Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Jul 2015, 10:19

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 01 Nov 2007
Posts: 101
Followers: 1

Kudos [?]: 29 [1] , given: 0

Circle A, center X. XB is the radius. There is a chord which [#permalink]  21 Mar 2008, 23:36
1
KUDOS
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?
SVP
Joined: 04 May 2006
Posts: 1936
Schools: CBS, Kellogg
Followers: 19

Kudos [?]: 500 [0], given: 1

Re: PS (geometry) [#permalink]  22 Mar 2008, 00:16
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: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 422

Kudos [?]: 2291 [0], given: 359

Re: PS (geometry) [#permalink]  22 Mar 2008, 00:34
Expert's post
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 990 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+ | PrepGame

Intern
Joined: 10 Jan 2008
Posts: 39
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: PS (geometry) [#permalink]  13 May 2008, 12:09
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: 180
Followers: 1

Kudos [?]: 34 [0], given: 0

Re: PS (geometry) [#permalink]  13 May 2008, 16:25
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: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 422

Kudos [?]: 2291 [1] , given: 359

Re: PS (geometry) [#permalink]  13 May 2008, 18:40
1
KUDOS
Expert's post
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+ | PrepGame

Intern
Joined: 10 Jan 2008
Posts: 39
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: PS (geometry) [#permalink]  14 May 2008, 08:47
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
Manager
Joined: 26 Aug 2005
Posts: 60
Followers: 1

Kudos [?]: 4 [0], given: 0

Re: PS (geometry) [#permalink]  14 May 2008, 19:34
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: 39
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: PS (geometry) [#permalink]  15 May 2008, 06:41
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
Re: PS (geometry)   [#permalink] 15 May 2008, 06:41
Display posts from previous: Sort by