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GRE Math Challenge #24-three circles of radius 1 are tangent
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07 Sep 2014, 06:41
07 Sep 2014, 06:41
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Question Stats:
82% (01:25) correct
17% (00:57) wrong based on 23 sessions
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In the figure, three circles of radius 1 are tangent to one another. What is theea of shaded region between them?
A. \(\frac{\pi}{2} - \sqrt{3}\)
B. \(1.5\)
C. \(\pi - \sqrt{3}\)
D. \(\sqrt{3} - \frac{\pi}{2}\)
E. \(2 - \frac{\pi}{2}\)
Re: GRE Math Challenge #24-three circles of radius 1 are tangent
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16 Jan 2018, 01:25
16 Jan 2018, 01:25
(sqrt 3 * 2)/2- pi 1^2*1/6*3
=sqrt 3 - pi/2
=sqrt 3 - pi/2
Re: GRE Math Challenge #24-three circles of radius 1 are tangent
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16 Jan 2018, 02:27
16 Jan 2018, 02:27
could not come up with the right logic in time but nonetheless:
In the given figure extend the center point from each circle to the other 2 circle
forming an equilateral triangle with side of 2 each
area of the triangle =\sqrt{3}/4*s^2 = \sqrt{3}/4*4=\sqrt{3}
Since the triangle is a equilateral triangle each angle is 60degree
each angle is formed right at the center of each circle(sector)
area of sector is given by \frac{60}{360}*area of the circle
area of 1 circle = pi*1^2 = pi
area of 1 sector = pi*60/360 = 1/6*pi
area of 3 sector = 3/6*pi = 1/2 pi
area of the shaded region = \sqrt{3}-1/2*pi
(d)
In the given figure extend the center point from each circle to the other 2 circle
forming an equilateral triangle with side of 2 each
area of the triangle =\sqrt{3}/4*s^2 = \sqrt{3}/4*4=\sqrt{3}
Since the triangle is a equilateral triangle each angle is 60degree
each angle is formed right at the center of each circle(sector)
area of sector is given by \frac{60}{360}*area of the circle
area of 1 circle = pi*1^2 = pi
area of 1 sector = pi*60/360 = 1/6*pi
area of 3 sector = 3/6*pi = 1/2 pi
area of the shaded region = \sqrt{3}-1/2*pi
(d)
