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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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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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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
00:00

Difficulty:

45% (medium)

Question Stats:

72% (02:14) correct 28% (02:32) wrong based on 81 sessions

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geometry.png [ 3.58 KiB | Viewed 1447 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. $$160π$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 19 Apr 2016 Posts: 274 Location: India GMAT 1: 570 Q48 V22 GMAT 2: 640 Q49 V28 GPA: 3.5 WE: Web Development (Computer Software) Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 16 Mar 2017, 00:52 1 1 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 Intern Joined: 08 Sep 2016 Posts: 35 Location: Georgia Concentration: Finance, International Business GPA: 3.75 WE: Analyst (Investment Banking) Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 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 can you explain why lines are perpendicular? Manager Joined: 17 Apr 2016 Posts: 89 Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 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 Hi 0akshay0 Can you explain how did you assume AC parallel to XB. Director Status: Come! Fall in Love with Learning! Joined: 05 Jan 2017 Posts: 515 Location: India Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 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 _________________ GMAT Mentors Intern Joined: 08 Sep 2016 Posts: 35 Location: Georgia Concentration: Finance, International Business GPA: 3.75 WE: Analyst (Investment Banking) Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 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 ? Manager Joined: 03 Jan 2016 Posts: 59 Location: India WE: Engineering (Energy and Utilities) Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6656 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If DB=2, AC=12, and X is the center of the circle shown as the above f [#permalink] ### Show Tags 19 Mar 2017, 17:06 Attachment: 23.png [ 5.03 KiB | Viewed 1137 times ] 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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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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20 Jul 2018, 21:01
1
The solution assumes that line XB is perpendicular to line AC.

Can we assume that 2 lines are perpendicular without any information provided by the question?
Re: If DB=2, AC=12, and X is the center of the circle shown as the above f &nbs [#permalink] 20 Jul 2018, 21:01
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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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