53

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 (180-30-30)

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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