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If DB=2, AC=12, and X is the center of the circle shown as the above f

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If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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Attachment:
geometry.png
geometry.png [ 3.58 KiB | Viewed 940 times ]


If DB=2, AC=12, and X is the center of the circle shown as the above figure, what is the area of the circle?

A. \(20π\)
B. \(80π\)
C. \(100π\)
D. \(120π\)
E. \(160π\)
[Reveal] Spoiler: OA

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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 16 Mar 2017, 00:52
MathRevolution wrote:
Attachment:
geometry.png


If DB=2, AC=12, and X is the center of the circle shown as the above figure, what is the area of the circle?

A. \(20π\)
B. \(80π\)
C. \(100π\)
D. \(120π\)
E. \(160π\)


let XC = r (radius), so XD = r - 2

\(r^2 = (r-2)^2 + 6^2\)

\(r^2 = r^2 - 4r + 4 + 36\)

4r = 40

r = 10

Area = pi * r * r = pi*10*10 = 100pi

Hence option C is correct
Hit Kudos if you liked it 8-)
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 16 Mar 2017, 04:03
0akshay0 wrote:
MathRevolution wrote:
Attachment:
geometry.png


If DB=2, AC=12, and X is the center of the circle shown as the above figure, what is the area of the circle?

A. \(20π\)
B. \(80π\)
C. \(100π\)
D. \(120π\)
E. \(160π\)


let XC = r (radius), so XD = r - 2

\(r^2 = (r-2)^2 + 6^2\)

\(r^2 = r^2 - 4r + 4 + 36\)

4r = 40

r = 10

Area = pi * r * r = pi*10*10 = 100pi

Hence option C is correct
Hit Kudos if you liked it 8-)



can you explain why lines are perpendicular?
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 16 Mar 2017, 08:54
0akshay0 wrote:
MathRevolution wrote:
Attachment:
geometry.png


If DB=2, AC=12, and X is the center of the circle shown as the above figure, what is the area of the circle?

A. \(20π\)
B. \(80π\)
C. \(100π\)
D. \(120π\)
E. \(160π\)


let XC = r (radius), so XD = r - 2

\(r^2 = (r-2)^2 + 6^2\)

\(r^2 = r^2 - 4r + 4 + 36\)

4r = 40

r = 10

Area = pi * r * r = pi*10*10 = 100pi

Hence option C is correct
Hit Kudos if you liked it 8-)


Hi 0akshay0

Can you explain how did you assume AC parallel to XB.
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 17 Mar 2017, 02:05
radius = XA = XB = XC= r
XD = XB-BD = r-2
DA = AC/2 = 12/2 = 6
applying pythagoras theorem in triangle xad we get
r^2 = (r-2)^2 + 36

r^2 = r^2 - 4r +4 +36
r = 10

area = pi *10*10 = 100pi
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 17 Mar 2017, 02:19
ByjusGMATapp wrote:
radius = XA = XB = XC= r
XD = XB-BD = r-2
DA = AC/2 = 12/2 = 6
applying pythagoras theorem in triangle xad we get
r^2 = (r-2)^2 + 36

r^2 = r^2 - 4r +4 +36
r = 10

area = pi *10*10 = 100pi


how do you know that DA =AC/2 ?
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 18 Mar 2017, 10:00
Any Perpendicular Drawn from the Center of the circle to a Chord always bisects the Chord.

So AD=DC=AC/2.

Hope this Helps!!!!!!!!!!


Regards
G V V S Narayana Raju


Give +1 Kudos if u like :-D
Expert Post
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink]

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New post 19 Mar 2017, 17:06
Attachment:
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You get the figure shown above, and according to the Pythagorean Theorem, you get \(r^2=(r-2)^2+6^2=r^2-4r+4+36\), and if you get rid of \(r^2\) on both sides and simplify the equation, from 4r=40, you get r=10. Thus, it becomes the area of the circle, which is \(π10^2=100π\).

The answer is C.
Answer: C
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Re: If DB=2, AC=12, and X is the center of the circle shown as the above f   [#permalink] 19 Mar 2017, 17:06
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If DB=2, AC=12, and X is the center of the circle shown as the above f

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