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SC Moderator V
Joined: 25 Sep 2018
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Location: United States (CA)
Concentration: Finance, Strategy
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:   55% (hard)

Question Stats: 70% (02:50) correct 30% (02:46) wrong based on 29 sessions

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Attachment: GC.png [ 24.15 KiB | Viewed 649 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: 205
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
Posts: 978
Location: India
In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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Attachment: Figure 1.png [ 3.69 KiB | Viewed 558 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

Let me know if this approach is lacking in some way.
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Originally posted by KanishkM on 17 Oct 2018, 10:11.
Last edited by KanishkM on 17 Oct 2018, 10:18, edited 1 time in total.
Senior Manager  P
Joined: 10 Dec 2017
Posts: 278
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,
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: 30 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: 7
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: 30 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, 10:28

# In the above figure , Triangle ACD is inscribed in the circle   