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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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GMAT 1: 800 Q59 V59
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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, 01:32
00:00

Difficulty:

55% (hard)

Question Stats:

69% (01:55) correct 31% (02:52) wrong based on 65 sessions

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geometry.png [ 3.58 KiB | Viewed 1076 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. $$160π$$

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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: 275 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, 01:52 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 Intern Joined: 09 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, 05:03 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: 100 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, 09:54 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. Senior Manager Joined: 05 Jan 2017 Posts: 427 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, 03:05 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 Intern Joined: 09 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, 03:19 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, 11:00 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: 5585 GMAT 1: 800 Q59 V59 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, 18:06 Attachment: 23.png [ 5.03 KiB | Viewed 767 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] 19 Mar 2017, 18:06
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