GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 09 Dec 2019, 18:40 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the

Author Message
TAGS:

### Hide Tags

Senior Manager  Joined: 25 Oct 2008
Posts: 446
Location: Kolkata,India
AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

4
22 00:00

Difficulty:

(N/A)

Question Stats: 78% (02:02) correct 22% (03:04) wrong based on 87 sessions

### HideShow timer Statistics

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?

Attachments #10.JPG [ 11.11 KiB | Viewed 6943 times ]

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.
Senior Manager  B
Joined: 31 Aug 2009
Posts: 338
Location: Sydney, Australia
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

7
1
- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- 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
##### General Discussion
Senior Manager  Joined: 01 Mar 2009
Posts: 305
Location: PDX
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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.
Manager  Joined: 22 Mar 2009
Posts: 56
Schools: Darden:Tepper:UCUIC:Kenan Flager:Nanyang:NUS:ISB:UCI Merage:Emory
WE 1: 3
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

1
1
yangsta8 wrote:
- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- 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. Senior Manager  Joined: 01 Mar 2009
Posts: 305
Location: PDX
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

Thanks guys - both the methods are very quick ways of solving this problem.
Manager  Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04-30-2015
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

1
yangsta8 wrote:
- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- 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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59622
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

1
earnit wrote:
yangsta8 wrote:
- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- 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 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°.
_________________
Manager  Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04-30-2015
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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
- 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....
Math Expert V
Joined: 02 Sep 2009
Posts: 59622
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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...
_________________
Manager  B
Joined: 20 Apr 2014
Posts: 88
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

kindly illustrate your approach to solve such problem. I think it looks complex but it is simple if one recognize the concept behind it.
Math Expert V
Joined: 02 Sep 2009
Posts: 59622
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

hatemnag wrote:
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.
_________________
Manager  B
Joined: 20 Apr 2014
Posts: 88
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- 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.
Manager  B
Joined: 20 Apr 2014
Posts: 88
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59622
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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: Untitled.png [ 20.65 KiB | Viewed 3526 times ]

_________________
Intern  Joined: 29 Aug 2016
Posts: 3
AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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?
Non-Human User Joined: 09 Sep 2013
Posts: 13730
Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the  [#permalink]

### Show Tags

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.
_________________ Re: AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the   [#permalink] 26 Nov 2019, 11:48
Display posts from previous: Sort by

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