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

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.

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°.

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.

Thanks guys - both the methods are very quick ways of solving this problem.

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...

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: ae-bd-ad-is-also-the-diameter-if-the-length-of-arc-bcd-is-4pi-the-83893.html#p628625 Please specify which part needs to be elaborated. Thank you.

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.

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)).

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?

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.