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The sides of equilateral triangle ABD are tangent to a circle with cen
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Bunuel
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The sides of equilateral triangle ABD are tangent to a circle with cen [#permalink] New post  18 Feb 2021, 00:15
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The sides of equilateral \( \triangle ABD\) are tangent to a circle with center O. If \(AB = 12 \sqrt{3}\) and \(OD = 12\), the circumference of the circle is

A. \(6 \pi\)

B. \(12 \pi\)

C. \(6 \pi \sqrt{3}\)

D. \(12 \pi \sqrt{3}\)

E. \(24 \pi\)


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The sides of equilateral triangle ABD are tangent to a circle with cen [#permalink] New post  18 Feb 2021, 00:26
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Bunuel wrote:
Image
The sides of equilateral \( \triangle ABD\) are tangent to a circle with center O. If \(AB = 12 \sqrt{3}\) and \(OD = 12\), the circumference of the circle is

A. \(6 \pi\)

B. \(12 \pi\)

C. \(6 \pi \sqrt{3}\)

D. \(12 \pi \sqrt{3}\)

E. \(24 \pi\)


Side of equilateral ∆ = 12√3

i.e Height of Equilateral ∆ i.e. \(AC= \frac{√3}{2}*Side = = \frac{√3}{2}*12√3 = 18\)

AO = OD = 12 (Given
AC = AO + OC
i.e 18 = 12+OC
i.e. OC = 6 = Radius of Incircle

Circumference \(= 2πr = 12π\)

Answer: Option B

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Re: The sides of equilateral triangle ABD are tangent to a circle with cen [#permalink] New post  18 Feb 2021, 00:46
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OD = OA
Side of the triangle = a = 12√3
Height of the triangle = √3a/2
= 12√3*√3/2
= 18 = AC

OC = AC - OA
= 18 - 12
= 6 = r

Circumference = 2πr
= 2π*6
= 12π

Option B

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TarunKumar1234
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Re: The sides of equilateral triangle ABD are tangent to a circle with cen [#permalink] New post  18 Feb 2021, 10:14
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from triangle, OC = sqrt (OD^2 - CD^2) = sqrt [12^2 - (6√3)^2] = sqrt[144 -108] = 6

So, circumference of the circle = 2π*r = 12π

I think B. :)
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rm201888
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Re: The sides of equilateral triangle ABD are tangent to a circle with cen [#permalink] New post  18 Feb 2021, 14:52
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If a circle is inscribed in equilateral triangle then "length of OA/OB/OD is equivalent to the diameter of circle".
Since OD is given i.e. 12 so radius is 6 and circumference is 12pi.

Option B
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