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

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Manager
Joined: 01 Nov 2007
Posts: 98
Circle A, center X. XB is the radius. There is a chord which [#permalink]

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22 Mar 2008, 00:36
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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?

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Joined: 04 May 2006
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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!
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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. $$Area = \frac{\pi*BY^2}{4}=100\pi$$
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Intern
Joined: 10 Jan 2008
Posts: 39

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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: 176

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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!
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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:

$$\frac{BY}{AB}=\frac{AY}{AD}=\frac{AB}{BD}$$
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Intern
Joined: 10 Jan 2008
Posts: 39

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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: 48

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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
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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
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Re: Circle A, center X. XB is the radius. There is a chord which [#permalink]

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19 Aug 2017, 12:18
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Re: Circle A, center X. XB is the radius. There is a chord which   [#permalink] 19 Aug 2017, 12:18
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