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Senior Manager  V
Joined: 25 Sep 2018
Posts: 423
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30 GPA: 3.97
WE: Investment Banking (Investment Banking)
In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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1 00:00

Difficulty:   45% (medium)

Question Stats: 73% (02:50) correct 27% (02:48) wrong based on 28 sessions

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Attachment: GC.png [ 24.15 KiB | Viewed 540 times ]

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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Intern  B
Joined: 04 Apr 2017
Posts: 4
GMAT 1: 640 Q47 V31 Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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Can anyone explain this pls?

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Intern  B
Joined: 08 Oct 2018
Posts: 7
Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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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
Intern  B
Joined: 15 Oct 2016
Posts: 1
Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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It's C. Solved by trigonometry and Pythagoras theorem

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Manager  G
Joined: 14 Jun 2018
Posts: 218
In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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1
$$BC^2 = 10^2 - 6^2 = 8$$

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
Director  G
Joined: 09 Mar 2018
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In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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Attachment: Figure 1.png [ 3.69 KiB | Viewed 449 times ]

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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Originally posted by KanishkM on 17 Oct 2018, 11:11.
Last edited by KanishkM on 17 Oct 2018, 11:18, edited 1 time in total.
Manager  B
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Location: India
Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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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 (90-x)
so In triangle BDC, angle D is X
hence,
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 Intern  Joined: 31 Dec 2017
Posts: 2
Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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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

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Intern  B
Joined: 09 Jun 2018
Posts: 8
Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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We first find BC:

\sqrt{100-36}=\sqrt{64}=8

Now we can find AB. We know that:

We also know that:

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

Intern  Joined: 31 Dec 2017
Posts: 2
Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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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   [#permalink] 17 Oct 2018, 11:28
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In the above figure , Triangle ACD is inscribed in the circle

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