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Circle O is inscribed in equilateral triangle ABC. If the ar
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Updated on: 22 Sep 2009, 06:23
Question Stats:
65% (02:34) correct 35% (02:41) wrong based on 104 sessions
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24\sqrt{3}\), what is the area of circle O? (A) \(2\pi\sqrt{3}\) (B) \(4\pi\) (C) \(4\pi \sqrt{3}\) (D) \(8\pi\) (E) \(12\pi\) Source: Jeff Sackmann's GMAT Extreme Challenge
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Originally posted by powerka on 21 Sep 2009, 10:19.
Last edited by powerka on 22 Sep 2009, 06:23, edited 1 time in total.



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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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21 Sep 2009, 10:29
Area equilateral triangle = (\sqrt{3}/4) x a^2 = 24 sqrt(3) a = 4\sqrt{6} hence inradius = a / (2\sqrt{3}) = 2\sqrt{2} Area of circle O = pi * (2\sqrt{2})^2 = 8*pi.... OA D
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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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22 Sep 2009, 08:56
bhushan252 wrote: Area equilateral triangle = (\sqrt{3}/4) x a^2 = 24 sqrt(3) a = 4\sqrt{6} hence inradius = a / (2\sqrt{3}) = 2\sqrt{2} Area of circle O = pi * (2\sqrt{2})^2 = 8*pi....OA D The answer is correct, but you have some calculation mistakes {eg: (2\sqrt{2})^2 = 2, not 8} Here's a more thorough approach: 1. Divide equilateral in 2 isosceles triangles 2. s/2 : s/2 sqrt(3) : s 3. area equilateral = s^2 sqrt(3) / 4 4. => s^2 sqrt(3) / 4 = 24 sqrt(3) => s = 4 sqrt(6) 5. => height = s/2 sqrt(3) = 2 * sqrt(6) * sqrt(3) 6. => diameter = 2/3 height = 4/3 * sqrt(6) * sqrt(3) 7. => radius = 2/3 * sqrt(6) * sqrt(3) 7. => area circle = Pi r^2 = Pi 4/9 * 6 * 3 = Pi 8
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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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25 Oct 2009, 00:43
e = a side of the triangle: 306090
½ * e * e sqrt(3)/2 e = 4sqrt(6)
Find radius of the circle (30 degree) 2sqrt(6)/ 2sqrt(3) = x / 1 x = 2sqrt(2)
Thus, the circle area is 8pi



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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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25 Feb 2010, 22:52
A trignometric approach: sqrt(3)/4 * s^2 =24sqrt(3) s=4*sqrt(6) Tan30=rad/(s/2) = r/2sqrt(6) = 1/(sqrt(3) Radius is perpendicular to tangent  the angle bisector/altitude are the same r=2sqrt(2) area=pi(2sqrt(2))^2 = 8pi



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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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29 Feb 2016, 19:38
I clearly need to brush up my fundamentals for inscribed circles in the triangle... so...we have an equilateral triangle...area is 24 sqrt 3. the area of an equilateral triangle is s^2*sqrt(3)/4. s=4*sqrt(6) now, we can draw another equilateral triangle inside the circle. each side will be half of the side of the big triangle, because the points of intersection of the big triangle and circle is right at the middle of each sides of the big triangle. so we have inside an equilateral triangle with the side 2*sqrt(6). now..we can draw inside this small equilateral triangle, 2 306090 triangles, which will have the hypotenuse as the radius of the circle. each leg opposed to 60 degrees angle will be sqrt(6). now, applying the 306090 rule, the hypotenuse will be 2*sqrt(2).
area of the circle = pi*r^2 > 8pi.



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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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29 Feb 2016, 20:11
mvictor wrote: I clearly need to brush up my fundamentals for inscribed circles in the triangle... so...we have an equilateral triangle...area is 24 sqrt 3. the area of an equilateral triangle is s^2*sqrt(3)/4. s=4*sqrt(6) now, we can draw another equilateral triangle inside the circle. each side will be half of the side of the big triangle, because the points of intersection of the big triangle and circle is right at the middle of each sides of the big triangle. so we have inside an equilateral triangle with the side 2*sqrt(6). now..we can draw inside this small equilateral triangle, 2 306090 triangles, which will have the hypotenuse as the radius of the circle. each leg opposed to 60 degrees angle will be sqrt(6). now, applying the 306090 rule, the hypotenuse will be 2*sqrt(2).
area of the circle = pi*r^2 > 8pi. You do not have to draw another equilateral triangle and can still solve without one as shown below. Refer below for the figure. Attachment:
20160229_212448.jpg [ 11.63 KiB  Viewed 1809 times ]
Once you calculate \(s=4*\sqrt{6}\) >\(s/2=2*\sqrt{6}\) Now, consider triangle OAB, right angled at B, it is a 306090 triangle with side \(AB= s/2=2*\sqrt{6}\) (either use this or apply trigonometry, whatever is easier for you!). Now you can clearly calculate radius of the incircle = OB. Hope this helps.



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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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29 Feb 2016, 21:29
Engr2012 wrote: mvictor wrote: I clearly need to brush up my fundamentals for inscribed circles in the triangle... so...we have an equilateral triangle...area is 24 sqrt 3. the area of an equilateral triangle is s^2*sqrt(3)/4. s=4*sqrt(6) now, we can draw another equilateral triangle inside the circle. each side will be half of the side of the big triangle, because the points of intersection of the big triangle and circle is right at the middle of each sides of the big triangle. so we have inside an equilateral triangle with the side 2*sqrt(6). now..we can draw inside this small equilateral triangle, 2 306090 triangles, which will have the hypotenuse as the radius of the circle. each leg opposed to 60 degrees angle will be sqrt(6). now, applying the 306090 rule, the hypotenuse will be 2*sqrt(2).
area of the circle = pi*r^2 > 8pi. You do not have to draw another equilateral triangle and can still solve without one as shown below. Refer below for the figure. Attachment: 20160229_212448.jpg Once you calculate \(s=4*\sqrt{6}\) >\(s/2=2*\sqrt{6}\) Now, consider triangle OAB, right angled at B, it is a 306090 triangle with side \(AB= s/2=2*\sqrt{6}\) (either use this or apply trigonometry, whatever is easier for you!). Now you can clearly calculate radius of the incircle = OB. Hope this helps. yep, definitely a faster way...but if you don't visualize it..you go with the way you visualize



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Circle O is inscribed in equilateral triangle ABC. If the ar
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29 Feb 2016, 22:13
powerka wrote: Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24\sqrt{3}\), what is the area of circle O?
(A) \(2\pi\sqrt{3}\) (B) \(4\pi\) (C) \(4\pi \sqrt{3}\) (D) \(8\pi\) (E) \(12\pi\)
Source: Jeff Sackmann's GMAT Extreme Challenge For a circle inscribed in an equilateral triangle, Side of the triangle = \(2\sqrt{3}\) * Radius of the circle Check: http://www.veritasprep.com/blog/2013/07 ... otherway/Area of triangle \(= (\sqrt{3}/4) * side^2 = 24*\sqrt{3}\) \(Side = 4*\sqrt{6}\) Radius of the circle = \(Side/(2*\sqrt{3}) = (4*\sqrt{6})/(2\sqrt{3}) = 2*\sqrt{2}\) Area of the circle \(= \pi * r^2 = \pi * (2*\sqrt{2})^2 = 8\pi\) Answer (D)
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Re: Circle O is inscribed in equilateral triangle ABC. If the ar
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29 Feb 2016, 22:23
Engr2012 wrote: mvictor wrote: I clearly need to brush up my fundamentals for inscribed circles in the triangle... so...we have an equilateral triangle...area is 24 sqrt 3. the area of an equilateral triangle is s^2*sqrt(3)/4. s=4*sqrt(6) now, we can draw another equilateral triangle inside the circle. each side will be half of the side of the big triangle, because the points of intersection of the big triangle and circle is right at the middle of each sides of the big triangle. so we have inside an equilateral triangle with the side 2*sqrt(6). now..we can draw inside this small equilateral triangle, 2 306090 triangles, which will have the hypotenuse as the radius of the circle. each leg opposed to 60 degrees angle will be sqrt(6). now, applying the 306090 rule, the hypotenuse will be 2*sqrt(2).
area of the circle = pi*r^2 > 8pi. You do not have to draw another equilateral triangle and can still solve without one as shown below. Refer below for the figure. Attachment: 20160229_212448.jpg Once you calculate \(s=4*\sqrt{6}\) >\(s/2=2*\sqrt{6}\) Now, consider triangle OAB, right angled at B, it is a 306090 triangle with side \(AB= s/2=2*\sqrt{6}\) (either use this or apply trigonometry, whatever is easier for you!). Now you can clearly calculate radius of the incircle = OB. Hope this helps. the radius shall be the one third of the height of equilateral triangle.



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