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

 It is currently 23 May 2019, 04:01

### 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

# Circle A, center X. XB is the radius. There is a chord which

Author Message
Manager
Joined: 01 Nov 2007
Posts: 90
Circle A, center X. XB is the radius. There is a chord which  [#permalink]

### Show Tags

22 Mar 2008, 00:36
1
2
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions

### HideShow timer Statistics

Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
SVP
Joined: 04 May 2006
Posts: 1608
Schools: CBS, Kellogg

### Show Tags

22 Mar 2008, 01:16
az780 wrote:
Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

My computer does not support drawing, pls get it by yoursefl
1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6

2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36
==> XD=8 --> r = 8+2 = 10 --> Area of cirle = 100pi

EXPECT A LESS TIME-CONSUMING APPROACH!
_________________
CEO
Joined: 17 Nov 2007
Posts: 3408
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

22 Mar 2008, 01:34
Another way:

1. ABY is geometrically similar to DBA

2. $$\frac{AB}{2}=\frac{BY}{AB}$$ -->$$BY=\frac{AB^2}{2}=\frac{6^2+2^2}{2}=20$$

3. $$Area = \frac{\pi*BY^2}{4}=100\pi$$
Attachments

t61632.png [ 6.7 KiB | Viewed 2202 times ]

_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | Limited GMAT/GRE Math tutoring in Chicago
Intern
Joined: 10 Jan 2008
Posts: 38

### Show Tags

13 May 2008, 13:09
I think I need some serious work in geometry.

Could someone please explain the following...
2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36
==> XD=8 --> r = 8+2 = 10 -->

or walker's method? Thanks
Manager
Joined: 12 Feb 2008
Posts: 169

### Show Tags

13 May 2008, 17:25
walker wrote:
Another way:

1. ABY is geometrically similar to DBA

2. $$\frac{AB}{2}=\frac{BY}{AB}$$ -->$$BY=\frac{AB^2}{2}=\frac{6^2+2^2}{2}=20$$

3. $$Area = \frac{\pi*BY^2}{4}=100\pi$$

the first answer is straight forward Pythagorean theorem.
i like the walkers answer, but don’t really get it.

walker could you please shed some light!
CEO
Joined: 17 Nov 2007
Posts: 3408
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

13 May 2008, 19:40
1
the first step is the most important in my approach:

the triangles DBA and ABY are right triangles, because angles ADB and YAB are right angles.
the triangles DBA and ABY have the same angle ABY=ABD and, therefore, third angles in triangles are equal: AYB=DAB
So,
1. ABY is geometrically similar to DBA

Next step is easer, because I only use the formula for similar triangles:

$$\frac{BY}{AB}=\frac{AY}{AD}=\frac{AB}{BD}$$
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | Limited GMAT/GRE Math tutoring in Chicago
Intern
Joined: 10 Jan 2008
Posts: 38

### Show Tags

14 May 2008, 09:47
Thanks Walker.

If anyone is wondering how we know that angle BAY is right, look up Thales' theorem. Until I looked at walker's method, I had no idea this theorem existed. I really don't remember this from high school. Did I simply forget it, or is it not taught in US public schools?

Also, if anyone is confused by the first (simple) method, the trick is to recognize that XB = XA = XD + DB
Intern
Joined: 26 Aug 2005
Posts: 42

### Show Tags

14 May 2008, 20:34
sondenso wrote:
az780 wrote:
Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

My computer does not support drawing, pls get it by yoursefl
1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6

2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36
==> XD=8 --> r = 8+2 = 10 --> Area of cirle = 100pi

EXPECT A LESS TIME-CONSUMING APPROACH!

sorry. i dont see how you get XD=8.
Intern
Joined: 10 Jan 2008
Posts: 38

### Show Tags

15 May 2008, 07:41
seongbae wrote:
sondenso wrote:
az780 wrote:
Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

My computer does not support drawing, pls get it by yoursefl
1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6

2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36
==> XD=8 --> r = 8+2 = 10 --> Area of cirle = 100pi

EXPECT A LESS TIME-CONSUMING APPROACH!

sorry. i dont see how you get XD=8.

(XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36

take one thing at a time.

(XA)^2 = (XD +2)^2 (Because XA and (XD + 2) are both equal to the radius of the circle)

(XD)^2 + (DA)^2 (Imagine a triangle ADX, with angle ADX a right angle. This is the pythagorean theorem. Leg AD squared plus leg XD squared equals hypotenuse AX squared. This equals XD squared plus 36.

So AX^2 equals both (XD +2)^2 and (XD)^2 + 36, and since (XD +2)^2 = (XD)^2 + 36, we can solve for XD

(XD)^2 + 2 XD + 2 XD + 4 = (XD)^2 + 36
4 XD = 32
XD = 8
Non-Human User
Joined: 09 Sep 2013
Posts: 10996
Re: Circle A, center X. XB is the radius. There is a chord which  [#permalink]

### Show Tags

19 Aug 2017, 12:18
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.

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________
Re: Circle A, center X. XB is the radius. There is a chord which   [#permalink] 19 Aug 2017, 12:18
Display posts from previous: Sort by