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

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

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New post 17 Oct 2018, 10:30
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A
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C
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E

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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  [#permalink]

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New post 17 Oct 2018, 10:51
Can anyone explain this pls?


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Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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New post 17 Oct 2018, 10:56
Abhi077 wrote:
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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  [#permalink]

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New post 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  [#permalink]

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New post 17 Oct 2018, 11:08
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\(BC^2 = 10^2 - 6^2 = 8\)

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  [#permalink]

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New post 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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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.
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Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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New post 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 (90-x)
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  [#permalink]

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New post 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


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Re: In the above figure , Triangle ACD is inscribed in the circle  [#permalink]

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New post 17 Oct 2018, 11:28
We first find BC:

\sqrt{100-36}=\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  [#permalink]

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New post 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


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Actually never, I think it was just a system error.


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