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In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 10:30
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In the above figure , Triangle ACD is inscribed in the circle. if side BD = 6 and CD = 10, What is the circumference of the circle? A)8.5 \(\pi\) B)10.5 \(\pi\) C)12.5\(\pi\) D)14\(\pi\) E)16\(\pi\)
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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 10:51
Can anyone explain this pls? Sent from my iPhone using GMAT Club Forum mobile app



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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 10:56
Abhi077 wrote: Attachment: GC.png In the above figure , Triangle ACD is inscribed in the circle. if side BD = 6 and CD = 10, What is the circumference of the circle? A)8.5 \(\pi\) B)10.5 \(\pi\) C)12.5\(\pi\) D)14\(\pi\) E)16\(\pi\) is it D



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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:05
It's C. Solved by trigonometry and Pythagoras theorem
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In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:08
\(BC^2 = 10^2  6^2 = 8\)
Let AB = k
In ADC , \(AD^2 = (8+k)^2  10^2\) In ADB , \(AD^2 = 6^2 + k^2\)
\((8+k)^2  10^2 = 6^2 + k^2\) k = 4.5 Diameter = 8+4.5 = 12.5 Circumference = 12.5π



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In the above figure , Triangle ACD is inscribed in the circle
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Updated on: 17 Oct 2018, 11:18
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I did an approximation in this question(After an unsuccessful previous attempt) We can calculate BC as 8, because triangle DBC is a right angled triangle. Now join a line parallel to DC making a rectangle AECD. Join the diagonal DE. Since in a rectangle diagonals bisect each other, length of OC will be approximately 6(if you can visualize this in the figure) OC as radius will give circumference as approximately 12 π Giving answer as C Let me know if this approach is lacking in some way.
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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:16
Abhi077 wrote: Attachment: GC.png In the above figure , Triangle ACD is inscribed in the circle. if side BD = 6 and CD = 10, What is the circumference of the circle? A)8.5 \(\pi\) B)10.5 \(\pi\) C)12.5\(\pi\) D)14\(\pi\) E)16\(\pi\) As the inscribed angle is 90 degree, AC is the diameter of the circle. In triangle DBC since angle B is right angle (10)2= (6)2+ (BC)2 BC= 8 Triangle ADB and BDC are similar( Why) In triangle ADC angle C is (90x) so In triangle BDC, angle D is X hence, BC/10= 6/AD BC=8 , so AD=15/2 then in triangle ADC (AC)2= (10)2 + (AD)2 AC= 25/2 AC is the diameter, so 2R=25/2 R=25/4 So Circumference= 2*pie*25/4 =12.5 pie C



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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:23
pandeyashwin wrote: \(BC^2 = 10^2  6^2 = 8\)
Let AB = k
In ADC , \(AD^2 = (8+k)^2  10^2\) In ADB , \(AD^2 = 6^2 + k^2\)
\((8+k)^2  10^2 = 6^2 + k^2\) k = 4.5 Diameter = 8+4.5 = 12.5 Circumference = 12.5π How did you get AB=8? I understand BC=8 Sent from my iPhone using GMAT Club Forum



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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:28
We first find BC:
\sqrt{10036}=\sqrt{64}=8
Now we can find AB. We know that:
\sqrt{AB^2+36}=AD^2
We also know that:
\sqrt{AD^2+100}=(8+AB)^2
Replacing AD^2 in the second equation, we got:
AB^2+36+100=64+AB^2+16*AB > AB=4
So we can now calculate the diameter AC, which is AB+BC=12
The circumference is 12pi
answer C



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Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:28
Imhotese8991 wrote: pandeyashwin wrote: \(BC^2 = 10^2  6^2 = 8\)
Let AB = k
In ADC , \(AD^2 = (8+k)^2  10^2\) In ADB , \(AD^2 = 6^2 + k^2\)
\((8+k)^2  10^2 = 6^2 + k^2\) k = 4.5 Diameter = 8+4.5 = 12.5 Circumference = 12.5π How did you get AB=8? I understand BC=8 Sent from my iPhone using GMAT Club Forum[/quote Actually never, I think it was just a system error. Sent from my iPhone using GMAT Club Forum




Re: In the above figure , Triangle ACD is inscribed in the circle
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17 Oct 2018, 11:28






