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An equilateral triangle that has an area of 16√ 3 is inscribed in a ci

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An equilateral triangle that has an area of 16√ 3 is inscribed in a ci  [#permalink]

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New post 05 Oct 2018, 00:46
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An equilateral triangle that has an area of 16√3 is inscribed in a circle. What is the area of the circle?


(A) 3π

(B) 16π/3

(C) 64π/3

(D) 12√3*π

(E) 20√3*π

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Re: An equilateral triangle that has an area of 16√ 3 is inscribed in a ci  [#permalink]

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New post 05 Oct 2018, 01:38
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Bunuel wrote:
An equilateral triangle that has an area of 16√3 is inscribed in a circle. What is the area of the circle?


(A) 3π

(B) 16π/3

(C) 64π/3

(D) 12√3*π

(E) 20√3*π


Area of equilateral triangle = \(16√3 = √3/4 a^2\)
Reducing, \(a = 8\)

Radius of Inscribed circle in a equilateral triangle,\(r = √3/6 a\)
\(r = 4/√3\)

Area of circle,\(A = πr^2\)
\(A = π*16/3\)

+1 for B
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An equilateral triangle that has an area of 16√ 3 is inscribed in a ci  [#permalink]

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New post Updated on: 05 Oct 2018, 03:19
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Area of an equilateral triangle is = √3/4a2 = 16√3
this gives a=8
We also know that half of an equilateral triangle is a 30-60-90 triangle with the longest side 'a', and sides of this triangle are in the ratio of x:x√3: 2x.
so a=2x
X=a/2
Hence, x√3=√3a/2 this is nothing but the median(altitude) of an equilateral triangle.
The important thing to remember is that the altitude will also be the median and will pass through the center of the circle (since it is an equilateral triangle). We know that centroid divides the median in the ratio 2:1. The centroid will be the center of the circle as each median will pass through it due to the symmetry.
the radius of the circle will be 2/3( median)= 2/3* √3a/2=√3a/3
a=8,this gives us radius= 8*√3/3
Area= πr2=π∗64/3
C:)

Originally posted by satya2029 on 05 Oct 2018, 02:56.
Last edited by satya2029 on 05 Oct 2018, 03:19, edited 1 time in total.
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Re: An equilateral triangle that has an area of 16√ 3 is inscribed in a ci  [#permalink]

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New post 05 Oct 2018, 03:10
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Akash720 wrote:
Bunuel wrote:
An equilateral triangle that has an area of 16√3 is inscribed in a circle. What is the area of the circle?


(A) 3π

(B) 16π/3

(C) 64π/3

(D) 12√3*π

(E) 20√3*π


Area of equilateral triangle = \(16√3 = √3/4 a^2\)
Reducing, \(a = 8\)

Radius of Inscribed circle in a equilateral triangle,\(r = √3/6 a\)
\(r = 4/√3\)

Area of circle,\(A = πr^2\)
\(A = π*16/3\)

+1 for B



Its the triangle that is inscribed in the circle and not vice versa.
The answer is C.
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Re: An equilateral triangle that has an area of 16√ 3 is inscribed in a ci  [#permalink]

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New post 05 Oct 2018, 19:49
Bunuel wrote:
An equilateral triangle that has an area of 16√3 is inscribed in a circle. What is the area of the circle?


(A) 3π

(B) 16π/3

(C) 64π/3

(D) 12√3*π

(E) 20√3*π


ABC equatorial triangle inscribed in the circle with radius r is \((3\sqrt{3}/4)r^2\)

\((3\sqrt{3}/4)r^2 = 16\sqrt{3}\)

\(r^2=64/3\)

Area of circle will be 64π/3

Hence C
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Re: An equilateral triangle that has an area of 16√ 3 is inscribed in a ci  [#permalink]

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New post 05 Oct 2018, 20:04
Bunuel wrote:
An equilateral triangle that has an area of 16√3 is inscribed in a circle. What is the area of the circle?


(A) 3π

(B) 16π/3

(C) 64π/3

(D) 12√3*π

(E) 20√3*π


Alternate Solution

Area of equilateral triangle = \((\sqrt{3}/4)a^2\)

\((\sqrt{3}/4)a^2 = 16\sqrt{3}\)

Side (a) = 8

height \(h = (\sqrt{3}/2)*8\)

Incentre divides height in the ration 2:1

Hence radius = 2/3(height of equilateral triangle)

\(r=2/3(4\sqrt{3})\)

Hence \(r= (8/3)\sqrt{3}\)

Area \(πr^2 = 64π/3\)

Hence C
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Re: An equilateral triangle that has an area of 16√ 3 is inscribed in a ci &nbs [#permalink] 05 Oct 2018, 20:04
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