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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?

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13 Sep 2009, 22:00

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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

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14 Sep 2009, 05:50

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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.

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05 Nov 2014, 22:19

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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?

- 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 2997 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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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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06 May 2016, 00:19

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).

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15 May 2016, 07:36

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 (180-30-30)

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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15 May 2016, 07:39

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.

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)).

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20 Sep 2016, 06:20

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 30-60-90 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?

gmatclubot

AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the
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