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655-705 (Hard)|   Geometry|      
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Bunuel
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC 15 Nov 2019, 01:28
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65% (hard)

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62% (02:38) correct 38% (02:49) wrong based on 160 sessions

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Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is area of circle O?


A. \(2\pi\sqrt{3}\)

B. \(4\pi\)

C. \(4\pi\sqrt{3}\)

D. \(8\pi\)

E. \(12\pi\)


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D
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC 15 Nov 2019, 01:41
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side of ∆ = 24√3 = √3*s^2/4
s= 4√6
radius of inscribed circle = s*√3/6
r= 4√6*√3/6
r= 2√18/3
area of circle = pi * ( 2√18/3)^2 = \(8\pi\)
IMO D


Bunuel
Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is area of circle O?


A. \(2\pi\sqrt{3}\)

B. \(4\pi\)

C. \(4\pi\sqrt{3}\)

D. \(8\pi\)

E. \(12\pi\)


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rajatchopra1994
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC 17 Nov 2019, 21:51
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Bunuel
Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is area of circle O?


A. \(2\pi\sqrt{3}\)

B. \(4\pi\)

C. \(4\pi\sqrt{3}\)

D. \(8\pi\)

E. \(12\pi\)


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Area of Equilateral Triangle: (√3/4)*a^2
therefore, (√3/4)*a^2= 24√3; a=4√6

The radius of a circle when inscribed in Equilateral Triangle: r = (√3a )/ 6
r= (√3*4*√6)/6
r= 2√2

Area of circle= pi*r^2
therefore, the area of a circle: pi* (2√2)^2 = 8pi

please provide the kudus :thumbup: :thumbup:
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC 13 Feb 2020, 04:00
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Since GMAT doesn't really care how we get the answer, I tried approximation here. For those who did not remember the side to radius relations,
Area for triangle= 24*3^1/3 = 40 approx.
Area of an inscribed circle in an equilateral triangle is roughly 60% of the traingles area = 40*3/5 = 24
Of the given options, 8pi is closest to 24 hence
IMO D
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC 12 Jan 2022, 07:53
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Let us keep in mind two important formulas of equilateral triangles, and we'll be good to go.

Area = (√3/4)*(side length)^2

Inradius = (side length)/(2√3)

Using both these formulas, the question can be solved easily.

*Another useful formula -
Circumradius of equilateral triangle = (side length)/√3
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC 20 Apr 2025, 00:28
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