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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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16 Mar 2017, 00:32
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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. \(160π\)
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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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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 \(r^2 = (r2)^2 + 6^2\) \(r^2 = r^2  4r + 4 + 36\) 4r = 40 r = 10 Area = pi * r * r = pi*10*10 = 100pi Hence option C is correct Hit Kudos if you liked it



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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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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 \(r^2 = (r2)^2 + 6^2\) \(r^2 = r^2  4r + 4 + 36\) 4r = 40 r = 10 Area = pi * r * r = pi*10*10 = 100pi Hence option C is correct Hit Kudos if you liked it 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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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 \(r^2 = (r2)^2 + 6^2\) \(r^2 = r^2  4r + 4 + 36\) 4r = 40 r = 10 Area = pi * r * r = pi*10*10 = 100pi Hence option C is correct Hit Kudos if you liked it Hi 0akshay0Can 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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17 Mar 2017, 02:05
radius = XA = XB = XC= r XD = XBBD = r2 DA = AC/2 = 12/2 = 6 applying pythagoras theorem in triangle xad we get r^2 = (r2)^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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17 Mar 2017, 02:19
ByjusGMATapp wrote: radius = XA = XB = XC= r XD = XBBD = r2 DA = AC/2 = 12/2 = 6 applying pythagoras theorem in triangle xad we get r^2 = (r2)^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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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



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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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19 Mar 2017, 17:06
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You get the figure shown above, and according to the Pythagorean Theorem, you get \(r^2=(r2)^2+6^2=r^24r+4+36\), 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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