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505-555 (Easy)|   Geometry|      
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Bunuel
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The sides of equilateral triangle ABD are tangent to a circle with cen Thu Feb 18, 2021 12:15 am
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83% (01:53) correct 17% (02:37) wrong based on 86 sessions

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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 Thu Feb 18, 2021 12:26 am
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Bunuel

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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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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The sides of equilateral triangle ABD are tangent to a circle with cen Thu Feb 18, 2021 12:46 am
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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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The sides of equilateral triangle ABD are tangent to a circle with cen Thu Feb 18, 2021 10:14 am
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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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The sides of equilateral triangle ABD are tangent to a circle with cen Thu Feb 18, 2021 2:52 pm
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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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The sides of equilateral triangle ABD are tangent to a circle with cen Tue May 03, 2022 5:04 am
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Don't think OD is even required to solve this.

In an equilateral triangle, circumcenter/incenter are the same, which in this case is O.

From AB, AC can be deduced as √3/2*AB = 12*3/2 = 18.

Now circumcenter divides the median or in this case altitude in 2:1 ratio. Hence, OA = 18*1/3 = 6, and circumference of the circle = 2*6*π = 12π

Option B

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The sides of equilateral triangle ABD are tangent to a circle with cen Wed Mar 08, 2023 6:33 pm
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Median drawn from a vertex to the opposite vertex in a triangle is in the ratio 2:1.
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